Linear differential equations Questions in English

(A) The given differential equation is $(x^2+1) \frac{dy}{dx} - 2xy = (x^2+1)^2 \cos x$.
Dividing by $(x^2+1)$,we get $\frac{dy}{dx} - \frac{2x}{x^2+1} y = (x^2+1) \cos x$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{2x}{x^2+1}$ and $Q(x) = (x^2+1) \cos x$.
The integrating factor $I.F. = e^{\int P(x) dx} = e^{\int -\frac{2x}{x^2+1} dx} = e^{-\ln(x^2+1)} = \frac{1}{x^2+1}$.
The solution is $y \cdot I.F. = \int Q(x) \cdot I.F. dx + C$.
$y \cdot \frac{1}{x^2+1} = \int (x^2+1) \cos x \cdot \frac{1}{x^2+1} dx = \int \cos x dx = \sin x + C$.
Given $y(0) = 1$,we have $\frac{1}{0^2+1} = \sin(0) + C \Rightarrow 1 = 0 + C \Rightarrow C = 1$.
Thus,$y = (x^2+1)(\sin x + 1)$.
Now,$\int_{-3}^3 y(x) dx = \int_{-3}^3 (x^2+1)(\sin x + 1) dx = \int_{-3}^3 (x^2 \sin x + x^2 + \sin x + 1) dx$.
Since $x^2 \sin x$ and $\sin x$ are odd functions,their integral over $[-3, 3]$ is $0$.
So,$\int_{-3}^3 y(x) dx = \int_{-3}^3 x^2 dx + \int_{-3}^3 1 dx = 2 \int_{0}^3 x^2 dx + 2 \int_{0}^3 1 dx = 2 [\frac{x^3}{3}]_0^3 + 2[x]_0^3 = 2(9) + 2(3) = 18 + 6 = 24$.

(C) Given $f(x)=x-1$ and $g(x)=e^x$.
First,find $g(f(f(x)))$:
$f(f(x)) = f(x-1) = (x-1)-1 = x-2$.
$g(f(f(x))) = g(x-2) = e^{x-2}$.
The differential equation is $\frac{d y}{d x} + \frac{y}{\sqrt{x}} = e^{-2 \sqrt{x}} \cdot e^{x-2} = e^{x-2 \sqrt{x}-2}$.
This is a linear differential equation of the form $\frac{d y}{d x} + P(x)y = Q(x)$,where $P(x) = \frac{1}{\sqrt{x}}$ and $Q(x) = e^{x-2 \sqrt{x}-2}$.
The integrating factor ($I$.$F$.) is $e^{\int P(x) dx} = e^{\int \frac{1}{\sqrt{x}} dx} = e^{2 \sqrt{x}}$.
The solution is $y \cdot e^{2 \sqrt{x}} = \int Q(x) \cdot (I.F.) dx + C$.
$y \cdot e^{2 \sqrt{x}} = \int e^{x-2 \sqrt{x}-2} \cdot e^{2 \sqrt{x}} dx + C = \int e^{x-2} dx + C = e^{x-2} + C$.
Using $y(0)=0$: $0 \cdot e^0 = e^{0-2} + C \implies 0 = e^{-2} + C \implies C = -e^{-2}$.
So,$y \cdot e^{2 \sqrt{x}} = e^{x-2} - e^{-2}$.
At $x=1$: $y \cdot e^2 = e^{1-2} - e^{-2} = e^{-1} - e^{-2} = \frac{1}{e} - \frac{1}{e^2} = \frac{e-1}{e^2}$.
Therefore,$y(1) = \frac{e-1}{e^4}$.

(A) Given the differential equations:
$1) \frac{dy_1}{y_1} = \sin^2 x dx \implies \ln y_1 = \int \sin^2 x dx + C_1$
$2) \frac{dy_2}{y_2} = \cos^2 x dx \implies \ln y_2 = \int \cos^2 x dx + C_2$
$3) \frac{dy_3}{y_3} = \left(\frac{2}{x^3} - 1\right) dx \implies \ln y_3 = \int (2x^{-3} - 1) dx + C_3 = -x^{-2} - x + C_3$
Summing the equations: $\ln(y_1 y_2 y_3) = \int (\sin^2 x + \cos^2 x + \frac{2}{x^3} - 1) dx + C = \int (1 + \frac{2}{x^3} - 1) dx + C = \int \frac{2}{x^3} dx + C = -x^{-2} + C$.
Using initial conditions at $x=1$: $\ln(5 \cdot \frac{1}{3} \cdot \frac{3}{5e}) = \ln(e^{-1}) = -1$.
So,$-1 = -(1)^{-2} + C \implies C = 0$.
Thus,$y_1 y_2 y_3 = e^{-1/x^2}$.
Now,evaluate $\lim_{x \rightarrow 0^+} \frac{e^{-1/x^2} + 2x}{e^{3x} \sin x} = \lim_{x \rightarrow 0^+} \frac{e^{-1/x^2}}{e^{3x} \sin x} + \lim_{x \rightarrow 0^+} \frac{2x}{e^{3x} \sin x}$.
Since $\lim_{x \rightarrow 0^+} \frac{e^{-1/x^2}}{\sin x} = 0$ and $\lim_{x \rightarrow 0^+} \frac{2x}{\sin x} = 2$,the limit is $0 + 2 = 2$.

(C) The slope of the tangent is given by $\frac{dy}{dx} = -y + e^{-x}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 1$ and $Q = e^{-x}$.
The Integrating Factor ($I$.$F$.) is $e^{\int P dx} = e^{\int 1 dx} = e^x$.
The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$.
$y e^x = \int e^{-x} \cdot e^x dx + C$.
$y e^x = \int 1 dx + C$.
$y e^x = x + C$.
Since the curve passes through the origin $(0,0)$,we substitute $x=0$ and $y=0$:
$0 \cdot e^0 = 0 + C \Rightarrow C = 0$.
Thus,the equation of the curve is $y e^x = x$,which can be written as $y e^x - x = 0$.

(B) Let the point on the curve be $(x, y)$. The slope of the tangent is $\frac{dy}{dx}$.
According to the problem,the sum of the ordinate $(y)$ and abscissa $(x)$ exceeds the slope by $5$,so:
$y + x = \frac{dy}{dx} + 5$
Rearranging the terms,we get the linear differential equation:
$\frac{dy}{dx} - y = x - 5$
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -1$ and $Q = x - 5$.
The integrating factor $(IF)$ is $e^{\int P dx} = e^{\int -1 dx} = e^{-x}$.
The general solution is $y \cdot IF = \int Q \cdot IF dx + C$:
$y e^{-x} = \int (x - 5) e^{-x} dx + C$
Using integration by parts for $\int (x - 5) e^{-x} dx$:
$= (x - 5)(-e^{-x}) - \int (1)(-e^{-x}) dx$
$= -(x - 5)e^{-x} - e^{-x} + C$
$= (-x + 5 - 1)e^{-x} + C = (4 - x)e^{-x} + C$
So,$y e^{-x} = (4 - x)e^{-x} + C$,which implies $y = 4 - x + C e^x$.
Given the curve passes through $(0, 2)$:
$2 = 4 - 0 + C e^0 \implies 2 = 4 + C \implies C = -2$.
Substituting $C = -2$ into the equation:
$y = 4 - x - 2 e^x$.

(A) Given the differential equation: $(2y - x) \frac{dy}{dx} = 1$.
Rearranging the equation,we get: $\frac{dx}{dy} = 2y - x$.
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = 1$ and $Q(y) = 2y$.
The integrating factor $(IF)$ is given by $e^{\int P(y) dy} = e^{\int 1 dy} = e^y$.
Multiplying both sides by the $IF$: $e^y \frac{dx}{dy} + x e^y = 2y e^y$.
This simplifies to: $\frac{d}{dy}(x e^y) = 2y e^y$.
Integrating both sides with respect to $y$: $x e^y = \int 2y e^y dy$.
Using integration by parts: $\int 2y e^y dy = 2(y e^y - e^y) + c = 2e^y(y - 1) + c$.
Thus,$x e^y = 2e^y(y - 1) + c$.
Dividing by $e^y$,we get $x = 2(y - 1) + ce^{-y}$.

(B) Given the differential equation: $x^2 y - x^3 \frac{dy}{dx} = y^4 \cos x$.
Divide both sides by $x^3 y^4$: $\frac{1}{x^2 y^3} - \frac{1}{y^4} \frac{dy}{dx} = \frac{\cos x}{x^3}$.
Let $v = y^{-3}$,then $\frac{dv}{dx} = -3 y^{-4} \frac{dy}{dx}$,which implies $-\frac{1}{3} \frac{dv}{dx} = y^{-4} \frac{dy}{dx}$.
Substituting this into the equation: $\frac{1}{x^3} v + \frac{1}{3} \frac{dv}{dx} = \frac{\cos x}{x^3}$.
Multiply by $3$: $\frac{dv}{dx} + \frac{3}{x^3} v = \frac{3 \cos x}{x^3}$.
This is a linear differential equation of the form $\frac{dv}{dx} + P(x)v = Q(x)$.
Integrating factor $IF = e^{\int \frac{3}{x^3} dx} = e^{-\frac{3}{2x^2}}$.
However,checking the original equation structure,it is a Bernoulli equation.
Rearranging $x^3 \frac{dy}{dx} - x^2 y = -y^4 \cos x \implies \frac{dy}{dx} - \frac{1}{x} y = -\frac{y^4 \cos x}{x^3}$.
Divide by $y^4$: $y^{-4} \frac{dy}{dx} - \frac{1}{x} y^{-3} = -\frac{\cos x}{x^3}$.
Let $v = y^{-3}$,then $\frac{dv}{dx} = -3 y^{-4} \frac{dy}{dx} \implies y^{-4} \frac{dy}{dx} = -\frac{1}{3} \frac{dv}{dx}$.
Substituting: $-\frac{1}{3} \frac{dv}{dx} - \frac{1}{x} v = -\frac{\cos x}{x^3} \implies \frac{dv}{dx} + \frac{3}{x} v = \frac{3 \cos x}{x^3}$.
$IF = e^{\int \frac{3}{x} dx} = e^{3 \ln x} = x^3$.
Solution: $v \cdot x^3 = \int x^3 \cdot \frac{3 \cos x}{x^3} dx = \int 3 \cos x dx = 3 \sin x + C$.
$y^{-3} x^3 = 3 \sin x + C$.
Given $y(0) = 1$,but the equation is singular at $x=0$. Assuming the form $x^3 = 3 y^3 \sin x$ as a potential solution.

(C) The given differential equation is $\left(1+x^2\right) \frac{dy}{dx} + 2xy = 4x^2$.
Dividing by $(1+x^2)$,we get $\frac{dy}{dx} + \frac{2x}{1+x^2}y = \frac{4x^2}{1+x^2}$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{2x}{1+x^2}$ and $Q(x) = \frac{4x^2}{1+x^2}$.
The integrating factor $(IF)$ is $e^{\int P(x) dx} = e^{\int \frac{2x}{1+x^2} dx} = e^{\ln(1+x^2)} = 1+x^2$.
The solution is $y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$.
$y(1+x^2) = \int \frac{4x^2}{1+x^2} \cdot (1+x^2) dx + C$.
$y(1+x^2) = \int 4x^2 dx + C = \frac{4x^3}{3} + C$.
Since the curve passes through the origin $(0,0)$,we substitute $x=0, y=0$: $0(1+0) = 0 + C \implies C = 0$.
Thus,$y(1+x^2) = \frac{4x^3}{3}$,which simplifies to $3y(1+x^2) = 4x^3$.

(C) Given the differential equation: $e^{y-x} \frac{dy}{dx} = \frac{y(\sin x + \cos x)}{1 + y \log y}$
Separate the variables: $e^y \frac{1 + y \log y}{y} dy = e^x (\sin x + \cos x) dx$
Simplify the left side: $e^y (\log y + \frac{1}{y}) dy = e^x (\sin x + \cos x) dx$
Integrate both sides: $\int e^y (\log y + \frac{1}{y}) dy = \int e^x (\sin x + \cos x) dx$
Using the standard integral formula $\int e^t (f(t) + f'(t)) dt = e^t f(t) + c$,where $f(y) = \log y$ and $f(x) = \sin x$:
$e^y \log y = e^x \sin x + c$
Thus,the correct option is $C$.

(A) Given the differential equation: $(1+y^{2})+(x-e^{\tan ^{-1} y}) \frac{dy}{dx}=0$
Rearranging the terms: $(x-e^{\tan ^{-1} y}) \frac{dy}{dx} = -(1+y^{2})$
Taking the reciprocal: $\frac{dx}{dy} = \frac{-(x-e^{\tan ^{-1} y})}{1+y^{2}} = \frac{-x}{1+y^{2}} + \frac{e^{\tan ^{-1} y}}{1+y^{2}}$
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = \frac{1}{1+y^{2}}$ and $Q(y) = \frac{e^{\tan ^{-1} y}}{1+y^{2}}$.
The Integrating Factor ($I$.$F$.) is $e^{\int P(y) dy} = e^{\int \frac{1}{1+y^{2}} dy} = e^{\tan ^{-1} y}$.
The general solution is given by $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + c$.
$x \cdot e^{\tan ^{-1} y} = \int \frac{e^{\tan ^{-1} y}}{1+y^{2}} \cdot e^{\tan ^{-1} y} dy + c$.
Let $t = e^{\tan ^{-1} y}$,then $dt = \frac{e^{\tan ^{-1} y}}{1+y^{2}} dy$.
Substituting this into the integral: $x \cdot e^{\tan ^{-1} y} = \int t dt + c = \frac{t^{2}}{2} + c$.
Thus,the general solution is $x \cdot e^{\tan ^{-1} y} = \frac{(e^{\tan ^{-1} y})^{2}}{2} + c$.

(C) Given the differential equation: $x(x-1) \frac{dy}{dx} = x^3(2x-1) + (x-2)y$.
Divide by $x(x-1)$ to write it in the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$:
$\frac{dy}{dx} - \frac{x-2}{x(x-1)}y = \frac{x^3(2x-1)}{x(x-1)} = \frac{x^2(2x-1)}{x-1}$.
Here,$P(x) = -\frac{x-2}{x(x-1)} = -(\frac{2}{x} - \frac{1}{x-1}) = \frac{1}{x-1} - \frac{2}{x}$.
The integrating factor $IF = e^{\int P(x) dx} = e^{\int (\frac{1}{x-1} - \frac{2}{x}) dx} = e^{\ln|x-1| - 2\ln|x|} = \frac{x-1}{x^2}$.
Multiplying the linear equation by $IF$:
$\frac{d}{dx} [y \cdot \frac{x-1}{x^2}] = \frac{x^2(2x-1)}{x-1} \cdot \frac{x-1}{x^2} = 2x-1$.
Integrating both sides with respect to $x$:
$y \cdot \frac{x-1}{x^2} = \int (2x-1) dx = x^2 - x + c$.
$y \cdot \frac{x-1}{x^2} = x(x-1) + c$.
Multiplying by $x^2$:
$y(x-1) = x^3(x-1) + cx^2$.
Thus,the correct option is $C$.

(A) Given the differential equation: $(1+y^2)+(x-e^{\tan ^{-1} y}) \frac{dy}{dx}=0$.
Rearranging the equation: $(x-e^{\tan ^{-1} y}) \frac{dy}{dx} = -(1+y^2)$.
Taking the reciprocal: $\frac{dx}{dy} = -\frac{x-e^{\tan ^{-1} y}}{1+y^2} = -\frac{x}{1+y^2} + \frac{e^{\tan ^{-1} y}}{1+y^2}$.
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = \frac{1}{1+y^2}$ and $Q(y) = \frac{e^{\tan ^{-1} y}}{1+y^2}$.
The integrating factor $(IF)$ is $e^{\int P(y) dy} = e^{\int \frac{1}{1+y^2} dy} = e^{\tan ^{-1} y}$.
The solution is $x \cdot (IF) = \int Q(y) \cdot (IF) dy + k$.
$x \cdot e^{\tan ^{-1} y} = \int \frac{e^{\tan ^{-1} y}}{1+y^2} \cdot e^{\tan ^{-1} y} dy + k$.
Let $u = \tan ^{-1} y$,then $du = \frac{1}{1+y^2} dy$.
$x \cdot e^{\tan ^{-1} y} = \int e^{2u} du + k = \frac{1}{2} e^{2u} + k = \frac{1}{2} e^{2 \tan ^{-1} y} + k$.
Multiplying by $2$: $2x e^{\tan ^{-1} y} = e^{2 \tan ^{-1} y} + 2k$.
Since $2k$ is also a constant,we can write it as $k$.
Thus,$2x e^{\tan ^{-1} y} = e^{2 \tan ^{-1} y} + k$.

(A) The given differential equation is $(1+x^2) \frac{dy}{dx} + 2xy = 4x^2$.
Dividing by $(1+x^2)$,we get $\frac{dy}{dx} + \frac{2x}{1+x^2}y = \frac{4x^2}{1+x^2}$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{2x}{1+x^2}$ and $Q(x) = \frac{4x^2}{1+x^2}$.
The integrating factor $(IF)$ is $e^{\int P(x) dx} = e^{\int \frac{2x}{1+x^2} dx} = e^{\ln(1+x^2)} = 1+x^2$.
The general solution is $y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$.
$y(1+x^2) = \int \frac{4x^2}{1+x^2} \cdot (1+x^2) dx + C$.
$y(1+x^2) = \int 4x^2 dx + C$.
$y(1+x^2) = \frac{4x^3}{3} + C$.
Since the curve passes through the origin $(0,0)$,we substitute $x=0$ and $y=0$: $0(1+0) = 0 + C \implies C = 0$.
Thus,the equation is $y(1+x^2) = \frac{4x^3}{3}$,which simplifies to $3(1+x^2)y = 4x^3$.

(C) Given equation is $y + \frac{d}{dx}(xy) = x(\sin x + \log x)$.
Expanding the derivative: $y + y + x \frac{dy}{dx} = x \sin x + x \log x$.
$2y + x \frac{dy}{dx} = x \sin x + x \log x$.
Dividing by $x$: $\frac{dy}{dx} + \frac{2}{x} y = \sin x + \log x$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{2}{x}$ and $Q(x) = \sin x + \log x$.
Integrating factor $IF = e^{\int P(x) dx} = e^{\int \frac{2}{x} dx} = e^{2 \log x} = x^2$.
The solution is $y \cdot IF = \int Q(x) \cdot IF dx + c$.
$y \cdot x^2 = \int x^2(\sin x + \log x) dx + c$.
$y x^2 = \int x^2 \sin x dx + \int x^2 \log x dx + c$.
Using integration by parts for $\int x^2 \sin x dx$: $-x^2 \cos x + 2x \sin x + 2 \cos x$.
Using integration by parts for $\int x^2 \log x dx$: $\frac{x^3}{3} \log x - \frac{x^3}{9}$.
So,$y x^2 = -x^2 \cos x + 2x \sin x + 2 \cos x + \frac{x^3}{3} \log x - \frac{x^3}{9} + c$.
Dividing by $x^2$: $y = -\cos x + \frac{2 \sin x}{x} + \frac{2 \cos x}{x^2} + \frac{x}{3} \log x - \frac{x}{9} + \frac{c}{x^2}$.
This matches option $C$.

(C) Given the differential equation: $y + \frac{d}{dx}(xy) = x(\sin x + \log x)$.
Expanding the derivative term: $y + y + x \frac{dy}{dx} = x(\sin x + \log x)$.
This simplifies to: $2y + x \frac{dy}{dx} = x(\sin x + \log x)$.
Dividing the entire equation by $x$ (assuming $x \neq 0$): $\frac{dy}{dx} + \frac{2}{x}y = \sin x + \log x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{2}{x}$ and $Q = \sin x + \log x$.
The integrating factor $(IF)$ is given by $e^{\int P dx}$.
$IF = e^{\int \frac{2}{x} dx} = e^{2 \log x} = e^{\log x^2} = x^2$.
Therefore,the correct option is $C$.

(A) Given the differential equation: $(1+x) \frac{dy}{dx} - xy = 1-x$.
Divide by $(1+x)$ to write it in the standard form $\frac{dy}{dx} + P(x)y = Q(x)$:
$\frac{dy}{dx} - \frac{x}{1+x}y = \frac{1-x}{1+x}$.
Here,$P(x) = -\frac{x}{1+x} = -\frac{x+1-1}{1+x} = -1 + \frac{1}{1+x}$ and $Q(x) = \frac{1-x}{1+x} = \frac{2-(1+x)}{1+x} = \frac{2}{1+x} - 1$.
The integrating factor $IF = e^{\int P(x) dx} = e^{\int (-1 + \frac{1}{1+x}) dx} = e^{-x + \ln(1+x)} = (1+x)e^{-x}$.
The general solution is $y \cdot IF = \int Q(x) \cdot IF dx + c$.
$y(1+x)e^{-x} = \int (\frac{1-x}{1+x}) (1+x)e^{-x} dx + c = \int (1-x)e^{-x} dx + c$.
Using integration by parts: $\int (1-x)e^{-x} dx = (1-x)(-e^{-x}) - \int (-1)(-e^{-x}) dx = -(1-x)e^{-x} - \int e^{-x} dx = (x-1)e^{-x} + e^{-x} + c = xe^{-x} + c$.
So,$y(1+x)e^{-x} = xe^{-x} + c$.
Multiplying by $e^x$,we get $y(1+x) = x + ce^x$.

(C) The given differential equation is $x \frac{dy}{dx} + y \log x = x e^x \cdot x^{-1/2} \log x$.
Dividing both sides by $x$,we get:
$\frac{dy}{dx} + y \frac{\log x}{x} = e^x \cdot x^{-1/2} \log x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{\log x}{x}$ and $Q = e^x \cdot x^{-1/2} \log x$.
The integrating factor $(IF)$ is given by $IF = e^{\int P dx} = e^{\int \frac{\log x}{x} dx}$.
Let $u = \log x$,then $du = \frac{1}{x} dx$.
So,$\int \frac{\log x}{x} dx = \int u du = \frac{u^2}{2} = \frac{(\log x)^2}{2}$.
Therefore,$IF = e^{\frac{(\log x)^2}{2}} = (e^{\log x})^{\frac{1}{2} \log x} = x^{\frac{1}{2} \log x}$.

(D) The given differential equation is $\sin x \frac{dy}{dx} + y \cos x = 4x$.
Dividing by $\sin x$,we get $\frac{dy}{dx} + y \cot x = \frac{4x}{\sin x}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \cot x$ and $Q = \frac{4x}{\sin x}$.
The integrating factor ($I$.$F$.) is $e^{\int P dx} = e^{\int \cot x dx} = e^{\ln|\sin x|} = \sin x$.
The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$.
Substituting the values,$y \sin x = \int \frac{4x}{\sin x} \cdot \sin x dx + C = \int 4x dx + C = 2x^2 + C$.
Thus,$y = \frac{2x^2 + C}{\sin x}$.
Given $y(\frac{\pi}{2}) = 0$,we have $0 = \frac{2(\frac{\pi}{2})^2 + C}{\sin(\frac{\pi}{2})} = \frac{\pi^2}{2} + C$,so $C = -\frac{\pi^2}{2}$.
The particular solution is $y = \frac{2x^2 - \frac{\pi^2}{2}}{\sin x}$.
Evaluating at $x = \frac{\pi}{6}$,$y(\frac{\pi}{6}) = \frac{2(\frac{\pi}{6})^2 - \frac{\pi^2}{2}}{\sin(\frac{\pi}{6})} = \frac{\frac{2\pi^2}{36} - \frac{\pi^2}{2}}{1/2} = 2 \left( \frac{\pi^2}{18} - \frac{\pi^2}{2} \right) = 2 \left( \frac{\pi^2 - 9\pi^2}{18} \right) = 2 \left( \frac{-8\pi^2}{18} \right) = -\frac{8}{9} \pi^2$.

(D) Given the differential equation: $y \, dx - (x + 3y^2) \, dy = 0$
Rearranging the terms: $y \, dx = (x + 3y^2) \, dy$
Dividing by $y \, dy$: $\frac{dx}{dy} = \frac{x + 3y^2}{y} = \frac{x}{y} + 3y$
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = -\frac{1}{y}$ and $Q(y) = 3y$.
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: $x \cdot \frac{1}{y} = \int 3y \cdot \frac{1}{y} \, dy + c$
$\frac{x}{y} = \int 3 \, dy + c = 3y + c$
So,$x = 3y^2 + cy$.
Since the curve passes through $(1, 1)$,we substitute $x=1$ and $y=1$: $1 = 3(1)^2 + c(1) \Rightarrow 1 = 3 + c \Rightarrow c = -2$.
The equation of the curve is $x = 3y^2 - 2y$.
Checking option $(D)$ $(-\frac{1}{3}, \frac{1}{3})$: $x = 3(\frac{1}{3})^2 - 2(\frac{1}{3}) = 3(\frac{1}{9}) - \frac{2}{3} = \frac{1}{3} - \frac{2}{3} = -\frac{1}{3}$.
Since the point satisfies the equation,the correct option is $(D)$.

(D) The given differential equation is $x \frac{dy}{dx} + 2y = x^2$.
Dividing by $x$,we get $\frac{dy}{dx} + \left(\frac{2}{x}\right)y = x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{2}{x}$ and $Q = x$.
The Integrating Factor ($I$.$F$.) is $e^{\int P dx} = e^{\int \frac{2}{x} dx} = e^{2 \ln|x|} = x^2$.
The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$.
$y \cdot x^2 = \int x \cdot x^2 dx + C = \int x^3 dx + C = \frac{x^4}{4} + C$.
Given $y(1) = 1$,we substitute $x = 1$ and $y = 1$: $1(1)^2 = \frac{1^4}{4} + C \implies 1 = \frac{1}{4} + C \implies C = \frac{3}{4}$.
Thus,the particular solution is $y x^2 = \frac{x^4}{4} + \frac{3}{4}$,or $y = \frac{x^2}{4} + \frac{3}{4x^2}$.
Now,we find $y\left(\frac{1}{2}\right) = \frac{(\frac{1}{2})^2}{4} + \frac{3}{4(\frac{1}{2})^2} = \frac{1/4}{4} + \frac{3}{4(1/4)} = \frac{1}{16} + 3 = \frac{1 + 48}{16} = \frac{49}{16}$.

(D) Given differential equation is $\frac{dy}{dx} + y = \frac{1+y}{x}$.
Rearranging the terms,we get $\frac{dy}{dx} + y = \frac{1}{x} + \frac{y}{x}$.
Grouping the $y$ terms,we have $\frac{dy}{dx} + y - \frac{y}{x} = \frac{1}{x}$,which simplifies to $\frac{dy}{dx} + \left(1 - \frac{1}{x}\right)y = \frac{1}{x}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 1 - \frac{1}{x}$ and $Q = \frac{1}{x}$.
The integrating factor ($I$.$F$.) is given by $e^{\int P dx}$.
$I$.$F$. $= e^{\int (1 - \frac{1}{x}) dx} = e^{x - \log x} = e^x \cdot e^{-\log x} = e^x \cdot e^{\log(x^{-1})} = e^x \cdot \frac{1}{x} = \frac{e^x}{x}$.

(D) Given the differential equation: $\cos x \frac{dy}{dx} - y \sin x = 6 x$.
Divide the entire equation by $\cos x$:
$\frac{dy}{dx} - y \tan x = 6 x \sec x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -\tan x$ and $Q = 6 x \sec x$.
Calculate the Integrating Factor $(IF)$:
$IF = e^{\int P dx} = e^{-\int \tan x dx} = e^{\ln(\cos x)} = \cos x$.
The general solution is given by:
$y \cdot (IF) = \int Q \cdot (IF) dx + c$.
Substituting the values:
$y \cdot \cos x = \int (6 x \sec x) \cdot \cos x dx + c$.
Since $\sec x \cdot \cos x = 1$,we have:
$y \cos x = \int 6 x dx + c$.
Integrating $6x$ with respect to $x$:
$y \cos x = 3 x^2 + c$.

(D) The given differential equation is $x \frac{dy}{dx}+y=x \log x$.
Dividing by $x$,we get $\frac{dy}{dx}+\frac{1}{x} y=\log x$.
This is a linear differential equation of the form $\frac{dy}{dx}+P(x)y=Q(x)$,where $P(x)=\frac{1}{x}$ and $Q(x)=\log x$.
The integrating factor $(I.F.)$ is $e^{\int P(x) dx} = e^{\int \frac{1}{x} dx} = e^{\log x} = x$.
The general solution is given by $y(I.F.) = \int Q(I.F.) dx + c$.
Substituting the values,$xy = \int x \log x dx + c$.
Using integration by parts,$\int x \log x dx = \log x \cdot \frac{x^2}{2} - \int \frac{1}{x} \cdot \frac{x^2}{2} dx = \frac{x^2}{2} \log x - \frac{x^2}{4} + c$.
So,$xy = \frac{x^2}{2} \log x - \frac{x^2}{4} + c$.
Given $2(y(2)) = \log 4 - 1$,which implies $y(2) = \frac{1}{2} \log 4 - \frac{1}{2} = \log 2 - \frac{1}{2}$.
Substituting $x=2$ in the general solution: $2(\log 2 - \frac{1}{2}) = \frac{4}{2} \log 2 - \frac{4}{4} + c$.
$2 \log 2 - 1 = 2 \log 2 - 1 + c$,which gives $c=0$.
Thus,$y = \frac{x}{2} \log x - \frac{x}{4}$.
For $x=e$,$y(e) = \frac{e}{2} \log e - \frac{e}{4} = \frac{e}{2} - \frac{e}{4} = \frac{e}{4}$.

(C) Given differential equation: $\frac{dy}{dx} = y \tan x - y^2 \sec x$
Divide by $y^2$: $\frac{1}{y^2} \frac{dy}{dx} - \frac{1}{y} \tan x = -\sec x$
Let $v = -\frac{1}{y}$,then $\frac{dv}{dx} = \frac{1}{y^2} \frac{dy}{dx}$.
Substituting this into the equation: $\frac{dv}{dx} + v \tan x = -\sec x$.
This is a linear differential equation of the form $\frac{dv}{dx} + P(x)v = Q(x)$,where $P(x) = \tan x$ and $Q(x) = -\sec x$.
Integrating factor ($I$.$F$.) $= e^{\int \tan x dx} = e^{\ln |\sec x|} = \sec x$.
The solution is $v \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + c$.
$v \sec x = \int -\sec x \cdot \sec x dx + c = -\int \sec^2 x dx + c$.
$v \sec x = -\tan x + c$.
Substituting $v = -\frac{1}{y}$: $-\frac{1}{y} \sec x = -\tan x + c$.
Multiplying by $-y$: $\sec x = y(\tan x - c)$.
Replacing $-c$ with a new constant $c$: $\sec x = y(\tan x + c)$.

(B) The given differential equation is of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{3x^2}{1+x^3}$ and $Q = \frac{1}{1+x^3}$.
First,we find the Integrating Factor ($I$.$F$.):
$\text{I.F.} = e^{\int P dx} = e^{\int \frac{3x^2}{1+x^3} dx} = e^{\ln(1+x^3)} = 1+x^3$.
The general solution is given by $y \cdot (\text{I.F.}) = \int Q \cdot (\text{I.F.}) dx + c$.
Substituting the values:
$y(1+x^3) = \int \left(\frac{1}{1+x^3}\right) \cdot (1+x^3) dx + c$
$y(1+x^3) = \int 1 dx + c$
$y(1+x^3) = x + c$.

(B) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x}$ and $Q = \sin x$.
First,we find the Integrating Factor ($I$.$F$.):
$I.F. = e^{\int P dx} = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$.
The general solution is given by $y \cdot (I.F.) = \int (Q \cdot I.F.) dx + c$.
Substituting the values:
$yx = \int (\sin x \cdot x) dx + c$.
Using integration by parts $(\int u v dx = u \int v dx - \int (u' \int v dx) dx)$:
Let $u = x$ and $v = \sin x$.
$yx = x(-\cos x) - \int (1 \cdot -\cos x) dx + c$.
$yx = -x \cos x + \int \cos x dx + c$.
$yx = -x \cos x + \sin x + c$.
Rearranging the terms:
$yx + x \cos x = \sin x + c$.
$x(y + \cos x) = \sin x + c$.

(B) Given equation: $\frac{dx}{dy} + \frac{x}{y} = x^2$
Divide both sides by $x^2$: $\frac{1}{x^2} \frac{dx}{dy} + \frac{1}{xy} = 1 \dots (i)$
Let $\frac{1}{x} = t$. Then $-\frac{1}{x^2} \frac{dx}{dy} = \frac{dt}{dy}$,so $\frac{1}{x^2} \frac{dx}{dy} = -\frac{dt}{dy}$.
Substituting into $(i)$: $-\frac{dt}{dy} + \frac{t}{y} = 1$,which simplifies to $\frac{dt}{dy} - \frac{t}{y} = -1$.
This is a linear differential equation in $t$. The Integrating Factor ($I$.$F$.) is $e^{\int -\frac{1}{y} dy} = e^{-\log y} = \frac{1}{y}$.
The solution is $t(I.F.) = \int (-1)(I.F.) dy + c$.
$t(\frac{1}{y}) = \int -\frac{1}{y} dy + c = -\log y + c$.
Substituting $t = \frac{1}{x}$: $\frac{1}{xy} = -\log y + c$.
Therefore,$\frac{1}{x} = cy - y \log y$.

(C) Given equation: $x \, dy = y(dx + y \, dy)$
Rearranging the terms: $x \, dy = y \, dx + y^2 \, dy$
$y \, dx = (x - y^2) \, dy$
Dividing by $y \, dy$: $\frac{dx}{dy} = \frac{x}{y} - y$
$\frac{dx}{dy} - \frac{1}{y} x = -y$
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = -\frac{1}{y}$ and $Q(y) = -y$.
Integrating Factor ($I$.$F$.) $= e^{\int P(y) \, dy} = e^{\int -\frac{1}{y} \, dy} = e^{-\ln y} = \frac{1}{y}$.
The solution is given by $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) \, dy + C$.
$x \cdot \frac{1}{y} = \int (-y) \cdot \frac{1}{y} \, dy + C$
$\frac{x}{y} = \int -1 \, dy + C$
$\frac{x}{y} = -y + C$
Given $y(1) = 1$,substitute $x = 1$ and $y = 1$:
$\frac{1}{1} = -1 + C \Rightarrow C = 2$.
So,$\frac{x}{y} = -y + 2$.
To find $y(-3)$,substitute $x = -3$:
$\frac{-3}{y} = -y + 2$
$-3 = -y^2 + 2y$
$y^2 - 2y - 3 = 0$
$(y - 3)(y + 1) = 0$
Since $y(x) > 0$,we have $y = 3$.

(A) Given the differential equation: $\sin^{2} y \frac{dx}{dy} + x = \cot y$.
Dividing by $\sin^{2} y$,we get: $\frac{dx}{dy} + (\operatorname{cosec}^{2} y)x = \cot y \operatorname{cosec}^{2} y$.
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = \operatorname{cosec}^{2} y$ and $Q(y) = \cot y \operatorname{cosec}^{2} y$.
The Integrating Factor ($I$.$F$.) is $e^{\int P(y) dy} = e^{\int \operatorname{cosec}^{2} y dy} = e^{-\cot y}$.
The solution is given by $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + C$.
$x e^{-\cot y} = \int \cot y \operatorname{cosec}^{2} y e^{-\cot y} dy + C$.
Let $t = -\cot y$,then $dt = \operatorname{cosec}^{2} y dy$.
$x e^{-\cot y} = \int (-t) e^{t} dt + C = -(t e^{t} - e^{t}) + C = e^{t}(1 - t) + C$.
Substituting back $t = -\cot y$: $x e^{-\cot y} = e^{-\cot y}(1 + \cot y) + C$.
Given $x = 0$ at $y = \frac{3\pi}{4}$,we have $0 = e^{-\cot(3\pi/4)}(1 + \cot(3\pi/4)) + C$.
Since $\cot(3\pi/4) = -1$,$0 = e^{1}(1 - 1) + C \implies C = 0$.
Thus,$x e^{-\cot y} = e^{-\cot y}(1 + \cot y)$,which simplifies to $x = 1 + \cot y$.

(C) The given differential equation is $\frac{dy}{dx} + \frac{1}{x}y = x^3 - 3$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x}$ and $Q = x^3 - 3$.
The integrating factor ($I$.$F$.) is given by the formula $I.F. = e^{\int P dx}$.
Substituting $P = \frac{1}{x}$,we get $I.F. = e^{\int \frac{1}{x} dx} = e^{\ln|x|} = x$.
Therefore,the integrating factor is $x$.

(D) Given the differential equation $r dx + (x - r^2) dr = 0$.
Rearranging the terms,we get $r dx = -(x - r^2) dr$.
Dividing by $dr$,we have $r \frac{dx}{dr} = r^2 - x$.
Rearranging into the standard linear form $\frac{dx}{dr} + P(r)x = Q(r)$,we get $\frac{dx}{dr} + \frac{1}{r}x = r$.
The Integrating Factor ($I$.$F$.) is $e^{\int \frac{1}{r} dr} = e^{\log r} = r$.
The general solution is given by $x \cdot (I.F.) = \int Q(r) \cdot (I.F.) dr + c$.
Substituting the values,$x \cdot r = \int r \cdot r dr + c$.
Integrating the right side,$xr = \int r^2 dr + c$.
Thus,$xr = \frac{r^3}{3} + c$.

(B) The given differential equation is $\frac{dy}{dx} + 2y = e^{-x}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 2$ and $Q = e^{-x}$.
The Integrating Factor ($I$.$F$.) is given by $I.F. = e^{\int P dx} = e^{\int 2 dx} = e^{2x}$.
The general solution is given by $y \cdot (I.F.) = \int (Q \cdot I.F.) dx + c$.
Substituting the values,we get $y e^{2x} = \int (e^{-x} \cdot e^{2x}) dx + c$.
$y e^{2x} = \int e^{x} dx + c$.
$y e^{2x} = e^{x} + c$.

(A) Given the differential equation: $(1-x^{2}) \frac{dy}{dx} + 2xy = x(1-x^{2})^{\frac{1}{2}}$
Divide by $(1-x^{2})$ to get the standard form $\frac{dy}{dx} + Py = Q$:
$\frac{dy}{dx} + \frac{2x}{1-x^{2}}y = \frac{x}{\sqrt{1-x^{2}}}$
Here,$P = \frac{2x}{1-x^{2}}$ and $Q = \frac{x}{\sqrt{1-x^{2}}}$.
Integrating Factor $(I.F.) = e^{\int P dx} = e^{\int \frac{2x}{1-x^{2}} dx} = e^{-\ln(1-x^{2})} = e^{\ln(\frac{1}{1-x^{2}})} = \frac{1}{1-x^{2}}$.
The general solution is $y \times (I.F.) = \int Q \times (I.F.) dx + c$.
$y \times \frac{1}{1-x^{2}} = \int \frac{x}{\sqrt{1-x^{2}}} \times \frac{1}{1-x^{2}} dx + c$
$y \times \frac{1}{1-x^{2}} = \int x(1-x^{2})^{-\frac{3}{2}} dx + c$
Let $u = 1-x^{2}$,then $du = -2x dx$,so $x dx = -\frac{1}{2} du$.
$y \times \frac{1}{1-x^{2}} = -\frac{1}{2} \int u^{-\frac{3}{2}} du + c = -\frac{1}{2} \left( \frac{u^{-\frac{1}{2}}}{-\frac{1}{2}} \right) + c = u^{-\frac{1}{2}} + c = \frac{1}{\sqrt{1-x^{2}}} + c$.
Multiplying both sides by $(1-x^{2})$:
$y = \frac{1-x^{2}}{\sqrt{1-x^{2}}} + c(1-x^{2}) = \sqrt{1-x^{2}} + c(1-x^{2})$.

(D) Given the differential equation: $(1+x^{2}) dt = (\tan^{-1} x - t) dx$.
Divide both sides by $(1+x^{2}) dx$:
$\frac{dt}{dx} = \frac{\tan^{-1} x - t}{1+x^{2}}$
Rearrange the equation into the standard linear form $\frac{dt}{dx} + P(x)t = Q(x)$:
$\frac{dt}{dx} + \frac{1}{1+x^{2}}t = \frac{\tan^{-1} x}{1+x^{2}}$
Here,$P(x) = \frac{1}{1+x^{2}}$.
The integrating factor ($I$.$F$.) is given by $e^{\int P(x) dx}$:
$I.F. = e^{\int \frac{1}{1+x^{2}} dx} = e^{\tan^{-1} x}$.

(A) Given differential equation: $\sin y \frac{d y}{d x} = \cos y (1 - x \cos y)$.
Expanding the right side: $\sin y \frac{d y}{d x} = \cos y - x \cos^2 y$.
Dividing both sides by $\cos^2 y$: $\frac{\sin y}{\cos^2 y} \frac{d y}{d x} = \frac{1}{\cos y} - x$.
This simplifies to: $\sec y \tan y \frac{d y}{d x} = \sec y - x$.
Rearranging the terms: $\sec y \tan y \frac{d y}{d x} - \sec y = -x$.
Let $v = \sec y$. Then $\frac{d v}{d x} = \sec y \tan y \frac{d y}{d x}$.
Substituting into the equation: $\frac{d v}{d x} - v = -x$.
This is a linear differential equation of the form $\frac{d v}{d x} + P(x)v = Q(x)$,where $P(x) = -1$.
The integrating factor ($I$.$F$.) is given by $e^{\int P(x) dx} = e^{\int -1 dx} = e^{-x}$.

(B) Given differential equation is $(1+y+x^{2}y)dx + (x+x^{3})dy = 0$.
Rearranging the terms,we have $(x+x^{3})dy = -(1+y(1+x^{2}))dx$.
Dividing by $dx(x+x^{3})$,we get $\frac{dy}{dx} = -\frac{1+y(1+x^{2})}{x(1+x^{2})}$.
This simplifies to $\frac{dy}{dx} = -\frac{1}{x(1+x^{2})} - \frac{y(1+x^{2})}{x(1+x^{2})}$.
Thus,$\frac{dy}{dx} + \left(\frac{1}{x}\right)y = -\frac{1}{x(1+x^{2})}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x}$ and $Q = -\frac{1}{x(1+x^{2})}$.
The integrating factor ($I$.$F$.) is given by $e^{\int P dx} = e^{\int \frac{1}{x} dx} = e^{\log x} = x$.

(D) The given differential equation is $x \frac{dy}{dx} + y \log x = x^2$.
Dividing both sides by $x$,we get:
$\frac{dy}{dx} + \left(\frac{\log x}{x}\right) y = x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{\log x}{x}$ and $Q = x$.
The integrating factor ($I$.$F$.) is given by $e^{\int P dx}$.
$I$.$F$. $= e^{\int \frac{\log x}{x} dx}$.
Let $u = \log x$,then $du = \frac{1}{x} dx$.
So,$\int \frac{\log x}{x} dx = \int u du = \frac{u^2}{2} = \frac{(\log x)^2}{2}$.
Therefore,$I$.$F$. $= e^{\frac{(\log x)^2}{2}} = (e^{\log x})^{\frac{\log x}{2}} = x^{\frac{\log x}{2}} = x^{\log (\sqrt{x})}$.
Thus,the correct option is $D$.

(B) The given differential equation is $y \log y \left(\frac{dx}{dy}\right) + x = \log y$.
Dividing both sides by $y \log y$,we get:
$\frac{dx}{dy} + \frac{x}{y \log y} = \frac{\log y}{y \log y} = \frac{1}{y}$.
This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$,where $P = \frac{1}{y \log y}$ and $Q = \frac{1}{y}$.
The integrating factor ($I$.$F$.) is given by $e^{\int P dy}$.
$I$.$F$. $= e^{\int \frac{1}{y \log y} dy}$.
Let $u = \log y$,then $du = \frac{1}{y} dy$.
So,$\int \frac{1}{y \log y} dy = \int \frac{1}{u} du = \log u = \log(\log y)$.
Therefore,$I$.$F$. $= e^{\log(\log y)} = \log y$.

(D) Given the differential equation: $\frac{dy}{dx}(x \log x) + y = 4 \log x$.
Divide the entire equation by $(x \log x)$ to bring it into the standard form $\frac{dy}{dx} + Py = Q$:
$\frac{dy}{dx} + \frac{1}{x \log x} y = \frac{4 \log x}{x \log x} = \frac{4}{x}$.
Here,$P = \frac{1}{x \log x}$.
The integrating factor ($I$.$F$.) is given by $e^{\int P dx}$:
$I.F. = e^{\int \frac{1}{x \log x} dx}$.
Let $u = \log x$,then $du = \frac{1}{x} dx$.
So,$\int \frac{1}{x \log x} dx = \int \frac{1}{u} du = \log u = \log(\log x)$.
Therefore,$I.F. = e^{\log(\log x)} = \log x$.

(D) The integrating factor $(I.F.)$ of the linear differential equation $\frac{dy}{dx} + Py = Q$ is given by $e^{\int P dx}$.
Given that the integrating factor is $\sin x$,we have:
$e^{\int P dx} = \sin x$
Taking the natural logarithm on both sides:
$\int P dx = \ln(\sin x)$
Differentiating both sides with respect to $x$:
$P = \frac{d}{dx}[\ln(\sin x)]$
Using the chain rule:
$P = \frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x)$
$P = \frac{1}{\sin x} \cdot \cos x$
$P = \cot x$
Therefore,the correct option is $D$.

(B) The given differential equation is $\frac{dy}{dx}(x \log x) + y = 2 \log x$.
Dividing both sides by $(x \log x)$,we get:
$\frac{dy}{dx} + \frac{y}{x \log x} = \frac{2 \log x}{x \log x} = \frac{2}{x}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x \log x}$ and $Q = \frac{2}{x}$.
The integrating factor $(IF)$ is given by $e^{\int P dx}$.
$IF = e^{\int \frac{1}{x \log x} dx}$.
Let $u = \log x$,then $du = \frac{1}{x} dx$.
$IF = e^{\int \frac{1}{u} du} = e^{\log u} = u = \log x$.
Thus,the integrating factor is $\log x$.

(A) Given the differential equation: $(x \log x) \frac{dy}{dx} + y = 2x \log x$.
Divide both sides by $(x \log x)$ to get the linear form $\frac{dy}{dx} + P(x)y = Q(x)$:
$\frac{dy}{dx} + \frac{1}{x \log x} y = 2$.
Here,$P(x) = \frac{1}{x \log x}$ and $Q(x) = 2$.
The Integrating Factor ($I$.$F$.) is given by:
$I.F. = e^{\int P(x) dx} = e^{\int \frac{1}{x \log x} dx} = e^{\log(\log x)} = \log x$.
The general solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$:
$y \log x = \int 2 \log x dx + C$.
Using integration by parts,$\int \log x dx = x \log x - x$:
$y \log x = 2(x \log x - x) + C$.
Since the equation is defined for $x \geq 1$,at $x=1$,$\log(1)=0$. Substituting $x=1$ into the equation:
$y(1) \cdot 0 = 2(1 \cdot 0 - 1) + C \implies 0 = -2 + C \implies C = 2$.
Thus,the solution is $y \log x = 2x \log x - 2x + 2$.
To find $y(e)$,substitute $x=e$:
$y(e) \log(e) = 2(e) \log(e) - 2(e) + 2$.
Since $\log(e) = 1$:
$y(e) \cdot 1 = 2e - 2e + 2 = 2$.
Therefore,$y(e) = 2$.

(A) Given the differential equation: $(\tan ^{-1} y - x) dy = (1 + y^2) dx$
Rearranging the terms to the form $\frac{dx}{dy} + P(y)x = Q(y)$:
$\frac{dx}{dy} = \frac{\tan ^{-1} y - x}{1 + y^2}$
$\frac{dx}{dy} = \frac{\tan ^{-1} y}{1 + y^2} - \frac{x}{1 + y^2}$
$\frac{dx}{dy} + \frac{1}{1 + y^2} x = \frac{\tan ^{-1} y}{1 + y^2}$
Here,$P(y) = \frac{1}{1 + y^2}$.
The integrating factor $(IF)$ is given by $IF = e^{\int P(y) dy}$.
$IF = e^{\int \frac{1}{1 + y^2} dy} = e^{\tan ^{-1} y}$.
Thus,the correct option is $A$.

(A) The given differential equation is of the form $\frac{dy}{dx} + Py = Q$,where $P = \tan x$ and $Q = \sec x$.
The integrating factor $(IF)$ is given by the formula $IF = e^{\int P dx}$.
Substituting $P = \tan x$,we get $IF = e^{\int \tan x dx}$.
Since $\int \tan x dx = \ln|\sec x|$,we have $IF = e^{\ln|\sec x|}$.
Using the property $e^{\ln f(x)} = f(x)$,we get $IF = \sec x$.
Therefore,the correct option is $A$.

(B) To find the integrating factor,we first write the differential equation in the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$.
Dividing the given equation $x \frac{dy}{dx} - y = x^2$ by $x$,we get:
$\frac{dy}{dx} - \frac{1}{x}y = x$.
Here,$P(x) = -\frac{1}{x}$.
The integrating factor $(IF)$ is given by the formula $IF = e^{\int P(x) dx}$.
$IF = e^{\int -\frac{1}{x} dx} = e^{-\ln|x|} = e^{\ln|x^{-1}|} = x^{-1} = \frac{1}{x}$.
Thus,the correct option is $B$.

(D) The given differential equation is $(\tan ^{-1} y - x) dy = (1+y^2) dx$.
Rearranging the terms,we get $\frac{dx}{dy} = \frac{\tan ^{-1} y - x}{1+y^2}$.
This can be written as $\frac{dx}{dy} + \frac{x}{1+y^2} = \frac{\tan ^{-1} y}{1+y^2}$.
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = \frac{1}{1+y^2}$ and $Q(y) = \frac{\tan ^{-1} y}{1+y^2}$.
The integrating factor $(IF)$ is given by $IF = e^{\int P(y) dy}$.
$IF = e^{\int \frac{1}{1+y^2} dy} = e^{\tan ^{-1} y}$.
Thus,the correct option is $D$.