Linear differential equations Questions in English

(C) The given differential equation is $2(y + 2) \ln(y + 2) dx + (x + 4 - 2 \ln(y + 2)) dy = 0$.
Rearranging,we get $\frac{dx}{dy} = -\frac{x + 4 - 2 \ln(y + 2)}{2(y + 2) \ln(y + 2)}$.
This can be written as $2(y + 2) \ln(y + 2) \frac{dx}{dy} + x = 2 \ln(y + 2) - 4$.
Let $t = \ln(y + 2)$,then $dt = \frac{1}{y + 2} dy$,so $dy = (y + 2) dt$.
The equation becomes $2t \frac{dx}{dt} + x = 2t - 4$,or $\frac{dx}{dt} + \frac{x}{2t} = 1 - \frac{2}{t}$.
This is a linear differential equation in $x$ with respect to $t$.
The integrating factor is $IF = e^{\int \frac{1}{2t} dt} = e^{\frac{1}{2} \ln t} = \sqrt{t}$.
Multiplying by $IF$,we get $\frac{d}{dt}(x \sqrt{t}) = \sqrt{t} - \frac{2}{\sqrt{t}}$.
Integrating both sides,$x \sqrt{t} = \int (t^{1/2} - 2t^{-1/2}) dt = \frac{2}{3} t^{3/2} - 4t^{1/2} + C$.
Dividing by $\sqrt{t}$,$x = \frac{2}{3} t - 4 + \frac{C}{\sqrt{t}}$.
Substituting $t = \ln(y + 2)$,$x = \frac{2}{3} \ln(y + 2) - 4 + \frac{C}{\sqrt{\ln(y + 2)}}$.
Given $x(e^4 - 2) = 1$,so $y = e^4 - 2$,$t = \ln(e^4) = 4$,$x = 1$.
$1 = \frac{2}{3}(4) - 4 + \frac{C}{\sqrt{4}} \implies 1 = \frac{8}{3} - 4 + \frac{C}{2} \implies 1 = -\frac{4}{3} + \frac{C}{2} \implies \frac{C}{2} = \frac{7}{3} \implies C = \frac{14}{3}$.
Now find $x(e^9 - 2)$,so $y = e^9 - 2$,$t = \ln(e^9) = 9$.
$x = \frac{2}{3}(9) - 4 + \frac{14/3}{\sqrt{9}} = 6 - 4 + \frac{14/3}{3} = 2 + \frac{14}{9} = \frac{32}{9}$.

(A) The given differential equation is $(1 + \cos^2 x) \frac{dy}{dx} - (\sin 2x) y = \sin x$.
Dividing by $(1 + \cos^2 x)$,we get $\frac{dy}{dx} - \left( \frac{\sin 2x}{1 + \cos^2 x} \right) y = \frac{\sin x}{1 + \cos^2 x}$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{\sin 2x}{1 + \cos^2 x}$ and $Q(x) = \frac{\sin x}{1 + \cos^2 x}$.
The integrating factor $I.F. = e^{\int P(x) dx} = e^{-\int \frac{\sin 2x}{1 + \cos^2 x} dx}$.
Let $u = 1 + \cos^2 x$,then $du = -2 \cos x \sin x dx = -\sin 2x dx$.
So,$I.F. = e^{\int \frac{du}{u}} = e^{\ln(u)} = 1 + \cos^2 x$.
The general solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$.
$y(1 + \cos^2 x) = \int \left( \frac{\sin x}{1 + \cos^2 x} \right) (1 + \cos^2 x) dx = \int \sin x dx = -\cos x + C$.
Given $f(0) = 0$,at $x = 0$,$y(1 + \cos^2 0) = -\cos 0 + C \implies 0(2) = -1 + C \implies C = 1$.
Thus,$y(1 + \cos^2 x) = 1 - \cos x$.
At $x = \frac{\pi}{2}$,$y(1 + \cos^2 \frac{\pi}{2}) = 1 - \cos \frac{\pi}{2} \implies y(1 + 0) = 1 - 0 \implies y = 1$.

(A) The given differential equation is $\sec x \, dy = -\{2(1-x) \tan x + x(2-x)\} \, dx$.
Dividing by $\sec x$,we get $\frac{dy}{dx} = -\{2(1-x) \sin x + x(2-x) \cos x\}$.
This simplifies to $\frac{dy}{dx} = 2(x-1) \sin x + (x^2-2x) \cos x$.
Integrating both sides with respect to $x$:
$y(x) = \int 2(x-1) \sin x \, dx + \int (x^2-2x) \cos x \, dx$.
Using integration by parts on the second term: $\int (x^2-2x) \cos x \, dx = (x^2-2x) \sin x - \int (2x-2) \sin x \, dx$.
Substituting this back: $y(x) = \int 2(x-1) \sin x \, dx + (x^2-2x) \sin x - \int 2(x-1) \sin x \, dx + C$.
Thus,$y(x) = (x^2-2x) \sin x + C$.
Given $y(0)=2$,we have $2 = (0^2-2(0)) \sin(0) + C$,which implies $C=2$.
So,$y(x) = (x^2-2x) \sin x + 2$.
For $x=2$,$y(2) = (2^2-2(2)) \sin(2) + 2 = (4-4) \sin(2) + 2 = 2$.

(C) The given differential equation is $(1-x^2) dy = [xy + (x^3+2) \sqrt{3(1-x^2)}] dx$.
Dividing by $(1-x^2) dx$,we get $\frac{dy}{dx} - \frac{x}{1-x^2} y = \frac{(x^3+2) \sqrt{3(1-x^2)}}{1-x^2}$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{x}{1-x^2}$ and $Q(x) = \frac{(x^3+2) \sqrt{3}}{\sqrt{1-x^2}}$.
The integrating factor $IF = e^{\int P(x) dx} = e^{-\int \frac{x}{1-x^2} dx} = e^{\frac{1}{2} \ln(1-x^2)} = \sqrt{1-x^2}$.
The general solution is $y \cdot IF = \int Q(x) \cdot IF dx + C$.
$y \sqrt{1-x^2} = \int \frac{(x^3+2) \sqrt{3}}{\sqrt{1-x^2}} \cdot \sqrt{1-x^2} dx + C = \sqrt{3} \int (x^3+2) dx + C$.
$y \sqrt{1-x^2} = \sqrt{3} (\frac{x^4}{4} + 2x) + C$.
Given $y(0)=0$,we have $0 = \sqrt{3}(0) + C$,so $C=0$.
Thus,$y \sqrt{1-x^2} = \sqrt{3} (\frac{x^4}{4} + 2x)$.
For $x = 1/2$,$y \sqrt{1 - 1/4} = \sqrt{3} (\frac{(1/2)^4}{4} + 2(1/2)) = \sqrt{3} (\frac{1}{64} + 1) = \sqrt{3} (\frac{65}{64})$.
$y \sqrt{3/4} = y \frac{\sqrt{3}}{2} = \sqrt{3} \frac{65}{64}$.
$y = \frac{65}{64} \times 2 = \frac{65}{32}$.
Since $m=65$ and $n=32$ are coprime,$m+n = 65+32 = 97$.

(A) The equation of the tangent at $(x, y)$ is $Y-y=Y^{\prime}(x)(X-x)$.
For $X=0$,$Y=y-x Y^{\prime}(x)$.
For $Y=0$,$X=x-\frac{y}{Y^{\prime}(x)}$.
The area of the triangle formed by the tangent and the coordinate axes is $A = \frac{1}{2} \left| x - \frac{y}{Y^{\prime}(x)} \right| \left| y - x Y^{\prime}(x) \right|$.
Since the curve is in the first quadrant and the area is given as $\frac{-y^2}{2 Y^{\prime}(x)}+1$,we have:
$A = \frac{1}{2} \left( \frac{x Y^{\prime}(x) - y}{Y^{\prime}(x)} \right) (y - x Y^{\prime}(x)) = \frac{-y^2}{2 Y^{\prime}(x)} + 1$.
Multiplying by $2 Y^{\prime}(x)$:
$-(y - x Y^{\prime}(x))^2 = -y^2 + 2 Y^{\prime}(x)$.
$-y^2 + 2xy Y^{\prime}(x) - x^2 (Y^{\prime}(x))^2 = -y^2 + 2 Y^{\prime}(x)$.
$2xy Y^{\prime}(x) - x^2 (Y^{\prime}(x))^2 = 2 Y^{\prime}(x)$.
Since $Y^{\prime}(x) \neq 0$,we divide by $Y^{\prime}(x)$:
$2xy - x^2 Y^{\prime}(x) = 2$.
$Y^{\prime}(x) = \frac{2xy - 2}{x^2} = \frac{2y}{x} - \frac{2}{x^2}$.
This is a linear differential equation: $\frac{dy}{dx} - \frac{2}{x} y = -\frac{2}{x^2}$.
Integrating factor $I.F. = e^{\int -\frac{2}{x} dx} = e^{-2 \ln x} = \frac{1}{x^2}$.
The solution is $y \cdot \frac{1}{x^2} = \int \left( -\frac{2}{x^2} \cdot \frac{1}{x^2} \right) dx = \int -2 x^{-4} dx = \frac{2}{3} x^{-3} + C$.
$y = \frac{2}{3x} + C x^2$.
Given $Y(1)=1$,$1 = \frac{2}{3} + C \Rightarrow C = \frac{1}{3}$.
So,$Y(x) = \frac{2}{3x} + \frac{x^2}{3}$.
$Y(2) = \frac{2}{3(2)} + \frac{4}{3} = \frac{1}{3} + \frac{4}{3} = \frac{5}{3}$.
$12 Y(2) = 12 \times \frac{5}{3} = 20$.

(A) The given differential equation is $\frac{d y}{d x}=\frac{\tan x+y}{\sin x(\sec x-\sin x \tan x)}$.
Simplifying the denominator: $\sin x(\frac{1}{\cos x}-\frac{\sin^2 x}{\cos x}) = \sin x(\frac{1-\sin^2 x}{\cos x}) = \sin x(\frac{\cos^2 x}{\cos x}) = \sin x \cos x$.
So,$\frac{d y}{d x} = \frac{\tan x + y}{\sin x \cos x} = \frac{\tan x}{\sin x \cos x} + \frac{y}{\sin x \cos x} = \sec^2 x + y(2 \csc 2x)$.
This is a linear differential equation of the form $\frac{d y}{d x} - (2 \csc 2x)y = \sec^2 x$.
The integrating factor $I.F. = e^{\int -2 \csc 2x dx} = e^{-\ln|\tan x|} = \frac{1}{\tan x}$ (since $x \in (0, \pi/2)$).
The general solution is $y \cdot \frac{1}{\tan x} = \int \sec^2 x \cdot \frac{1}{\tan x} dx + c$.
Let $\tan x = t$,then $\sec^2 x dx = dt$. So,$y \cot x = \int \frac{1}{t} dt + c = \ln|t| + c = \ln(\tan x) + c$.
Thus,$y = \tan x (\ln(\tan x) + c)$.
Given $y(\frac{\pi}{4}) = 2$,we have $2 = \tan(\frac{\pi}{4})(\ln(\tan(\frac{\pi}{4})) + c) = 1(0 + c)$,so $c = 2$.
The solution is $y = \tan x (\ln(\tan x) + 2)$.
For $x = \frac{\pi}{3}$,$y(\frac{\pi}{3}) = \tan(\frac{\pi}{3})(\ln(\tan(\frac{\pi}{3})) + 2) = \sqrt{3}(\ln \sqrt{3} + 2) = \sqrt{3}(\frac{1}{2} \ln 3 + 2) = \sqrt{3}(2 + \log_e \sqrt{3})$.

(A) Given the differential equation: $\sec^2 x dx + (e^{2y} \tan^2 x + \tan x) dy = 0$.
Dividing by $dy$,we get: $\sec^2 x \frac{dx}{dy} + e^{2y} \tan^2 x + \tan x = 0$.
Let $t = \tan x$,then $\frac{dt}{dy} = \sec^2 x \frac{dx}{dy}$.
The equation becomes: $\frac{dt}{dy} + t = -e^{2y} t^2$.
Dividing by $t^2$: $t^{-2} \frac{dt}{dy} + t^{-1} = -e^{2y}$.
Let $u = t^{-1} = \frac{1}{\tan x}$,then $\frac{du}{dy} = -t^{-2} \frac{dt}{dy}$.
Substituting this into the equation: $-\frac{du}{dy} + u = -e^{2y}$,or $\frac{du}{dy} - u = e^{2y}$.
This is a linear differential equation with Integrating Factor $I.F. = e^{\int -1 dy} = e^{-y}$.
The solution is $u e^{-y} = \int e^{2y} e^{-y} dy = \int e^y dy = e^y + C$.
So,$\frac{1}{\tan x} e^{-y} = e^y + C$.
Given $y(\frac{\pi}{4}) = 0$,we have $\frac{1}{\tan(\pi/4)} e^0 = e^0 + C \Rightarrow 1 = 1 + C \Rightarrow C = 0$.
Thus,$\frac{1}{\tan x} e^{-y} = e^y \Rightarrow e^{2y} = \frac{1}{\tan x} = \cot x$.
For $x = \frac{\pi}{6}$,$e^{2\alpha} = \cot(\frac{\pi}{6}) = \sqrt{3}$.
Therefore,$e^{8\alpha} = (e^{2\alpha})^4 = (\sqrt{3})^4 = 9$.

(A) Given the differential equation: $(t+1) dx = (2x + (t+1)^4) dt$.
Rearranging into the standard linear form $\frac{dx}{dt} + P(t)x = Q(t)$:
$\frac{dx}{dt} = \frac{2x + (t+1)^4}{t+1} = \frac{2x}{t+1} + (t+1)^3$.
$\frac{dx}{dt} - \frac{2}{t+1}x = (t+1)^3$.
The integrating factor $I.F. = e^{\int -\frac{2}{t+1} dt} = e^{-2 \ln(t+1)} = (t+1)^{-2} = \frac{1}{(t+1)^2}$.
Multiplying both sides by $I.F.$:
$\frac{d}{dt} \left( \frac{x}{(t+1)^2} \right) = (t+1)^3 \cdot \frac{1}{(t+1)^2} = (t+1)$.
Integrating both sides with respect to $t$:
$\frac{x}{(t+1)^2} = \int (t+1) dt = \frac{(t+1)^2}{2} + C$.
Using the initial condition $x(0) = 2$:
$\frac{2}{(0+1)^2} = \frac{(0+1)^2}{2} + C \Rightarrow 2 = \frac{1}{2} + C \Rightarrow C = \frac{3}{2}$.
Thus,the solution is $x = \frac{(t+1)^4}{2} + \frac{3}{2}(t+1)^2$.
For $t=1$:
$x(1) = \frac{(1+1)^4}{2} + \frac{3}{2}(1+1)^2 = \frac{16}{2} + \frac{3}{2}(4) = 8 + 6 = 14$.

(A) Given the linear differential equation $f'(x) - \alpha f(x) = 3$.
The integrating factor is $I.F. = e^{\int -\alpha dx} = e^{-\alpha x}$.
Multiplying both sides by the $I.F.$,we get $\frac{d}{dx} [f(x) e^{-\alpha x}] = 3 e^{-\alpha x}$.
Integrating both sides,$f(x) e^{-\alpha x} = \int 3 e^{-\alpha x} dx = -\frac{3}{\alpha} e^{-\alpha x} + C$.
Thus,$f(x) = -\frac{3}{\alpha} + C e^{\alpha x}$.
Given $f(0) = 2$,we have $2 = -\frac{3}{\alpha} + C$,so $C = 2 + \frac{3}{\alpha}$.
Given $\lim_{x \rightarrow -\infty} f(x) = 1$.
If $\alpha > 0$,then $e^{\alpha x} \rightarrow 0$ as $x \rightarrow -\infty$,so $f(x) \rightarrow -\frac{3}{\alpha} = 1$,which implies $\alpha = -3$. This contradicts $\alpha > 0$.
If $\alpha < 0$,then $e^{\alpha x} \rightarrow \infty$ as $x \rightarrow -\infty$. For the limit to be $1$,the coefficient of $e^{\alpha x}$ must be $0$.
So $C = 0$,which means $2 + \frac{3}{\alpha} = 0$,so $\alpha = -\frac{3}{2}$.
Then $f(x) = -\frac{3}{-3/2} = 2$. Since $f(x) = 2$ is a constant function,$f'(x) = 0$. The equation $f'(x) = \alpha f(x) + 3$ becomes $0 = (-3/2)(2) + 3 = 0$,which is consistent.
Thus $f(x) = 2$ for all $x$.
Therefore,$f(-\log_e 2) = 2$.

(A) Given the differential equation: $\frac{dx}{dy} = \frac{1+x-y^2}{y}$.
Rearranging the terms,we get: $\frac{dx}{dy} - \frac{x}{y} = \frac{1-y^2}{y}$.
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = -\frac{1}{y}$ and $Q(y) = \frac{1-y^2}{y}$.
The integrating factor $(IF)$ is given by $IF = e^{\int P(y) dy} = e^{\int -\frac{1}{y} dy} = e^{-\ln|y|} = \frac{1}{y}$.
The general solution is $x \cdot IF = \int Q(y) \cdot IF dy + C$.
Substituting the values: $x \cdot \frac{1}{y} = \int \left(\frac{1-y^2}{y}\right) \cdot \frac{1}{y} dy + C$.
$x \cdot \frac{1}{y} = \int \left(\frac{1}{y^2} - 1\right) dy + C$.
$x \cdot \frac{1}{y} = -\frac{1}{y} - y + C$.
Multiplying by $y$: $x = -1 - y^2 + Cy$.
Given the condition $x(1) = 1$,we substitute $y=1$ and $x=1$: $1 = -1 - (1)^2 + C(1) \Rightarrow 1 = -2 + C \Rightarrow C = 3$.
Thus,the particular solution is $x = -1 - y^2 + 3y$.
To find $5x(2)$,we substitute $y=2$: $x(2) = -1 - (2)^2 + 3(2) = -1 - 4 + 6 = 1$.
Therefore,$5x(2) = 5(1) = 5$.

(C) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -1$ and $Q(x) = 1 + 4 \sin x$.
The integrating factor $(IF)$ is $e^{\int P(x) dx} = e^{\int -1 dx} = e^{-x}$.
Multiplying both sides by the $IF$,we get: $e^{-x} \frac{dy}{dx} - e^{-x} y = e^{-x}(1 + 4 \sin x)$,which simplifies to $\frac{d}{dx}(y e^{-x}) = e^{-x} + 4 e^{-x} \sin x$.
Integrating both sides: $y e^{-x} = \int e^{-x} dx + 4 \int e^{-x} \sin x dx$.
Using the formula $\int e^{ax} \sin(bx) dx = \frac{e^{ax}}{a^2+b^2}(a \sin(bx) - b \cos(bx))$,we get $\int e^{-x} \sin x dx = \frac{e^{-x}}{(-1)^2+1^2}(-1 \sin x - 1 \cos x) = -\frac{e^{-x}}{2}(\sin x + \cos x)$.
Thus,$y e^{-x} = -e^{-x} + 4 \left( -\frac{e^{-x}}{2}(\sin x + \cos x) \right) + C = -e^{-x} - 2e^{-x}(\sin x + \cos x) + C$.
Dividing by $e^{-x}$,we get $y = -1 - 2(\sin x + \cos x) + C e^x$.
Given $y(\pi) = 1$: $1 = -1 - 2(\sin \pi + \cos \pi) + C e^{\pi} \Rightarrow 1 = -1 - 2(0 - 1) + C e^{\pi} \Rightarrow 1 = -1 + 2 + C e^{\pi} \Rightarrow 1 = 1 + C e^{\pi} \Rightarrow C = 0$.
So,$y(x) = -1 - 2(\sin x + \cos x)$.
Then $y\left(\frac{\pi}{2}\right) = -1 - 2(\sin \frac{\pi}{2} + \cos \frac{\pi}{2}) = -1 - 2(1 + 0) = -3$.
Finally,$y\left(\frac{\pi}{2}\right) + 10 = -3 + 10 = 7$.

(D) The given differential equation is $(x^2+4)^2 dy + (2x^3y+8xy-2) dx = 0$.
Rearranging the terms,we get $(x^2+4)^2 \frac{dy}{dx} + 2x(x^2+4)y = 2$.
Dividing by $(x^2+4)^2$,we obtain the linear differential equation: $\frac{dy}{dx} + \frac{2x}{x^2+4}y = \frac{2}{(x^2+4)^2}$.
The integrating factor $(IF)$ is $e^{\int \frac{2x}{x^2+4} dx} = e^{\ln(x^2+4)} = x^2+4$.
Multiplying both sides by the $IF$,we get $\frac{d}{dx} [y(x^2+4)] = \frac{2}{x^2+4}$.
Integrating both sides with respect to $x$,we get $y(x^2+4) = \int \frac{2}{x^2+2^2} dx = 2 \cdot \frac{1}{2} \tan^{-1}(\frac{x}{2}) + C = \tan^{-1}(\frac{x}{2}) + C$.
Given $y(0)=0$,we have $0(0+4) = \tan^{-1}(0) + C$,which implies $C=0$.
Thus,$y(x^2+4) = \tan^{-1}(\frac{x}{2})$.
At $x=2$,$y(2^2+4) = \tan^{-1}(\frac{2}{2}) = \tan^{-1}(1) = \frac{\pi}{4}$.
Therefore,$y(8) = \frac{\pi}{4}$,so $y(2) = \frac{\pi}{32}$.

(B) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P=2$ and $Q=\sin(2x)$.
The Integrating Factor ($I$.$F$.) is given by $e^{\int P dx} = e^{\int 2 dx} = e^{2x}$.
The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$.
$y \cdot e^{2x} = \int e^{2x} \sin(2x) dx + C$.
Using the formula $\int e^{ax} \sin(bx) dx = \frac{e^{ax}}{a^2+b^2} (a \sin(bx) - b \cos(bx))$,we get:
$y \cdot e^{2x} = \frac{e^{2x}}{2^2+2^2} (2 \sin(2x) - 2 \cos(2x)) + C = \frac{e^{2x}}{8} (2 \sin(2x) - 2 \cos(2x)) + C = \frac{e^{2x}}{4} (\sin(2x) - \cos(2x)) + C$.
Given $y(0) = \frac{3}{4}$,substitute $x=0$ and $y=\frac{3}{4}$:
$\frac{3}{4} \cdot e^0 = \frac{e^0}{4} (\sin(0) - \cos(0)) + C \Rightarrow \frac{3}{4} = \frac{1}{4} (0 - 1) + C \Rightarrow \frac{3}{4} = -\frac{1}{4} + C \Rightarrow C = 1$.
So,$y \cdot e^{2x} = \frac{e^{2x}}{4} (\sin(2x) - \cos(2x)) + 1$,which simplifies to $y = \frac{1}{4} (\sin(2x) - \cos(2x)) + e^{-2x}$.
Now,evaluate at $x = \frac{\pi}{8}$:
$y\left(\frac{\pi}{8}\right) = \frac{1}{4} (\sin(\frac{\pi}{4}) - \cos(\frac{\pi}{4})) + e^{-2(\frac{\pi}{8})} = \frac{1}{4} (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) + e^{-\pi/4} = 0 + e^{-\pi/4} = e^{-\pi/4}$.

(B) Given $\lim _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1$. Applying $L$'$H$ôpital's rule with respect to $t$:
$\lim _{t \rightarrow x} \frac{2t f(x)-x^2 f'(t)}{1}=1$
$2x f(x)-x^2 f'(x)=1$
$f'(x) - \frac{2}{x} f(x) = -\frac{1}{x^2}$
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{2}{x}$ and $Q(x) = -\frac{1}{x^2}$.
Integrating factor $I.F. = e^{\int -\frac{2}{x} dx} = e^{-2 \ln x} = \frac{1}{x^2}$.
The solution is $f(x) \cdot \frac{1}{x^2} = \int -\frac{1}{x^2} \cdot \frac{1}{x^2} dx + C = \int -x^{-4} dx + C = \frac{1}{3x^3} + C$.
$f(x) = \frac{1}{3x} + Cx^2$.
Given $f(1) = 1$,we have $1 = \frac{1}{3} + C$,so $C = \frac{2}{3}$.
Thus,$f(x) = \frac{1}{3x} + \frac{2x^2}{3} = \frac{1+2x^3}{3x}$.
$f(2) = \frac{1+2(8)}{3(2)} = \frac{17}{6}$.
$f(3) = \frac{1+2(27)}{3(3)} = \frac{55}{9}$.
$2f(2) + 3f(3) = 2(\frac{17}{6}) + 3(\frac{55}{9}) = \frac{17}{3} + \frac{55}{3} = \frac{72}{3} = 24$.

(B) The given differential equation is $(1+x^2) \frac{dy}{dx} + y = e^{\tan^{-1} x}$.
Dividing by $(1+x^2)$,we get $\frac{dy}{dx} + \frac{y}{1+x^2} = \frac{e^{\tan^{-1} x}}{1+x^2}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{1+x^2}$ and $Q = \frac{e^{\tan^{-1} x}}{1+x^2}$.
The integrating factor ($I$.$F$.) is $e^{\int P dx} = e^{\int \frac{1}{1+x^2} dx} = e^{\tan^{-1} x}$.
The solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$.
$y \cdot e^{\tan^{-1} x} = \int \frac{e^{\tan^{-1} x}}{1+x^2} \cdot e^{\tan^{-1} x} dx$.
Let $\tan^{-1} x = z$,then $\frac{1}{1+x^2} dx = dz$.
$y \cdot e^{\tan^{-1} x} = \int e^{2z} dz = \frac{e^{2z}}{2} + C = \frac{e^{2\tan^{-1} x}}{2} + C$.
Given $y(1) = 0$,so $0 \cdot e^{\tan^{-1}(1)} = \frac{e^{2\tan^{-1}(1)}}{2} + C \Rightarrow 0 = \frac{e^{\pi/2}}{2} + C \Rightarrow C = -\frac{e^{\pi/2}}{2}$.
Thus,$y \cdot e^{\tan^{-1} x} = \frac{e^{2\tan^{-1} x}}{2} - \frac{e^{\pi/2}}{2}$.
For $x=0$,$y \cdot e^{\tan^{-1}(0)} = \frac{e^{2\tan^{-1}(0)}}{2} - \frac{e^{\pi/2}}{2} \Rightarrow y \cdot 1 = \frac{1}{2} - \frac{e^{\pi/2}}{2}$.
Therefore,$y(0) = \frac{1}{2}(1 - e^{\pi/2})$.

(C) The given differential equation is $(2x \ln x) \frac{dy}{dx} + 2y = \frac{3}{x} \ln x$.
Dividing by $(2x \ln x)$,we get $\frac{dy}{dx} + \frac{y}{x \ln x} = \frac{3}{2x^2}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x \ln x}$ and $Q = \frac{3}{2x^2}$.
The integrating factor ($I$.$F$.) is $e^{\int P dx} = e^{\int \frac{1}{x \ln x} dx} = e^{\ln(\ln x)} = \ln x$.
The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$.
$y \ln x = \int \frac{3}{2x^2} \ln x dx$.
Using integration by parts,$\int \ln x \cdot (\frac{3}{2} x^{-2}) dx = \ln x \cdot (-\frac{3}{2x}) - \int \frac{1}{x} (-\frac{3}{2x}) dx = -\frac{3 \ln x}{2x} + \frac{3}{2} \int x^{-2} dx = -\frac{3 \ln x}{2x} - \frac{3}{2x} + C$.
Given $y(e^{-1}) = 0$,so $0 \cdot \ln(e^{-1}) = -\frac{3 \ln(e^{-1})}{2e^{-1}} - \frac{3}{2e^{-1}} + C$.
$0 = -\frac{3(-1)}{2e^{-1}} - \frac{3}{2e^{-1}} + C \Rightarrow 0 = \frac{3e}{2} - \frac{3e}{2} + C \Rightarrow C = 0$.
Thus,$y \ln x = -\frac{3 \ln x}{2x} - \frac{3}{2x} \Rightarrow y = -\frac{3}{2x} - \frac{3}{2x \ln x}$.
At $x = e$,$y(e) = -\frac{3}{2e} - \frac{3}{2e \ln e} = -\frac{3}{2e} - \frac{3}{2e} = -\frac{6}{2e} = -\frac{3}{e}$.

(C) Given the differential equation: $\sec y \frac{dy}{dx} + 2x \sin y = x^3 \cos y$.
Divide both sides by $\cos y$ (or multiply by $\sec y$):
$\sec^2 y \frac{dy}{dx} + 2x \tan y = x^3$.
Let $t = \tan y$,then $\frac{dt}{dx} = \sec^2 y \frac{dy}{dx}$.
The equation becomes a linear differential equation: $\frac{dt}{dx} + 2xt = x^3$.
The integrating factor $IF = e^{\int 2x dx} = e^{x^2}$.
Multiplying by $IF$: $\frac{d}{dx}(t e^{x^2}) = x^3 e^{x^2}$.
Integrating both sides: $t e^{x^2} = \int x^3 e^{x^2} dx + C$.
Let $u = x^2$,then $du = 2x dx$,so $\int x^3 e^{x^2} dx = \frac{1}{2} \int u e^u du = \frac{1}{2} (u e^u - e^u) + C = \frac{1}{2} e^{x^2} (x^2 - 1) + C$.
Thus,$\tan y \cdot e^{x^2} = \frac{1}{2} e^{x^2} (x^2 - 1) + C$.
Given $y(1) = 0$,so $\tan(0) \cdot e^1 = \frac{1}{2} e^1 (1 - 1) + C \implies 0 = 0 + C \implies C = 0$.
So,$\tan y = \frac{1}{2} (x^2 - 1)$.
For $x = \sqrt{3}$,$\tan y = \frac{1}{2} ((\sqrt{3})^2 - 1) = \frac{1}{2} (3 - 1) = 1$.
Since $\tan y = 1$,$y = \frac{\pi}{4}$.

(C) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} - 3y = \alpha$.
The integrating factor is $IF = e^{\int -3 dx} = e^{-3x}$.
Multiplying both sides by $IF$,we get $\frac{d}{dx}(y e^{-3x}) = \alpha e^{-3x}$.
Integrating both sides,$y e^{-3x} = \int \alpha e^{-3x} dx = \frac{\alpha e^{-3x}}{-3} + C$.
Thus,$y = -\frac{\alpha}{3} + C e^{3x}$.
Given $\lim_{x \rightarrow -\infty} f(x) = 7$,as $x \rightarrow -\infty$,$e^{3x} \rightarrow 0$. Therefore,$7 = -\frac{\alpha}{3}$,which implies $\alpha = -21$.
Substituting $\alpha = -21$ into the equation,$y = 7 + C e^{3x}$.
Using $f(0) = 1$,we get $1 = 7 + C$,so $C = -6$.
Thus,$f(x) = 7 - 6 e^{3x}$.
Now,calculate $f(-\log_e 3) = 7 - 6 e^{3(-\log_e 3)} = 7 - 6 e^{\log_e(3^{-3})} = 7 - 6(3^{-3}) = 7 - 6(\frac{1}{27}) = 7 - \frac{6}{27} = 7 - \frac{2}{9} = \frac{63-2}{9} = \frac{61}{9}$.
Therefore,$9 f(-\log_e 3) = 9 \times \frac{61}{9} = 61$.

(A) Given the limit $\lim _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1$.
Applying $L$'$H$ôpital's rule with respect to $t$:
$\lim _{t \rightarrow x} \frac{2t f(x) - x^2 f'(t)}{1} = 1$.
Substituting $t=x$,we get $2x f(x) - x^2 f'(x) = 1$.
Rearranging the terms: $f'(x) - \frac{2}{x} f(x) = -\frac{1}{x^2}$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{2}{x}$ and $Q(x) = -\frac{1}{x^2}$.
The integrating factor $IF = e^{\int P(x) dx} = e^{\int -\frac{2}{x} dx} = e^{-2 \ln x} = \frac{1}{x^2}$.
The solution is $f(x) \cdot \frac{1}{x^2} = \int (-\frac{1}{x^2}) \cdot \frac{1}{x^2} dx = \int -x^{-4} dx = \frac{x^{-3}}{3} + C = \frac{1}{3x^3} + C$.
Thus,$f(x) = \frac{1}{3x} + Cx^2$.
Given $f(1) = 1$,we have $1 = \frac{1}{3} + C$,which implies $C = \frac{2}{3}$.
Therefore,$f(x) = \frac{1}{3x} + \frac{2x^2}{3}$.

(A) Given the linear differential equation $f^{\prime}(x) + \frac{f(x)}{x} = 2$.
The integrating factor is $e^{\int \frac{1}{x} dx} = e^{\ln x} = x$.
Multiplying by the integrating factor,we get $\frac{d}{dx}(x f(x)) = 2x$.
Integrating both sides,$x f(x) = x^2 + c$,which gives $f(x) = x + \frac{c}{x}$ for all $x \in (0, \infty)$.
Given $f(1) \neq 1$,we have $1 + c \neq 1$,so $c \neq 0$.
Now,$f^{\prime}(x) = 1 - \frac{c}{x^2}$.
Evaluating option $A$: $\lim _{x \rightarrow 0+} f^{\prime}\left(\frac{1}{x}\right) = \lim _{x \rightarrow 0+} (1 - c x^2) = 1$.
Evaluating option $B$: $\lim _{x \rightarrow 0+} x f\left(\frac{1}{x}\right) = \lim _{x \rightarrow 0+} x \left(\frac{1}{x} + cx\right) = \lim _{x \rightarrow 0+} (1 + cx^2) = 1 \neq 2$.
Evaluating option $C$: $\lim _{x \rightarrow 0+} x^2 f^{\prime}(x) = \lim _{x \rightarrow 0+} x^2 (1 - \frac{c}{x^2}) = \lim _{x \rightarrow 0+} (x^2 - c) = -c \neq 0$.
Thus,option $A$ is correct.

(C) Given the equation: $3 \int_1^x f(t) dt = x f(x) - \frac{x^3}{3}$.
Differentiating both sides with respect to $x$ using the Leibniz rule:
$3 f(x) = f(x) + x f'(x) - x^2$.
Rearranging the terms:
$2 f(x) = x f'(x) - x^2 \implies x f'(x) - 2 f(x) = x^2$.
Dividing by $x$ $(x \geq 1)$:
$f'(x) - \frac{2}{x} f(x) = x$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{2}{x}$ and $Q(x) = x$.
Integrating Factor $(IF)$ = $e^{\int P(x) dx} = e^{\int -\frac{2}{x} dx} = e^{-2 \ln x} = x^{-2} = \frac{1}{x^2}$.
The solution is given by $f(x) \cdot IF = \int Q(x) \cdot IF dx + C$:
$f(x) \cdot \frac{1}{x^2} = \int x \cdot \frac{1}{x^2} dx = \int \frac{1}{x} dx = \ln x + C$.
So,$f(x) = x^2 \ln x + C x^2$.
Using the condition $f(1) = \frac{1}{3}$:
$f(1) = 1^2 \ln(1) + C(1)^2 = \frac{1}{3} \implies 0 + C = \frac{1}{3} \implies C = \frac{1}{3}$.
Thus,$f(x) = x^2 \ln x + \frac{x^2}{3}$.
Calculating $f(e)$:
$f(e) = e^2 \ln(e) + \frac{e^2}{3} = e^2(1) + \frac{e^2}{3} = \frac{3e^2 + e^2}{3} = \frac{4e^2}{3}$.

(B) The given differential equation is $(x^2-5) \frac{dy}{dx} - 2xy = -2x(x^2-5)^2$.
Dividing by $(x^2-5)$,we get $\frac{dy}{dx} - \frac{2x}{x^2-5} y = -2x(x^2-5)$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{2x}{x^2-5}$ and $Q(x) = -2x(x^2-5)$.
The integrating factor $IF = e^{\int P(x) dx} = e^{-\int \frac{2x}{x^2-5} dx} = e^{-\ln|x^2-5|} = \frac{1}{x^2-5}$.
The solution is $y \cdot IF = \int Q(x) \cdot IF dx + c$.
$y \cdot \frac{1}{x^2-5} = \int -2x(x^2-5) \cdot \frac{1}{x^2-5} dx + c$.
$\frac{y}{x^2-5} = \int -2x dx + c = -x^2 + c$.
Given $y(2) = 7$,we have $\frac{7}{2^2-5} = -2^2 + c \Rightarrow \frac{7}{-1} = -4 + c \Rightarrow c = -3$.
So,$\frac{y}{x^2-5} = -x^2 - 3$,which means $y = -(x^2-5)(x^2+3) = -(x^4 - 2x^2 - 15) = -x^4 + 2x^2 + 15$.
To find the maximum value,let $t = x^2$ where $t \ge 0$. Then $y = -t^2 + 2t + 15$.
This is a downward parabola with vertex at $t = -\frac{b}{2a} = -\frac{2}{2(-1)} = 1$.
Since $t=1$ is in the domain $t \ge 0$,the maximum value is $y(1) = -(1)^2 + 2(1) + 15 = -1 + 2 + 15 = 16$.

(C) Given $f^{\prime}(x) - 2f(x) > 0$.
Multiply by the integrating factor $e^{-2x}$:
$e^{-2x} f^{\prime}(x) - 2e^{-2x} f(x) > 0$.
This is equivalent to $\frac{d}{dx}(f(x) e^{-2x}) > 0$.
Let $g(x) = f(x) e^{-2x}$. Since $g^{\prime}(x) > 0$,$g(x)$ is a strictly increasing function.
For $x > 0$,$g(x) > g(0)$.
Since $g(0) = f(0) e^0 = 1 \cdot 1 = 1$,we have $f(x) e^{-2x} > 1$,which implies $f(x) > e^{2x}$ for all $x > 0$.
Since $f(x) > e^{2x} > 0$ for $x > 0$,and $f^{\prime}(x) > 2f(x)$,we have $f^{\prime}(x) > 2e^{2x} > 0$.
Since $f^{\prime}(x) > 0$,$f(x)$ is an increasing function in $(0, \infty)$.
Thus,$A$ and $C$ are correct.

(C) The equation of the tangent at point $P(x, y)$ is $Y - y = \frac{dy}{dx}(X - x)$.
To find the $y$-intercept,set $X = 0$,which gives $Y = y - x \frac{dy}{dx}$.
According to the problem,the $y$-intercept is equal to the cube of the abscissa,so $y - x \frac{dy}{dx} = x^3$.
Rearranging this gives $x \frac{dy}{dx} - y = -x^3$,or $\frac{dy}{dx} - \frac{y}{x} = -x^2$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{1}{x}$ and $Q(x) = -x^2$.
The integrating factor is $IF = e^{\int -\frac{1}{x} dx} = e^{-\ln x} = \frac{1}{x}$.
The solution is $y \cdot \frac{1}{x} = \int (-x^2) \cdot \frac{1}{x} dx = \int -x dx = -\frac{x^2}{2} + C$.
Thus,$f(x) = -\frac{x^3}{2} + Cx$.
Given $f(1) = 1$,we have $1 = -\frac{1}{2} + C$,which implies $C = \frac{3}{2}$.
So,$f(x) = -\frac{x^3}{2} + \frac{3}{2}x$.
Calculating $f(-3) = -\frac{(-3)^3}{2} + \frac{3}{2}(-3) = -\frac{-27}{2} - \frac{9}{2} = \frac{18}{2} = 9$.

(A) Given the equation $6 \int_1^x f(t) dt = 3x f(x) - x^3$.
Differentiating both sides with respect to $x$ using the Newton-Leibniz theorem:
$6 f(x) = 3 f(x) + 3x f'(x) - 3x^2$.
Rearranging the terms:
$3x f'(x) = 3 f(x) + 3x^2 \Rightarrow x f'(x) - f(x) = x^2$.
Dividing by $x^2$ (assuming $x \geq 1$):
$\frac{x f'(x) - f(x)}{x^2} = 1 \Rightarrow \frac{d}{dx} \left( \frac{f(x)}{x} \right) = 1$.
Integrating both sides with respect to $x$:
$\frac{f(x)}{x} = x + C$.
Given $f(1) = 2$,we substitute $x=1$:
$\frac{f(1)}{1} = 1 + C \Rightarrow 2 = 1 + C \Rightarrow C = 1$.
Thus,$f(x) = x^2 + x$.
Calculating $f(2)$:
$f(2) = 2^2 + 2 = 4 + 2 = 6$.

(B) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = g^{\prime}(x)$ and $Q(x) = g(x)g^{\prime}(x)$.
The integrating factor ($I$.$F$.) is given by $e^{\int P(x) dx} = e^{\int g^{\prime}(x) dx} = e^{g(x)}$.
The general solution is $y \cdot e^{g(x)} = \int Q(x) e^{g(x)} dx + C = \int g(x) g^{\prime}(x) e^{g(x)} dx + C$.
Let $u = g(x)$,then $du = g^{\prime}(x) dx$. The integral becomes $\int u e^u du = u e^u - e^u = e^{g(x)}(g(x) - 1)$.
Thus,$y e^{g(x)} = e^{g(x)}(g(x) - 1) + C$.
Given $y(0) = 0$ and $g(0) = 0$,we substitute these values: $0 \cdot e^0 = e^0(0 - 1) + C \implies 0 = -1 + C \implies C = 1$.
So,the solution is $y e^{g(x)} = e^{g(x)}(g(x) - 1) + 1$.
To find $y(2)$,substitute $x = 2$ and $g(2) = 0$: $y(2) e^0 = e^0(0 - 1) + 1 \implies y(2) = -1 + 1 = 0$.

(A) The given differential equation is $\frac{dy}{dx} - y \tan x = 2x \sec x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -\tan x$ and $Q = 2x \sec x$.
The integrating factor ($I$.$F$.) is $e^{\int P dx} = e^{-\int \tan x dx} = e^{-\ln(\sec x)} = \cos x$.
Multiplying both sides by the $I$.$F$.,we get $\frac{d}{dx}(y \cos x) = 2x \sec x \cdot \cos x = 2x$.
Integrating both sides,$y \cos x = x^2 + C$.
Given $y(0) = 0$,we have $0 = 0^2 + C$,so $C = 0$.
Thus,$y = x^2 \sec x$.
For option $(A)$: $y(\frac{\pi}{4}) = (\frac{\pi}{4})^2 \sec(\frac{\pi}{4}) = \frac{\pi^2}{16} \cdot \sqrt{2} = \frac{\pi^2}{8\sqrt{2}}$. This is true.
For option $(D)$: $y'(x) = 2x \sec x + x^2 \sec x \tan x$.
$y'(\frac{\pi}{3}) = 2(\frac{\pi}{3}) \sec(\frac{\pi}{3}) + (\frac{\pi}{3})^2 \sec(\frac{\pi}{3}) \tan(\frac{\pi}{3}) = 2(\frac{\pi}{3})(2) + \frac{\pi^2}{9}(2)(\sqrt{3}) = \frac{4\pi}{3} + \frac{2\pi^2\sqrt{3}}{9}$. This is true.
Therefore,options $(A)$ and $(D)$ are correct.

(B) Given $f(x+y)=f(x) f^{\prime}(y)+f^{\prime}(x) f(y)$ and $f(0)=1$.
Setting $x=0$ and $y=0$ in the given equation:
$f(0+0)=f(0)f^{\prime}(0)+f^{\prime}(0)f(0)$
$f(0)=2f(0)f^{\prime}(0)$
Since $f(0)=1$,we have $1=2(1)f^{\prime}(0)$,which implies $f^{\prime}(0)=\frac{1}{2}$.
Now,setting $y=0$ in the original equation:
$f(x+0)=f(x)f^{\prime}(0)+f^{\prime}(x)f(0)$
$f(x)=f(x) \cdot \frac{1}{2} + f^{\prime}(x) \cdot 1$
$f^{\prime}(x) = f(x) - \frac{1}{2}f(x) = \frac{1}{2}f(x)$.
This is a first-order linear differential equation:
$\frac{f^{\prime}(x)}{f(x)} = \frac{1}{2}$.
Integrating both sides with respect to $x$:
$\int \frac{f^{\prime}(x)}{f(x)} dx = \int \frac{1}{2} dx$
$\ln(f(x)) = \frac{x}{2} + C$.
Using $f(0)=1$,we get $\ln(1) = 0 + C$,so $C=0$.
Thus,$f(x) = e^{x/2}$.
Finally,we calculate $\log _e(f(4))$:
$f(4) = e^{4/2} = e^2$.
$\log _e(f(4)) = \log _e(e^2) = 2$.

(B) The given differential equation is $\frac{dy}{dx} + \alpha y = x e^{\beta x}$. This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \alpha$ and $Q(x) = x e^{\beta x}$.
The integrating factor is $IF = e^{\int \alpha dx} = e^{\alpha x}$.
Multiplying both sides by $IF$,we get $\frac{d}{dx}(y e^{\alpha x}) = x e^{(\alpha+\beta)x}$.
Case $I$: If $\alpha + \beta = 0$,then $\beta = -\alpha$. The equation becomes $\frac{d}{dx}(y e^{\alpha x}) = x$. Integrating both sides,$y e^{\alpha x} = \frac{x^2}{2} + C$. Using $y(1) = 1$,we get $1 \cdot e^{\alpha} = \frac{1}{2} + C$,so $C = e^{\alpha} - \frac{1}{2}$. Thus,$y = \frac{x^2}{2} e^{-\alpha x} + (e^{\alpha} - \frac{1}{2}) e^{-\alpha x}$. For $\alpha = 1$,$y = \frac{x^2}{2} e^{-x} + (e - \frac{1}{2}) e^{-x}$,which matches option $(A)$.
Case $II$: If $\alpha + \beta \neq 0$,integrating $\int x e^{(\alpha+\beta)x} dx$ by parts gives $\frac{x e^{(\alpha+\beta)x}}{\alpha+\beta} - \frac{e^{(\alpha+\beta)x}}{(\alpha+\beta)^2} + C$. Thus,$y e^{\alpha x} = \frac{x e^{(\alpha+\beta)x}}{\alpha+\beta} - \frac{e^{(\alpha+\beta)x}}{(\alpha+\beta)^2} + C$. Using $y(1) = 1$,$C = e^{\alpha} - \frac{e^{\alpha+\beta}}{\alpha+\beta} + \frac{e^{\alpha+\beta}}{(\alpha+\beta)^2}$. For $\alpha = -1, \beta = 2$,we have $\alpha+\beta = 1$. Substituting these values leads to the form in option $(C)$.
Therefore,both $(A)$ and $(C)$ belong to $S$.

(B) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{x}{x^2-1}$ and $Q(x) = \frac{x^4+2x}{\sqrt{1-x^2}}$.
Integrating factor ($I$.$F$.) $= e^{\int \frac{x}{x^2-1} dx} = e^{\frac{1}{2} \ln|x^2-1|} = e^{\frac{1}{2} \ln(1-x^2)} = \sqrt{1-x^2}$.
The general solution is $y \cdot \sqrt{1-x^2} = \int \frac{x^4+2x}{\sqrt{1-x^2}} \cdot \sqrt{1-x^2} dx + c = \int (x^4+2x) dx + c = \frac{x^5}{5} + x^2 + c$.
Given $f(0)=0$,we have $0 \cdot 1 = 0 + 0 + c$,so $c=0$.
Thus,$f(x) = \frac{x^5/5 + x^2}{\sqrt{1-x^2}}$.
We need to evaluate $I = \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) dx$. Since $f(x) = \frac{x^5/5}{\sqrt{1-x^2}} + \frac{x^2}{\sqrt{1-x^2}}$,the first part is an odd function,so its integral over the symmetric interval is $0$.
Thus,$I = 2 \int_{0}^{\frac{\sqrt{3}}{2}} \frac{x^2}{\sqrt{1-x^2}} dx$.
Let $x = \sin \theta$,then $dx = \cos \theta d\theta$. When $x=0, \theta=0$; when $x=\frac{\sqrt{3}}{2}, \theta=\frac{\pi}{3}$.
$I = 2 \int_{0}^{\frac{\pi}{3}} \frac{\sin^2 \theta}{\cos \theta} \cos \theta d\theta = 2 \int_{0}^{\frac{\pi}{3}} \sin^2 \theta d\theta = \int_{0}^{\frac{\pi}{3}} (1 - \cos 2\theta) d\theta$.
$I = [\theta - \frac{1}{2} \sin 2\theta]_{0}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{1}{2} \sin(\frac{2\pi}{3}) = \frac{\pi}{3} - \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\pi}{3} - \frac{\sqrt{3}}{4}$.

(A) The given differential equation is $(1+e^x) y^{\prime}+y e^x=1$. Dividing by $(1+e^x)$,we get $\frac{d y}{d x}+\frac{e^x}{1+e^x} y = \frac{1}{1+e^x}$.
This is a linear differential equation of the form $\frac{d y}{d x}+P(x)y=Q(x)$.
The integrating factor is $I.F. = e^{\int \frac{e^x}{1+e^x} dx} = e^{\ln(1+e^x)} = 1+e^x$.
The solution is $y(1+e^x) = \int 1 dx = x+c$.
Given $y(0)=2$,we have $2(1+e^0) = 0+c \Rightarrow c=4$.
Thus,$y(x) = \frac{x+4}{1+e^x}$.
For $(A)$,$y(-4) = \frac{-4+4}{1+e^{-4}} = 0$. So $(A)$ is true.
For $(B)$,$y(-2) = \frac{-2+4}{1+e^{-2}} = \frac{2}{1+e^{-2}} \neq 0$. So $(B)$ is false.
For $(C)$ and $(D)$,we find the critical points by setting $y^{\prime}(x) = 0$.
$y^{\prime}(x) = \frac{(1+e^x) - (x+4)e^x}{(1+e^x)^2} = \frac{1+e^x - xe^x - 4e^x}{(1+e^x)^2} = \frac{1-e^x(x+3)}{(1+e^x)^2}$.
Let $g(x) = 1-e^x(x+3)$.
$g(0) = 1-e^0(3) = 1-3 = -2$.
$g(-1) = 1-e^{-1}(2) = 1-\frac{2}{e} \approx 1-0.736 = 0.264$.
Since $g(x)$ is continuous and $g(-1) > 0$ and $g(0) < 0$,there exists a root in $(-1, 0)$.
Thus,$y(x)$ has a critical point in $(-1, 0)$. So $(C)$ is true.
The correct options are $(A)$ and $(C)$.

(C) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 12$ and $Q = \cos \left(\frac{\pi}{12} x\right)$.
The integrating factor $(I.F.)$ is $e^{\int 12 dx} = e^{12x}$.
The general solution is $y \cdot e^{12x} = \int e^{12x} \cos \left(\frac{\pi}{12} x\right) dx + C$.
Using the formula $\int e^{ax} \cos(bx) dx = \frac{e^{ax}}{a^2 + b^2} (a \cos(bx) + b \sin(bx))$,we get:
$y \cdot e^{12x} = \frac{e^{12x}}{12^2 + (\frac{\pi}{12})^2} \left(12 \cos \left(\frac{\pi}{12} x\right) + \frac{\pi}{12} \sin \left(\frac{\pi}{12} x\right)\right) + C$.
Simplifying,$y = \frac{1}{144 + \frac{\pi^2}{144}} \left(12 \cos \left(\frac{\pi}{12} x\right) + \frac{\pi}{12} \sin \left(\frac{\pi}{12} x\right)\right) + C e^{-12x}$.
Given $y(0) = 0$,we have $0 = \frac{12}{144 + \frac{\pi^2}{144}} + C$,so $C = -\frac{12}{144 + \frac{\pi^2}{144}}$.
Thus,$y(x) = \frac{1}{144 + \frac{\pi^2}{144}} \left(12 \cos \left(\frac{\pi}{12} x\right) + \frac{\pi}{12} \sin \left(\frac{\pi}{12} x\right) - 12 e^{-12x}\right)$.
As $x \to \infty$,$e^{-12x} \to 0$,so $y(x)$ approaches a periodic function $f(x) = A \cos \left(\frac{\pi}{12} x - \phi\right)$.
Since $y(x)$ approaches a periodic function,for a value $\beta$ within the range of this periodic function,the line $y = \beta$ will intersect the curve $y = y(x)$ at infinitely many points as $x \to \infty$. Thus,statement $C$ is $TRUE$.

(B) Given $\lim _{t \rightarrow x} \frac{t^{10} f(x)-x^{10} f(t)}{t^9-x^9}=1$.
Applying $L$'$H$ôpital's rule with respect to $t$:
$\lim _{t \rightarrow x} \frac{10 t^9 f(x)-x^{10} f^{\prime}(t)}{9 t^8}=1$
$\Rightarrow \frac{10 x^9 f(x)-x^{10} f^{\prime}(x)}{9 x^8}=1$
$\Rightarrow 10 x f(x)-x^2 f^{\prime}(x)=9 x^8 \cdot \frac{9}{x^8} = 9$
$\Rightarrow f^{\prime}(x)-\frac{10}{x} f(x)=-\frac{9}{x^2}$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{10}{x}$ and $Q(x) = -\frac{9}{x^2}$.
The integrating factor $IF = e^{\int P(x) dx} = e^{-10 \ln x} = x^{-10} = \frac{1}{x^{10}}$.
The solution is $y \cdot IF = \int Q(x) \cdot IF dx + C$.
$\frac{f(x)}{x^{10}} = \int -\frac{9}{x^2} \cdot \frac{1}{x^{10}} dx = -9 \int x^{-12} dx = -9 \left( \frac{x^{-11}}{-11} \right) + C = \frac{9}{11 x^{11}} + C$.
Given $f(1)=2$,we have $\frac{2}{1} = \frac{9}{11} + C \Rightarrow C = 2 - \frac{9}{11} = \frac{13}{11}$.
Thus,$f(x) = x^{10} \left( \frac{9}{11 x^{11}} + \frac{13}{11} \right) = \frac{9}{11 x} + \frac{13}{11} x^{10}$.
Therefore,option $(B)$ is correct.

(C) The given differential equation is $y^2 dx + (x - \frac{1}{y}) dy = 0$.
Dividing by $y^2 dy$,we get $\frac{dx}{dy} + \frac{x}{y^2} = \frac{1}{y^3}$.
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = \frac{1}{y^2}$ and $Q(y) = \frac{1}{y^3}$.
The integrating factor ($I$.$F$.) is $e^{\int P(y) dy} = e^{\int y^{-2} dy} = e^{-1/y}$.
The solution is $x \cdot e^{-1/y} = \int Q(y) \cdot e^{-1/y} dy + C$.
Let $t = -1/y$,then $dt = \frac{1}{y^2} dy$.
$x \cdot e^{-1/y} = \int (-t) e^t dt + C = -(t e^t - e^t) + C = e^t(1 - t) + C$.
Substituting $t = -1/y$,we get $x \cdot e^{-1/y} = e^{-1/y}(1 + \frac{1}{y}) + C$.
Given $x(1) = 1$,we have $1 \cdot e^{-1} = e^{-1}(1 + 1) + C$,so $e^{-1} = 2e^{-1} + C$,which implies $C = -e^{-1}$.
Thus,$x = 1 + \frac{1}{y} - e^{1/y} \cdot e^{-1} = 1 + \frac{1}{y} - e^{(1/y) - 1}$.
For $y = 1/2$,$x = 1 + \frac{1}{1/2} - e^{(1/(1/2)) - 1} = 1 + 2 - e^{2-1} = 3 - e$.

(D) The given differential equation is $(1+y^2)+(x-2 e^{\tan ^{-1} y}) \frac{d y}{d x}=0$.
Rearranging the terms,we get $\frac{d x}{d y} = -\frac{x-2 e^{\tan ^{-1} y}}{1+y^2} = \frac{2 e^{\tan ^{-1} y}-x}{1+y^2}$.
This can be written as $\frac{d x}{d y} + \frac{x}{1+y^2} = \frac{2 e^{\tan ^{-1} y}}{1+y^2}$.
This is a linear differential equation of the form $\frac{d x}{d y} + P(y)x = Q(y)$,where $P(y) = \frac{1}{1+y^2}$ and $Q(y) = \frac{2 e^{\tan ^{-1} y}}{1+y^2}$.
The integrating factor ($I$.$F$.) is $e^{\int P(y) dy} = e^{\int \frac{1}{1+y^2} dy} = e^{\tan ^{-1} y}$.
The solution is $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + C$.
$x e^{\tan ^{-1} y} = \int \frac{2 e^{\tan ^{-1} y}}{1+y^2} \cdot e^{\tan ^{-1} y} dy = \int \frac{2 e^{2 \tan ^{-1} y}}{1+y^2} dy$.
Let $t = \tan ^{-1} y$,then $dt = \frac{1}{1+y^2} dy$.
$x e^{\tan ^{-1} y} = \int 2 e^{2t} dt = e^{2t} + C = e^{2 \tan ^{-1} y} + C$.
Given $f(0)=1$,i.e.,$x=1$ when $y=0$:
$1 \cdot e^{\tan ^{-1} 0} = e^{2 \tan ^{-1} 0} + C \implies 1 \cdot 1 = 1 + C \implies C = 0$.
Thus,$x e^{\tan ^{-1} y} = e^{2 \tan ^{-1} y} \implies x = e^{\tan ^{-1} y}$.
For $y = \frac{1}{\sqrt{3}}$,$x = e^{\tan ^{-1}(1/\sqrt{3})} = e^{\pi / 6}$.

(A) The given differential equation is $\frac{dy}{dx} + \frac{x}{x^2-1}y = \frac{x^6+4x}{\sqrt{1-x^2}}$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$.
Integrating factor ($I$.$F$.) = $e^{\int P(x) dx} = e^{\int \frac{x}{x^2-1} dx} = e^{\frac{1}{2} \ln|x^2-1|} = \sqrt{1-x^2}$ (since $-1 < x < 1$,$x^2-1 < 0$,so $|x^2-1| = 1-x^2$).
The solution is $y \cdot \sqrt{1-x^2} = \int \frac{x^6+4x}{\sqrt{1-x^2}} \cdot \sqrt{1-x^2} dx = \int (x^6+4x) dx = \frac{x^7}{7} + 2x^2 + C$.
Given $f(0)=0$,we have $0 = 0 + 0 + C \Rightarrow C=0$.
Thus,$f(x) = \frac{x^7/7 + 2x^2}{\sqrt{1-x^2}}$.
We need to evaluate $6 \int_{-1/2}^{1/2} f(x) dx = 6 \int_{-1/2}^{1/2} \frac{x^7/7 + 2x^2}{\sqrt{1-x^2}} dx$.
Since $\frac{x^7/7}{\sqrt{1-x^2}}$ is an odd function,its integral over $[-1/2, 1/2]$ is $0$.
So,the expression becomes $6 \int_{-1/2}^{1/2} \frac{2x^2}{\sqrt{1-x^2}} dx = 24 \int_0^{1/2} \frac{x^2}{\sqrt{1-x^2}} dx$.
Let $x = \sin \theta$,then $dx = \cos \theta d\theta$. When $x=0, \theta=0$; when $x=1/2, \theta=\pi/6$.
Integral = $24 \int_0^{\pi/6} \frac{\sin^2 \theta}{\cos \theta} \cos \theta d\theta = 24 \int_0^{\pi/6} \sin^2 \theta d\theta = 24 \int_0^{\pi/6} \frac{1-\cos 2\theta}{2} d\theta = 12 [\theta - \frac{\sin 2\theta}{2}]_0^{\pi/6} = 12(\frac{\pi}{6} - \frac{\sqrt{3}}{4}) = 2\pi - 3\sqrt{3}$.
Comparing with $2\pi - \alpha$,we get $\alpha = 3\sqrt{3}$.
Therefore,$\alpha^2 = (3\sqrt{3})^2 = 27$.

(A) The given differential equation is $(1+x^2)dy = (5x^2\sqrt{1+x^2} - xy)dx$.
Rearranging,we get $(1+x^2)\frac{dy}{dx} + xy = 5x^2\sqrt{1+x^2}$.
Dividing by $(1+x^2)$,we get $\frac{dy}{dx} + \frac{x}{1+x^2}y = \frac{5x^2}{\sqrt{1+x^2}}$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{x}{1+x^2}$ and $Q(x) = \frac{5x^2}{\sqrt{1+x^2}}$.
Integrating factor $I.F. = e^{\int P(x)dx} = e^{\int \frac{x}{1+x^2}dx} = e^{\frac{1}{2}\ln(1+x^2)} = \sqrt{1+x^2}$.
The solution is $y \cdot I.F. = \int Q(x) \cdot I.F. dx + C$.
$y\sqrt{1+x^2} = \int \frac{5x^2}{\sqrt{1+x^2}} \cdot \sqrt{1+x^2} dx = \int 5x^2 dx = \frac{5x^3}{3} + C$.
Given $y(0)=0$,we have $0\sqrt{1+0} = \frac{5(0)^3}{3} + C$,so $C=0$.
Thus,$y = \frac{5x^3}{3\sqrt{1+x^2}}$.
For $x=\sqrt{3}$,$y(\sqrt{3}) = \frac{5(\sqrt{3})^3}{3\sqrt{1+(\sqrt{3})^2}} = \frac{5(3\sqrt{3})}{3\sqrt{4}} = \frac{15\sqrt{3}}{3(2)} = \frac{5\sqrt{3}}{2}$.

(A) Given the equation: $2(x+2)^2 f(x) - 3(x+2)^2 = 10 \int_0^x (t+2) f(t) dt$.
At $x=0$,$2(2)^2 f(0) - 3(2)^2 = 10(0) \implies 8 f(0) = 12 \implies f(0) = \frac{3}{2}$.
Differentiating both sides with respect to $x$ using the Leibniz rule:
$4(x+2) f(x) + 2(x+2)^2 f'(x) - 6(x+2) = 10(x+2) f(x)$.
Divide by $2(x+2)$ (assuming $x \geq 0$):
$2 f(x) + (x+2) f'(x) - 3 = 5 f(x)$.
$(x+2) f'(x) - 3 f(x) = 3$.
This is a linear differential equation: $\frac{df}{dx} - \frac{3}{x+2} f = \frac{3}{x+2}$.
Integrating factor $IF = e^{\int -\frac{3}{x+2} dx} = e^{-3 \ln(x+2)} = (x+2)^{-3}$.
Multiplying by $IF$: $\frac{d}{dx} [f(x) (x+2)^{-3}] = 3(x+2)^{-4}$.
Integrating both sides: $f(x) (x+2)^{-3} = \int 3(x+2)^{-4} dx = -(x+2)^{-3} + C$.
$f(x) = -1 + C(x+2)^3$.
Using $f(0) = \frac{3}{2}$: $\frac{3}{2} = -1 + C(2)^3 \implies \frac{5}{2} = 8C \implies C = \frac{5}{16}$.
Thus,$f(x) = \frac{5}{16}(x+2)^3 - 1$.
For $f(2)$: $f(2) = \frac{5}{16}(4)^3 - 1 = \frac{5}{16}(64) - 1 = 5(4) - 1 = 19$.

(C) Given the differential equation: $x^2 f^{\prime}(x) - 2 x f(x) = 3$.
Divide both sides by $x^4$ (where $x \neq 0$):
$\frac{x^2 f^{\prime}(x) - 2 x f(x)}{x^4} = \frac{3}{x^4}$
This simplifies to the derivative of a quotient:
$\frac{d}{dx} \left( \frac{f(x)}{x^2} \right) = 3 x^{-4}$
Integrating both sides with respect to $x$:
$\frac{f(x)}{x^2} = \int 3 x^{-4} dx = 3 \left( \frac{x^{-3}}{-3} \right) + C = -\frac{1}{x^3} + C$
Multiplying by $x^2$:
$f(x) = -\frac{1}{x} + C x^2$
Given $f(1) = 4$,substitute $x=1$:
$4 = -\frac{1}{1} + C(1)^2 \Rightarrow 4 = -1 + C \Rightarrow C = 5$.
Thus,$f(x) = 5x^2 - \frac{1}{x}$.
Now,calculate $2f(2)$:
$f(2) = 5(2)^2 - \frac{1}{2} = 20 - 0.5 = 19.5$.
$2f(2) = 2 \times 19.5 = 39$.

(C) Given the differential equation: $2 \cos x \frac{d y}{d x} = \sin 2x - 4y \sin x$.
Dividing by $2 \cos x$,we get: $\frac{d y}{d x} = \frac{2 \sin x \cos x}{2 \cos x} - \frac{4y \sin x}{2 \cos x} = \sin x - 2y \tan x$.
Rearranging gives the linear differential equation: $\frac{d y}{d x} + 2y \tan x = \sin x$.
The integrating factor $I.F. = e^{\int 2 \tan x \, dx} = e^{2 \ln |\sec x|} = \sec^2 x$.
Multiplying by $I.F.$,we have: $y \sec^2 x = \int \sin x \sec^2 x \, dx = \int \tan x \sec x \, dx = \sec x + C$.
Given $y(\frac{\pi}{3}) = 0$,we substitute $x = \frac{\pi}{3}$: $0 \cdot \sec^2(\frac{\pi}{3}) = \sec(\frac{\pi}{3}) + C \implies 0 = 2 + C \implies C = -2$.
Thus,$y \sec^2 x = \sec x - 2$,which simplifies to $y = \cos x - 2 \cos^2 x$.
Now,$y(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) - 2 \cos^2(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} - 2(\frac{1}{2}) = \frac{1}{\sqrt{2}} - 1$.
Next,$y^{\prime}(x) = -\sin x - 4 \cos x(-\sin x) = -\sin x + 4 \sin x \cos x = -\sin x + 2 \sin 2x$.
$y^{\prime}(\frac{\pi}{4}) = -\sin(\frac{\pi}{4}) + 2 \sin(\frac{\pi}{2}) = -\frac{1}{\sqrt{2}} + 2$.
Finally,$y^{\prime}(\frac{\pi}{4}) + y(\frac{\pi}{4}) = (-\frac{1}{\sqrt{2}} + 2) + (\frac{1}{\sqrt{2}} - 1) = 1$.

(A) The given differential equation is $\sqrt{4-x^2} \frac{dy}{dx} + y \sin^{-1}\left(\frac{x}{2}\right) = \left(\sin^{-1}\left(\frac{x}{2}\right)\right)^3$.
Dividing by $\sqrt{4-x^2}$,we get $\frac{dy}{dx} + \frac{\sin^{-1}(x/2)}{\sqrt{4-x^2}} y = \frac{(\sin^{-1}(x/2))^3}{\sqrt{4-x^2}}$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{\sin^{-1}(x/2)}{\sqrt{4-x^2}}$ and $Q(x) = \frac{(\sin^{-1}(x/2))^3}{\sqrt{4-x^2}}$.
The integrating factor $IF = e^{\int P(x) dx} = e^{\int \frac{\sin^{-1}(x/2)}{\sqrt{4-x^2}} dx}$.
Let $u = \sin^{-1}(x/2)$,then $du = \frac{1}{\sqrt{1-(x/2)^2}} \cdot \frac{1}{2} dx = \frac{1}{\sqrt{4-x^2}} dx$.
So,$IF = e^{\int u du} = e^{u^2/2} = e^{\frac{(\sin^{-1}(x/2))^2}{2}}$.
The solution is $y \cdot IF = \int Q(x) \cdot IF dx + C$.
$y e^{\frac{(\sin^{-1}(x/2))^2}{2}} = \int u^3 e^{u^2/2} du + C$.
Let $t = u^2/2$,then $dt = u du$. The integral becomes $\int 2t e^t dt = 2(t e^t - e^t) + C = 2e^t(t-1) + C$.
Substituting back,$y e^{u^2/2} = 2e^{u^2/2} (\frac{u^2}{2} - 1) + C = e^{u^2/2} (u^2 - 2) + C$.
$y = u^2 - 2 + C e^{-u^2/2}$.
Given $y(2) = \frac{\pi^2-8}{4}$,$u = \sin^{-1}(1) = \pi/2$.
$\frac{\pi^2-8}{4} = \frac{\pi^2}{4} - 2 + C e^{-\pi^2/8} \Rightarrow C = 0$.
Thus,$y = u^2 - 2 = (\sin^{-1}(x/2))^2 - 2$.
At $x=0$,$y(0) = (\sin^{-1}(0))^2 - 2 = -2$.
Therefore,$y^2(0) = (-2)^2 = 4$.

(D) The given differential equation is $\cos x(\ln(\cos x))^2 dy + (\sin x - 3y \sin x \ln(\cos x)) dx = 0$.
Dividing by $dx \cdot \cos x(\ln(\cos x))^2$,we get:
$\frac{dy}{dx} - \frac{3 \sin x \ln(\cos x)}{\cos x(\ln(\cos x))^2} y = -\frac{\sin x}{\cos x(\ln(\cos x))^2}$
$\frac{dy}{dx} - \frac{3 \tan x}{\ln(\cos x)} y = -\frac{\tan x}{(\ln(\cos x))^2}$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{3 \tan x}{\ln(\cos x)}$.
Integrating factor $I.F. = e^{\int P(x) dx} = e^{\int -\frac{3 \tan x}{\ln(\cos x)} dx}$.
Let $u = \ln(\cos x)$,then $du = \frac{1}{\cos x} (-\sin x) dx = -\tan x dx$.
So,$I.F. = e^{3 \int \frac{1}{u} du} = e^{3 \ln u} = u^3 = (\ln(\cos x))^3$.
The solution is $y \cdot I.F. = \int Q(x) \cdot I.F. dx + C$.
$y(\ln(\cos x))^3 = \int -\frac{\tan x}{(\ln(\cos x))^2} \cdot (\ln(\cos x))^3 dx + C$.
$y(\ln(\cos x))^3 = -\int \tan x \ln(\cos x) dx + C$.
Let $v = \ln(\cos x)$,then $dv = -\tan x dx$.
$y(\ln(\cos x))^3 = \int v dv + C = \frac{v^2}{2} + C = \frac{(\ln(\cos x))^2}{2} + C$.
Given $y(\frac{\pi}{4}) = -\frac{1}{\ln 2}$. Since $\ln(\cos \frac{\pi}{4}) = \ln(\frac{1}{\sqrt{2}}) = -\frac{1}{2} \ln 2$.
$(-\frac{1}{\ln 2})(-\frac{1}{2} \ln 2)^3 = \frac{1}{2} (-\frac{1}{2} \ln 2)^2 + C$.
$(-\frac{1}{\ln 2})(-\frac{1}{8} (\ln 2)^3) = \frac{1}{2} (\frac{1}{4} (\ln 2)^2) + C$.
$\frac{1}{8} (\ln 2)^2 = \frac{1}{8} (\ln 2)^2 + C \implies C = 0$.
Thus,$y(\ln(\cos x))^3 = \frac{(\ln(\cos x))^2}{2} \implies y = \frac{1}{2 \ln(\cos x)}$.
For $x = \frac{\pi}{6}$,$y(\frac{\pi}{6}) = \frac{1}{2 \ln(\cos \frac{\pi}{6})} = \frac{1}{2 \ln(\frac{\sqrt{3}}{2})} = \frac{1}{2(\frac{1}{2} \ln 3 - \ln 2)} = \frac{1}{\ln 3 - \ln 4}$.

(A) The given differential equation is $\frac{dy}{dx}+(\tan x)y=\frac{2+\sec x}{(1+2\sec x)^2}$.
This is a linear differential equation of the form $\frac{dy}{dx}+Py=Q$,where $P=\tan x$ and $Q=\frac{2+\sec x}{(1+2\sec x)^2}$.
The integrating factor $IF = e^{\int \tan x dx} = e^{\ln|\sec x|} = \sec x$.
The general solution is $y \cdot \sec x = \int Q \cdot IF dx + C$.
$y \sec x = \int \frac{2+\sec x}{(1+2\sec x)^2} \sec x dx = \int \frac{2\cos x+1}{(\cos x+2)^2} dx$.
Using the substitution $\cos x = \frac{1-t^2}{1+t^2}$ where $t = \tan(x/2)$,we get $dx = \frac{2dt}{1+t^2}$.
$y \sec x = \int \frac{2(\frac{1-t^2}{1+t^2})+1}{(\frac{1-t^2}{1+t^2}+2)^2} \frac{2dt}{1+t^2} = \int \frac{3-t^2}{(t^2+3)^2} 2dt$.
Let $u = t + \frac{3}{t}$,then $du = (1 - \frac{3}{t^2}) dt = \frac{t^2-3}{t^2} dt$. This integral simplifies to $y \sec x = \frac{2}{t+3/t} + C = \frac{2t}{t^2+3} + C$.
Given $f(\pi/3) = \sqrt{3}/10$,at $x=\pi/3$,$t = \tan(\pi/6) = 1/\sqrt{3}$.
$(\sqrt{3}/10) \cdot 2 = \frac{2(1/\sqrt{3})}{1/3+3} + C \implies \sqrt{3}/5 = \frac{2/\sqrt{3}}{10/3} + C = \frac{\sqrt{3}}{5} + C \implies C=0$.
Thus $y \sec x = \frac{2t}{t^2+3}$. At $x=\pi/4$,$t = \tan(\pi/8) = \sqrt{2}-1$.
$y \cdot \sqrt{2} = \frac{2(\sqrt{2}-1)}{(\sqrt{2}-1)^2+3} = \frac{2(\sqrt{2}-1)}{2-2\sqrt{2}+1+3} = \frac{2(\sqrt{2}-1)}{6-2\sqrt{2}} = \frac{\sqrt{2}-1}{3-\sqrt{2}}$.
$y = \frac{\sqrt{2}-1}{\sqrt{2}(3-\sqrt{2})} = \frac{\sqrt{2}-1}{3\sqrt{2}-2} \cdot \frac{3\sqrt{2}+2}{3\sqrt{2}+2} = \frac{6+\sqrt{2}-2}{18-4} = \frac{4+\sqrt{2}}{14}$.
Wait,re-evaluating the integral $\int \frac{3-t^2}{(t^2+3)^2} 2dt$: $y \sec x = \frac{2t}{t^2+3} + C$ is correct. With $C=0$,$y = \cos x \cdot \frac{2t}{t^2+3} = \frac{1-t^2}{1+t^2} \cdot \frac{2t}{t^2+3}$.
At $t=\sqrt{2}-1$,$t^2 = 3-2\sqrt{2}$,$y = \frac{1-(3-2\sqrt{2})}{1+3-2\sqrt{2}} \cdot \frac{2(\sqrt{2}-1)}{3-2\sqrt{2}+3} = \frac{2\sqrt{2}-2}{4-2\sqrt{2}} \cdot \frac{2\sqrt{2}-2}{6-2\sqrt{2}} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}-1}{3-\sqrt{2}} = \frac{4-\sqrt{2}}{14}$.

(B) Given $10 \int_1^x f(t) dt = 5x f(x) - x^5 - 9$.
Differentiating both sides with respect to $x$ using the Leibniz rule:
$10 f(x) = 5 f(x) + 5x f'(x) - 5x^4$.
Rearranging the terms:
$5 f(x) + 5x^4 = 5x f'(x)$
$f'(x) - \frac{1}{x} f(x) = x^3$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{1}{x}$ and $Q(x) = x^3$.
The integrating factor $IF = e^{\int P(x) dx} = e^{\int -\frac{1}{x} dx} = e^{-\ln x} = \frac{1}{x}$.
Multiplying by $IF$:
$\frac{1}{x} f'(x) - \frac{1}{x^2} f(x) = x^2$.
Integrating both sides:
$\frac{f(x)}{x} = \int x^2 dx = \frac{x^3}{3} + C$.
To find $C$,put $x=1$ in the original equation: $10 \int_1^1 f(t) dt = 5(1)f(1) - 1^5 - 9 \Rightarrow 0 = 5f(1) - 10 \Rightarrow f(1) = 2$.
Substituting $x=1$ in $\frac{f(x)}{x} = \frac{x^3}{3} + C$:
$\frac{2}{1} = \frac{1}{3} + C \Rightarrow C = 2 - \frac{1}{3} = \frac{5}{3}$.
Thus,$f(x) = \frac{x^4}{3} + \frac{5x}{3}$.
For $x=3$: $f(3) = \frac{3^4}{3} + \frac{5(3)}{3} = \frac{81}{3} + 5 = 27 + 5 = 32$.

(A) The given differential equation is $\frac{dy}{dx} + (2 \sec^2 x)y = 2 \sec^2 x + 3 \tan x \sec^2 x$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = 2 \sec^2 x$ and $Q(x) = 2 \sec^2 x + 3 \tan x \sec^2 x$.
Integrating factor ($I$.$F$.) $= e^{\int 2 \sec^2 x dx} = e^{2 \tan x}$.
The solution is $y \cdot e^{2 \tan x} = \int (2 \sec^2 x + 3 \tan x \sec^2 x) e^{2 \tan x} dx$.
Let $u = \tan x$,then $du = \sec^2 x dx$. The integral becomes $\int (2 + 3u) e^{2u} du$.
Using integration by parts: $\int (2 + 3u) e^{2u} du = (2 + 3u) \frac{e^{2u}}{2} - \int 3 \frac{e^{2u}}{2} du = (1 + \frac{3}{2}u) e^{2u} - \frac{3}{4} e^{2u} + C = (\frac{3}{2}u + \frac{1}{4}) e^{2u} + C$.
So,$y \cdot e^{2 \tan x} = (\frac{3}{2} \tan x + \frac{1}{4}) e^{2 \tan x} + C$.
Dividing by $e^{2 \tan x}$,we get $y = \frac{3}{2} \tan x + \frac{1}{4} + C e^{-2 \tan x}$.
Given $y(0) = \frac{5}{4}$,so $\frac{5}{4} = 0 + \frac{1}{4} + C \implies C = 1$.
Thus,$y(x) = \frac{3}{2} \tan x + \frac{1}{4} + e^{-2 \tan x}$.
At $x = \frac{\pi}{4}$,$y(\frac{\pi}{4}) = \frac{3}{2}(1) + \frac{1}{4} + e^{-2} = \frac{7}{4} + e^{-2}$.
Therefore,$12(y(\frac{\pi}{4}) - e^{-2}) = 12(\frac{7}{4}) = 21$.

(B) Given $\int_0^x g(t) dt = x - \int_0^x tg(t) dt$.
Differentiating both sides with respect to $x$ using the Leibniz rule:
$g(x) = 1 - xg(x)$.
Rearranging gives $g(x)(1+x) = 1$,so $g(x) = \frac{1}{1+x}$.
Substitute $g(x)$ into the differential equation:
$\frac{dy}{dx} - y \tan x = 2(x+1) \sec x \cdot \frac{1}{1+x} = 2 \sec x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -\tan x$ and $Q = 2 \sec x$.
The integrating factor $IF = e^{\int P dx} = e^{-\int \tan x dx} = e^{\ln(\cos x)} = \cos x$.
The solution is $y \cdot IF = \int Q \cdot IF dx + C$.
$y \cos x = \int 2 \sec x \cdot \cos x dx + C = \int 2 dx + C = 2x + C$.
Given $y(0) = 0$,we have $0 \cdot \cos(0) = 2(0) + C$,so $C = 0$.
Thus,$y \cos x = 2x$,which means $y = 2x \sec x$.
At $x = \frac{\pi}{3}$,$y(\frac{\pi}{3}) = 2 \cdot \frac{\pi}{3} \cdot \sec(\frac{\pi}{3}) = 2 \cdot \frac{\pi}{3} \cdot 2 = \frac{4 \pi}{3}$.

(B) The given differential equation is $\frac{dy}{dx} + 3(\tan^2 x + 1)y = \sec^2 x$.
Since $1 + \tan^2 x = \sec^2 x$,the equation simplifies to $\frac{dy}{dx} + 3(\sec^2 x)y = \sec^2 x$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = 3\sec^2 x$ and $Q(x) = \sec^2 x$.
The integrating factor $(IF)$ is $e^{\int P(x) dx} = e^{\int 3\sec^2 x dx} = e^{3\tan x}$.
The general solution is $y \cdot IF = \int Q(x) \cdot IF dx + C$.
$y \cdot e^{3\tan x} = \int \sec^2 x \cdot e^{3\tan x} dx + C$.
Let $u = 3\tan x$,then $du = 3\sec^2 x dx$,so $\sec^2 x dx = \frac{du}{3}$.
$y \cdot e^{3\tan x} = \int e^u \frac{du}{3} + C = \frac{1}{3}e^{3\tan x} + C$.
Given $y(0) = \frac{1}{3} + e^3$,at $x=0$,$\tan(0)=0$,so $y(0) \cdot e^0 = \frac{1}{3}e^0 + C \Rightarrow \frac{1}{3} + e^3 = \frac{1}{3} + C \Rightarrow C = e^3$.
Thus,$y \cdot e^{3\tan x} = \frac{1}{3}e^{3\tan x} + e^3$.
At $x = \frac{\pi}{4}$,$\tan\left(\frac{\pi}{4}\right) = 1$.
$y\left(\frac{\pi}{4}\right) \cdot e^3 = \frac{1}{3}e^3 + e^3 = \frac{4}{3}e^3$.
Therefore,$y\left(\frac{\pi}{4}\right) = \frac{4}{3}$.

(B) Given $f(x) = 1 - 2x + e^x \int_0^x e^{-t} f(t) dt$.
Differentiating both sides with respect to $x$ using the Leibniz rule:
$f'(x) = -2 + e^x \int_0^x e^{-t} f(t) dt + e^x (e^{-x} f(x)) = -2 + (f(x) - (1 - 2x)) + f(x) = 2f(x) + 2x - 3$.
This is a linear differential equation: $\frac{dy}{dx} - 2y = 2x - 3$.
The integrating factor is $IF = e^{\int -2 dx} = e^{-2x}$.
Multiplying by $IF$: $\frac{d}{dx}(y e^{-2x}) = (2x - 3) e^{-2x}$.
Integrating both sides: $y e^{-2x} = \int (2x - 3) e^{-2x} dx = (2x - 3) \frac{e^{-2x}}{-2} - \int 2 \cdot \frac{e^{-2x}}{-2} dx = -\frac{2x-3}{2} e^{-2x} - \frac{1}{2} e^{-2x} + C$.
$y = -\frac{2x-3}{2} - \frac{1}{2} + C e^{2x} = -x + \frac{3}{2} - \frac{1}{2} + C e^{2x} = -x + 1 + C e^{2x}$.
Since $f(0) = 1 - 0 + 0 = 1$,we have $1 = -0 + 1 + C e^0 \Rightarrow C = 0$.
Thus,$f(x) = 1 - x$.
The region bounded by $y = 1 - x$ and the coordinate axes is a triangle with vertices $(0,0), (1,0), (0,1)$.
The area is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}$.

(C) Given the differential equation: $\left(7 x^4 \cot y-e^x \operatorname{cosec} y\right) \frac{d x}{d y}=x^5$.
Rearranging the equation: $\frac{d x}{d y} = \frac{x^5}{7 x^4 \cot y - e^x \operatorname{cosec} y}$.
This is equivalent to $\frac{d y}{d x} = \frac{7 x^4 \cot y - e^x \operatorname{cosec} y}{x^5} = \frac{7 \cot y}{x} - \frac{e^x \operatorname{cosec} y}{x^5}$.
Multiplying by $\sin y$: $\sin y \frac{d y}{d x} - \frac{7 \cos y}{x} = -\frac{e^x}{x^5}$.
Let $t = \cos y$,then $\frac{d t}{d x} = -\sin y \frac{d y}{d x}$.
Substituting this into the equation: $-\frac{d t}{d x} - \frac{7 t}{x} = -\frac{e^x}{x^5}$,which simplifies to $\frac{d t}{d x} + \frac{7 t}{x} = \frac{e^x}{x^5}$.
This is a linear differential equation with Integrating Factor $I.F. = e^{\int \frac{7}{x} dx} = e^{7 \ln x} = x^7$.
The solution is $t \cdot x^7 = \int x^7 \cdot \frac{e^x}{x^5} dx = \int x^2 e^x dx$.
Using integration by parts: $\int x^2 e^x dx = x^2 e^x - 2 \int x e^x dx = x^2 e^x - 2(x e^x - e^x) + C = e^x(x^2 - 2x + 2) + C$.
So,$\cos y \cdot x^7 = e^x(x^2 - 2x + 2) + C$.
At $x=1, y=\frac{\pi}{2}$,we have $\cos(\frac{\pi}{2}) \cdot 1^7 = e^1(1-2+2) + C \implies 0 = e + C \implies C = -e$.
Thus,$\cos y = \frac{e^x(x^2 - 2x + 2) - e}{x^7}$.
At $x=2$,$\cos y = \frac{e^2(4 - 4 + 2) - e}{2^7} = \frac{2e^2 - e}{128}$.

(D) The given differential equation is $x(x^2 + e^x) dy + (e^x(x-2)y - x^3) dx = 0$.
Rearranging it into the linear form $\frac{dy}{dx} + P(x)y = Q(x)$:
$x(x^2 + e^x) \frac{dy}{dx} + e^x(x-2)y = x^3$
$\frac{dy}{dx} + \frac{e^x(x-2)}{x(x^2 + e^x)} y = \frac{x^2}{x^2 + e^x}$.
Integrating Factor ($I$.$F$.) $= e^{\int \frac{e^x(x-2)}{x(x^2 + e^x)} dx}$.
Let $u = 1 + \frac{e^x}{x^2}$. Then $du = \frac{x^2 e^x - e^x(2x)}{x^4} dx = \frac{e^x(x-2)}{x^3} dx$.
Thus,$I$.$F$. $= e^{\int \frac{1}{u} du} = u = 1 + \frac{e^x}{x^2}$.
The general solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$.
$y(1 + \frac{e^x}{x^2}) = \int \frac{x^2}{x^2 + e^x} \cdot (\frac{x^2 + e^x}{x^2}) dx + C$.
$y(1 + \frac{e^x}{x^2}) = \int 1 dx + C = x + C$.
Since the curve passes through $(1, 0)$,we have $0(1 + e) = 1 + C$,so $C = -1$.
Thus,$y = \frac{x-1}{1 + \frac{e^x}{x^2}}$.
For $x = 2$,$y(2) = \frac{2-1}{1 + \frac{e^2}{4}} = \frac{1}{\frac{4+e^2}{4}} = \frac{4}{4+e^2}$.