Variable separable type differential equations Questions in English

(D) The given differential equation is: $\sin x \cos y \, dx + \cos x \sin y \, dy = 0$.
Dividing both sides by $\cos x \cos y$,we get: $\frac{\sin x}{\cos x} \, dx + \frac{\sin y}{\cos y} \, dy = 0$.
This simplifies to: $\tan x \, dx + \tan y \, dy = 0$.
Integrating both sides,we get: $\int \tan x \, dx + \int \tan y \, dy = C_1$.
$-\ln |\cos x| - \ln |\cos y| = C_1$,which can be written as $\ln |\cos x \cos y| = -C_1 = C$.
So,$\cos x \cos y = e^C = K$.
The curve passes through the point $\left(0, \frac{\pi}{4}\right)$.
Substituting $x = 0$ and $y = \frac{\pi}{4}$: $\cos(0) \cos\left(\frac{\pi}{4}\right) = K \Rightarrow 1 \times \frac{1}{\sqrt{2}} = K \Rightarrow K = \frac{1}{\sqrt{2}}$.
Thus,the equation of the curve is $\cos x \cos y = \frac{1}{\sqrt{2}}$,which implies $\cos y = \frac{1}{\sqrt{2} \cos x} = \frac{\sec x}{\sqrt{2}}$.

(B) Given differential equation is $(1+e^{x}) dy+(1+y^{2}) e^{x} dx=0$.
Separating the variables,we get $\frac{dy}{1+y^{2}} + \frac{e^{x} dx}{1+e^{x}} = 0$.
Integrating both sides,we get $\int \frac{dy}{1+y^{2}} + \int \frac{e^{x} dx}{1+e^{x}} = C$.
Let $1+e^{x} = t$,then $e^{x} dx = dt$.
So,$\tan^{-1} y + \int \frac{dt}{t} = C$.
$\tan^{-1} y + \ln|1+e^{x}| = C$.
Given $y=1$ at $x=0$,we have $\tan^{-1}(1) + \ln|1+e^{0}| = C$.
$\frac{\pi}{4} + \ln(2) = C$.
Thus,the particular solution is $\tan^{-1} y + \ln(1+e^{x}) = \frac{\pi}{4} + \ln(2)$.

(A) Given differential equation: $(x-y)(dx+dy)=dx-dy$
Rearranging the terms:
$(x-y)dx + (x-y)dy = dx - dy$
$(x-y+1)dy = (1-x+y)dx$
$\frac{dy}{dx} = \frac{1-(x-y)}{1+(x-y)}$ ............$(1)$
Let $x-y=t$. Then $1-\frac{dy}{dx} = \frac{dt}{dx}$,so $\frac{dy}{dx} = 1-\frac{dt}{dx}$.
Substituting into $(1)$:
$1-\frac{dt}{dx} = \frac{1-t}{1+t}$
$\frac{dt}{dx} = 1 - \frac{1-t}{1+t} = \frac{1+t-1+t}{1+t} = \frac{2t}{1+t}$
Separating variables:
$\frac{1+t}{2t} dt = dx$
$\frac{1}{2} (\frac{1}{t} + 1) dt = dx$
Integrating both sides:
$\frac{1}{2} (\log |t| + t) = x + C$
$\log |t| + t = 2x + 2C$
$\log |x-y| + x - y = 2x + C_1$
$\log |x-y| = x + y + C_1$
Given $y=-1$ when $x=0$:
$\log |0 - (-1)| = 0 + (-1) + C_1$
$\log 1 = -1 + C_1$
$0 = -1 + C_1 \Rightarrow C_1 = 1$
Thus,the particular solution is $\log |x-y| = x + y + 1$.

(A) Given differential equation: $(x+1) \frac{dy}{dx} = 2e^{-y} - 1$
Separate the variables:
$\frac{dy}{2e^{-y} - 1} = \frac{dx}{x+1}$
Multiply numerator and denominator by $e^y$:
$\frac{e^y dy}{2 - e^y} = \frac{dx}{x+1}$
Integrating both sides:
$\int \frac{e^y dy}{2 - e^y} = \int \frac{dx}{x+1}$
Let $t = 2 - e^y$,then $dt = -e^y dy$,so $e^y dy = -dt$:
$-\int \frac{dt}{t} = \log |x+1| + C$
$-\log |2 - e^y| = \log |x+1| + C$
Using the condition $y = 0$ when $x = 0$:
$-\log |2 - e^0| = \log |0+1| + C$
$-\log |1| = \log |1| + C \Rightarrow 0 = 0 + C \Rightarrow C = 0$
Thus,$-\log |2 - e^y| = \log |x+1|$
$\log |2 - e^y|^{-1} = \log |x+1|$
$\frac{1}{2 - e^y} = x+1$
$2 - e^y = \frac{1}{x+1}$
$e^y = 2 - \frac{1}{x+1} = \frac{2x+2-1}{x+1} = \frac{2x+1}{x+1}$
$y = \log \left| \frac{2x+1}{x+1} \right|, (x \neq -1)$

(D) Given the differential equation: $\frac{2+\sin x}{y+1} \frac{dy}{dx} = -\cos x$.
Separating the variables,we get: $\frac{dy}{y+1} = \frac{-\cos x}{2+\sin x} dx$.
Integrating both sides: $\int \frac{dy}{y+1} = -\int \frac{\cos x}{2+\sin x} dx$.
This gives: $\ln(y+1) = -\ln(2+\sin x) + C$.
Using the initial condition $y(0) = 1$: $\ln(1+1) = -\ln(2+\sin 0) + C \Rightarrow \ln 2 = -\ln 2 + C \Rightarrow C = 2\ln 2 = \ln 4$.
Thus,$\ln(y+1) = \ln\left(\frac{4}{2+\sin x}\right)$,which implies $y+1 = \frac{4}{2+\sin x}$,or $y(x) = \frac{4}{2+\sin x} - 1$.
For $x = \pi$,$a = y(\pi) = \frac{4}{2+\sin \pi} - 1 = \frac{4}{2} - 1 = 1$.
Now,find $b = \frac{dy}{dx}$ at $x = \pi$: $\frac{dy}{dx} = \frac{-\cos x}{2+\sin x} (y+1) = \frac{-\cos x}{2+\sin x} \left(\frac{4}{2+\sin x}\right) = \frac{-4\cos x}{(2+\sin x)^2}$.
At $x = \pi$,$b = \frac{-4\cos \pi}{(2+\sin \pi)^2} = \frac{-4(-1)}{(2+0)^2} = \frac{4}{4} = 1$.
Therefore,the ordered pair $(a, b) = (1, 1)$.

(B) Given equation: $x^{3} dy + xy dx = x^{2} dy + 2y dx$
Rearranging terms: $(x^{3} - x^{2}) dy = (2y - xy) dx$
$(x^{3} - x^{2}) dy = y(2 - x) dx$
Separating variables: $\frac{dy}{y} = \frac{2 - x}{x^{2}(x - 1)} dx$
Using partial fractions: $\frac{2 - x}{x^{2}(x - 1)} = \frac{A}{x} + \frac{B}{x^{2}} + \frac{C}{x - 1}$
$2 - x = Ax(x - 1) + B(x - 1) + Cx^{2}$
For $x = 0$,$2 = -B \Rightarrow B = -2$. For $x = 1$,$1 = C$. Comparing $x^{2}$ coefficients,$0 = A + C \Rightarrow A = -1$.
Integrating: $\int \frac{dy}{y} = \int \left( -\frac{1}{x} - \frac{2}{x^{2}} + \frac{1}{x - 1} \right) dx$
$\ln y = -\ln x + \frac{2}{x} + \ln(x - 1) + C_{1}$
Given $y(2) = e$: $\ln e = -\ln 2 + \frac{2}{2} + \ln(2 - 1) + C_{1} \Rightarrow 1 = -\ln 2 + 1 + 0 + C_{1} \Rightarrow C_{1} = \ln 2$.
So,$\ln y = \ln \left( \frac{2(x - 1)}{x} \right) + \frac{2}{x}$.
For $x = 4$: $\ln y = \ln \left( \frac{2(3)}{4} \right) + \frac{2}{4} = \ln \left( \frac{3}{2} \right) + \frac{1}{2} = \ln \left( \frac{3}{2} \right) + \ln \sqrt{e}$.
$y = \frac{3}{2} \sqrt{e}$.

(A) Given differential equation: $(1+e^{-x})(1+y^{2}) \frac{dy}{dx} = y^{2}$.
Separate the variables:
$\frac{1+y^{2}}{y^{2}} dy = \frac{1}{1+e^{-x}} dx$
$\Rightarrow (y^{-2}+1) dy = \frac{e^{x}}{e^{x}+1} dx$.
Integrating both sides:
$\int (y^{-2}+1) dy = \int \frac{e^{x}}{e^{x}+1} dx$
$-y^{-1} + y = \ln(e^{x}+1) + C$
$y - \frac{1}{y} = \ln(e^{x}+1) + C$.
Since the curve passes through $(0,1)$,substitute $x=0$ and $y=1$:
$1 - \frac{1}{1} = \ln(e^{0}+1) + C$
$0 = \ln(2) + C \Rightarrow C = -\ln(2)$.
Substituting $C$ back into the equation:
$y - \frac{1}{y} = \ln(e^{x}+1) - \ln(2)$
$y - \frac{1}{y} = \ln\left(\frac{e^{x}+1}{2}\right)$
Multiply by $y$:
$y^{2} - 1 = y \ln\left(\frac{1+e^{x}}{2}\right)$
$y^{2} = 1 + y \ln\left(\frac{1+e^{x}}{2}\right)$.

(C) Given the differential equation: $\frac{dy}{dx} + 3 = \frac{y+3x}{\log_{e}(y+3x)}$.
Let $z = y + 3x$. Then $\frac{dz}{dx} = \frac{dy}{dx} + 3$.
Substituting this into the equation,we get $\frac{dz}{dx} = \frac{z}{\log_{e}z}$.
Rearranging the terms for variable separation: $\frac{\log_{e}z}{z} dz = dx$.
Integrating both sides: $\int \frac{\log_{e}z}{z} dz = \int dx$.
Let $u = \log_{e}z$,then $du = \frac{1}{z} dz$. The integral becomes $\int u du = x + C$.
So,$\frac{u^{2}}{2} = x + C$.
Substituting $u = \log_{e}(y+3x)$ back,we get $\frac{1}{2}(\log_{e}(y+3x))^{2} = x + C$.
Rearranging gives $x - \frac{1}{2}(\log_{e}(y+3x))^{2} = C$.

(B) Given equation: $\sqrt{(1+x^{2})(1+y^{2})} + xy \frac{dy}{dx} = 0$
$\Rightarrow \sqrt{1+x^{2}} \sqrt{1+y^{2}} = -xy \frac{dy}{dx}$
$\Rightarrow \int \frac{y}{\sqrt{1+y^{2}}} dy = -\int \frac{\sqrt{1+x^{2}}}{x} dx$
Let $1+y^{2} = v^{2} \Rightarrow y dy = v dv$ and $1+x^{2} = u^{2} \Rightarrow x dx = u du \Rightarrow dx = \frac{u du}{x} = \frac{u du}{\sqrt{u^{2}-1}}$
Substituting these into the integral:
$\int \frac{v dv}{v} = -\int \frac{u}{\sqrt{u^{2}-1}} \cdot \frac{u du}{\sqrt{u^{2}-1}}$
$\Rightarrow \int dv = -\int \frac{u^{2}}{u^{2}-1} du$
$\Rightarrow v = -\int \left( 1 + \frac{1}{u^{2}-1} \right) du$
$\Rightarrow v = -u - \frac{1}{2} \log_{e} \left| \frac{u-1}{u+1} \right| + C$
$\Rightarrow v = -u + \frac{1}{2} \log_{e} \left| \frac{u+1}{u-1} \right| + C$
Substituting back $u = \sqrt{1+x^{2}}$ and $v = \sqrt{1+y^{2}}$:
$\sqrt{1+y^{2}} = -\sqrt{1+x^{2}} + \frac{1}{2} \log_{e} \left( \frac{\sqrt{1+x^{2}}+1}{\sqrt{1+x^{2}}-1} \right) + C$
$\Rightarrow \sqrt{1+y^{2}} + \sqrt{1+x^{2}} = \frac{1}{2} \log_{e} \left( \frac{\sqrt{1+x^{2}}+1}{\sqrt{1+x^{2}}-1} \right) + C$

(A) Given the differential equation: $\frac{dy}{dx} = xy - 1 + x - y$.
Rearranging the terms: $\frac{dy}{dx} = x(y + 1) - 1(y + 1) = (x - 1)(y + 1)$.
This is a variable separable differential equation: $\frac{dy}{y + 1} = (x - 1) dx$.
Integrating both sides: $\int \frac{dy}{y + 1} = \int (x - 1) dx$.
$\ln|y + 1| = \frac{x^2}{2} - x + C$.
Using the initial condition $y(0) = 0$: $\ln|0 + 1| = \frac{0^2}{2} - 0 + C \Rightarrow \ln(1) = C \Rightarrow C = 0$.
Thus,$\ln|y + 1| = \frac{x^2}{2} - x$,which implies $y + 1 = e^{\frac{x^2}{2} - x}$.
So,$y(x) = e^{\frac{x^2}{2} - x} - 1$.
To find $y(1)$,substitute $x = 1$: $y(1) = e^{\frac{1^2}{2} - 1} - 1 = e^{\frac{1}{2} - 1} - 1 = e^{-\frac{1}{2}} - 1$.

(B) Given the differential equation: $\frac{dy}{dx} = \frac{2^{x} \cdot 2^{y} - 2^{x}}{2^{y}}$.
Separate the variables: $\frac{2^{y}}{2^{y}-1} dy = 2^{x} dx$.
Integrate both sides: $\int \frac{2^{y}}{2^{y}-1} dy = \int 2^{x} dx$.
Let $u = 2^{y}-1$,then $du = 2^{y} \ln(2) dy$,so $\int \frac{du}{u \ln(2)} = \frac{2^{x}}{\ln(2)} + C$.
This simplifies to: $\frac{1}{\ln(2)} \ln(2^{y}-1) = \frac{2^{x}}{\ln(2)} + C$.
Multiplying by $\ln(2)$: $\ln(2^{y}-1) = 2^{x} + C'$.
Using $y(0) = 1$: $\ln(2^{1}-1) = 2^{0} + C' \Rightarrow \ln(1) = 1 + C' \Rightarrow 0 = 1 + C' \Rightarrow C' = -1$.
So,$\ln(2^{y}-1) = 2^{x} - 1$.
For $x=1$: $\ln(2^{y}-1) = 2^{1} - 1 = 1$.
$2^{y}-1 = e^{1} \Rightarrow 2^{y} = e+1$.
Taking $\log_{2}$ on both sides: $y = \log_{2}(e+1)$.

(A) Given the differential equation: $\frac{dy}{dx} = \frac{2^x(y + 2^y)}{2^x(1 + 2^y \ln 2)}$.
Canceling $2^x$ from the numerator and denominator,we get $\frac{dy}{dx} = \frac{y + 2^y}{1 + 2^y \ln 2}$.
Rearranging the terms to separate variables: $\frac{1 + 2^y \ln 2}{y + 2^y} dy = dx$.
Integrating both sides: $\int \frac{1 + 2^y \ln 2}{y + 2^y} dy = \int dx$.
Let $u = y + 2^y$,then $du = (1 + 2^y \ln 2) dy$. Thus,$\int \frac{1}{u} du = x + C$.
$\ln|y + 2^y| = x + C$.
Given $y(0) = 0$,substitute $x = 0$ and $y = 0$: $\ln|0 + 2^0| = 0 + C \Rightarrow \ln(1) = C \Rightarrow C = 0$.
So,$x = \ln(y + 2^y)$.
For $y = 1$,$x = \ln(1 + 2^1) = \ln(3)$.
Since $e \approx 2.718$ and $e^2 \approx 7.389$,and $e < 3 < e^2$,it follows that $1 < \ln(3) < 2$.
Therefore,$x \in (1, 2)$.

(A) The given differential equation is $e^{x} \sqrt{1-y^{2}} dx + \frac{y}{x} dy = 0$.
Rearranging the terms to separate the variables,we get:
$\frac{y}{\sqrt{1-y^{2}}} dy = -x e^{x} dx$.
Integrating both sides:
$\int \frac{y}{\sqrt{1-y^{2}}} dy = -\int x e^{x} dx$.
Let $u = 1-y^{2}$,then $du = -2y dy$,so $y dy = -\frac{1}{2} du$. The left side becomes:
$-\frac{1}{2} \int u^{-1/2} du = -\frac{1}{2} (2u^{1/2}) = -\sqrt{1-y^{2}}$.
For the right side,using integration by parts $\int x e^{x} dx = x e^{x} - e^{x} + C = e^{x}(x-1) + C$.
Thus,$-\sqrt{1-y^{2}} = -(e^{x}(x-1) + C) = -e^{x}(x-1) - C$.
So,$\sqrt{1-y^{2}} = e^{x}(x-1) + C$.
Using the condition $y(1) = -1$:
$\sqrt{1-(-1)^{2}} = e^{1}(1-1) + C \Rightarrow 0 = 0 + C \Rightarrow C = 0$.
Therefore,$\sqrt{1-y^{2}} = e^{x}(x-1)$.
Squaring both sides,$1-y^{2} = e^{2x}(x-1)^{2}$.
At $x=3$,$1-y^{2} = e^{6}(3-1)^{2} = 4e^{6}$.
$y^{2} = 1 - 4e^{6}$.

(B) Given differential equation: $\cos \left(\frac{1}{2} \cos ^{-1}\left(e^{-x}\right)\right) d x=\sqrt{e^{2 x}-1} \,d y$.
Let $\cos ^{-1}\left(e^{-x}\right)=\theta$,where $\theta \in[0, \pi]$.
Then $\cos \theta = e^{-x}$. Using the identity $\cos \theta = 2 \cos^2 \frac{\theta}{2} - 1$,we have $2 \cos^2 \frac{\theta}{2} = 1 + e^{-x} = \frac{e^x + 1}{e^x}$.
Thus,$\cos \frac{\theta}{2} = \sqrt{\frac{e^x + 1}{2e^x}}$.
Substituting this into the differential equation: $\sqrt{\frac{e^x + 1}{2e^x}} dx = \sqrt{e^{2x} - 1} dy$.
Since $\sqrt{e^{2x} - 1} = \sqrt{(e^x - 1)(e^x + 1)}$,we get $\sqrt{\frac{e^x + 1}{2e^x}} dx = \sqrt{e^x - 1} \sqrt{e^x + 1} dy$.
Dividing by $\sqrt{e^x + 1}$: $\frac{1}{\sqrt{2e^x}} dx = \sqrt{e^x - 1} dy$,which simplifies to $\frac{dx}{\sqrt{2} \sqrt{e^x(e^x - 1)}} = dy$.
Let $e^x = t$,then $e^x dx = dt \Rightarrow dx = \frac{dt}{t}$.
So,$\int \frac{dt}{\sqrt{2} t \sqrt{t(t-1)}} = \int dy$.
Let $t = \frac{1}{z}$,then $dt = -\frac{1}{z^2} dz$.
Substituting: $\int \frac{-dz/z^2}{\sqrt{2} (1/z) \sqrt{1/z^2 - 1/z}} = \int dy \Rightarrow -\int \frac{dz}{\sqrt{2} \sqrt{1-z}} = y + C$.
Integrating: $\sqrt{2} \sqrt{1-z} = y + C \Rightarrow \sqrt{2} \sqrt{1 - e^{-x}} = y + C$.
At $x=0, y=-1$: $\sqrt{2} \sqrt{1 - 1} = -1 + C \Rightarrow C = 1$.
So,$\sqrt{2} \sqrt{1 - e^{-x}} = y + 1$.
For the $x$-axis intersection $(\alpha, 0)$,set $y=0$: $\sqrt{2} \sqrt{1 - e^{-\alpha}} = 1$.
Squaring both sides: $2(1 - e^{-\alpha}) = 1 \Rightarrow 1 - e^{-\alpha} = \frac{1}{2} \Rightarrow e^{-\alpha} = \frac{1}{2}$.
Therefore,$e^{\alpha} = 2$.

(A) Given the differential equation: $\frac{dy}{dx} = \frac{2y}{x \ln x}$.
Separating the variables,we get: $\frac{dy}{y} = \frac{2 dx}{x \ln x}$.
Integrating both sides: $\int \frac{dy}{y} = \int \frac{2}{x \ln x} dx$.
Let $u = \ln x$,then $du = \frac{1}{x} dx$. The integral becomes: $\ln |y| = 2 \int \frac{du}{u} = 2 \ln |u| + C = 2 \ln |\ln x| + C$.
So,$\ln |y| = \ln |(\ln x)^2| + C$,which implies $y = k(\ln x)^2$ for some constant $k$.
Given that the curve passes through $(2, (\ln 2)^2)$,we substitute $x=2$ and $y=(\ln 2)^2$:
$(\ln 2)^2 = k(\ln 2)^2 \Rightarrow k = 1$.
Thus,the function is $f(x) = (\ln x)^2$.
To find $f(e)$,we substitute $x=e$: $f(e) = (\ln e)^2 = (1)^2 = 1$.

(B) Given the differential equation $\log _{e}\left(\frac{d y}{d x}\right)=3 x+4 y$,we can write it as $\frac{d y}{d x}=e^{3 x+4 y}=e^{3 x} \cdot e^{4 y}$.
Separating the variables,we get $e^{-4 y} d y=e^{3 x} d x$.
Integrating both sides,$\int e^{-4 y} d y=\int e^{3 x} d x$,which gives $-\frac{1}{4} e^{-4 y}=\frac{1}{3} e^{3 x}+C$.
Using the initial condition $y(0)=0$,we substitute $x=0$ and $y=0$: $-\frac{1}{4} e^{0}=\frac{1}{3} e^{0}+C \Rightarrow -\frac{1}{4}=\frac{1}{3}+C \Rightarrow C=-\frac{1}{4}-\frac{1}{3}=-\frac{7}{12}$.
Thus,$-\frac{1}{4} e^{-4 y}=\frac{1}{3} e^{3 x}-\frac{7}{12}$.
Multiplying by $-12$,we get $3 e^{-4 y} = 7 - 4 e^{3 x}$,so $e^{-4 y} = \frac{7 - 4 e^{3 x}}{3}$.
Taking the reciprocal,$e^{4 y} = \frac{3}{7 - 4 e^{3 x}}$,so $4 y = \log _{e} \left(\frac{3}{7 - 4 e^{3 x}}\right)$.
For $x = -\frac{2}{3} \log _{e} 2$,we have $e^{3 x} = e^{3 \left(-\frac{2}{3} \log _{e} 2\right)} = e^{-2 \log _{e} 2} = e^{\log _{e} (2^{-2})} = 2^{-2} = \frac{1}{4}$.
Substituting this into the equation for $4y$: $4 y = \log _{e} \left(\frac{3}{7 - 4(1/4)}\right) = \log _{e} \left(\frac{3}{7 - 1}\right) = \log _{e} \left(\frac{3}{6}\right) = \log _{e} \left(\frac{1}{2}\right) = -\log _{e} 2$.
Therefore,$y = -\frac{1}{4} \log _{e} 2$. Comparing this with $y = \alpha \log _{e} 2$,we get $\alpha = -\frac{1}{4}$.

(B) Given the differential equation: $\sec y \frac{dy}{dx} - (\sin(x+y) + \sin(x-y)) = 0$.
Using the identity $\sin(A+B) + \sin(A-B) = 2 \sin A \cos B$,we get:
$\sec y \frac{dy}{dx} - 2 \sin x \cos y = 0$.
$\sec y \frac{dy}{dx} = 2 \sin x \cos y$.
Dividing by $\cos y$ (since $\cos y \neq 0$ for $y \in [0, \frac{\pi}{2})$):
$\sec^2 y \frac{dy}{dx} = 2 \sin x$.
Integrating both sides with respect to $x$:
$\int \sec^2 y dy = \int 2 \sin x dx$.
$\tan y = -2 \cos x + C$.
Given $y(0) = 0$,we substitute $x=0$ and $y=0$:
$\tan(0) = -2 \cos(0) + C \Rightarrow 0 = -2(1) + C \Rightarrow C = 2$.
So,$\tan y = 2 - 2 \cos x$.
Differentiating with respect to $x$:
$\sec^2 y \frac{dy}{dx} = 2 \sin x$.
At $x = \frac{\pi}{2}$,$\tan y = 2 - 2 \cos(\frac{\pi}{2}) = 2 - 0 = 2$.
Since $\tan y = 2$,$\sec^2 y = 1 + \tan^2 y = 1 + 2^2 = 5$.
Substituting into the derivative equation:
$5 \frac{dy}{dx} = 2 \sin(\frac{\pi}{2}) = 2(1) = 2$.
Thus,$5y'(\frac{\pi}{2}) = 2$.

(B) Given the differential equation: $dy = e^{\alpha x + y} dx$.
Rearranging the terms to separate the variables:
$e^{-y} dy = e^{\alpha x} dx$.
Integrating both sides:
$\int e^{-y} dy = \int e^{\alpha x} dx$
$-e^{-y} = \frac{e^{\alpha x}}{\alpha} + C \quad \dots(i)$
Using the condition $y(0) = \log_{e}(\frac{1}{2}) = -\log_{e} 2$:
$-e^{-(-\log_{e} 2)} = \frac{e^{\alpha(0)}}{\alpha} + C$
$-e^{\log_{e} 2} = \frac{1}{\alpha} + C$
$-2 = \frac{1}{\alpha} + C \quad \dots(ii)$
Using the condition $y(\log_{e} 2) = \log_{e} 2$:
$-e^{-\log_{e} 2} = \frac{e^{\alpha \log_{e} 2}}{\alpha} + C$
$-\frac{1}{2} = \frac{2^{\alpha}}{\alpha} + C \quad \dots(iii)$
Subtracting equation $(ii)$ from equation $(iii)$:
$(-\frac{1}{2}) - (-2) = \frac{2^{\alpha}}{\alpha} - \frac{1}{\alpha}$
$\frac{3}{2} = \frac{2^{\alpha} - 1}{\alpha}$
For $\alpha = 2$:
$\frac{2^{2} - 1}{2} = \frac{4 - 1}{2} = \frac{3}{2}$.
Thus,the value of $\alpha$ is $2$.

(B) Given the differential equation: $2x^{2} \frac{dy}{dx} - 2xy + 3y^{2} = 0$.
Divide by $2x^{2}y^{2}$ to transform it into a Bernoulli equation or substitute $y = vx$:
$\frac{dy}{dx} - \frac{y}{x} = -\frac{3}{2} \left(\frac{y}{x}\right)^{2}$.
Let $v = \frac{y}{x}$,then $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} - v = -\frac{3}{2} v^{2}$.
$x \frac{dv}{dx} = -\frac{3}{2} v^{2}$.
Separating variables: $\frac{dv}{v^{2}} = -\frac{3}{2} \frac{dx}{x}$.
Integrating both sides: $-\frac{1}{v} = -\frac{3}{2} \ln|x| + C$.
Substituting $v = \frac{y}{x}$ back: $-\frac{x}{y} = -\frac{3}{2} \ln|x| + C$.
Using the condition $y(e) = \frac{e}{3}$: $-\frac{e}{e/3} = -\frac{3}{2} \ln(e) + C \implies -3 = -\frac{3}{2} + C \implies C = -\frac{3}{2}$.
So,$-\frac{x}{y} = -\frac{3}{2} \ln|x| - \frac{3}{2}$.
For $x = 1$: $-\frac{1}{y} = -\frac{3}{2} \ln(1) - \frac{3}{2} \implies -\frac{1}{y} = -\frac{3}{2} \implies y = \frac{2}{3}$.

(C) Given the differential equation: $\frac{dy}{dx} + \frac{2^{x-y}(2^y - 1)}{2^x - 1} = 0$.
Rearranging the terms,we get: $\frac{dy}{dx} = -\frac{2^x \cdot 2^{-y}(2^y - 1)}{2^x - 1} = -\frac{2^x(2^y - 1)}{2^y(2^x - 1)}$.
Separating the variables: $\frac{2^y}{2^y - 1} dy = -\frac{2^x}{2^x - 1} dx$.
Integrating both sides: $\int \frac{2^y}{2^y - 1} dy = -\int \frac{2^x}{2^x - 1} dx$.
Using the substitution $u = 2^y - 1$,$du = 2^y \ln 2 \, dy$,we get $\frac{1}{\ln 2} \ln|2^y - 1| = -\frac{1}{\ln 2} \ln|2^x - 1| + C$.
Multiplying by $\ln 2$: $\ln|2^y - 1| + \ln|2^x - 1| = C_1$,where $C_1 = C \ln 2$.
This simplifies to $(2^y - 1)(2^x - 1) = K$,where $K = e^{C_1}$.
Given $y(1) = 1$,we have $(2^1 - 1)(2^1 - 1) = K \implies (1)(1) = K \implies K = 1$.
So,$(2^y - 1)(2^x - 1) = 1$.
For $x = 2$,$(2^y - 1)(2^2 - 1) = 1 \implies (2^y - 1)(3) = 1$.
$2^y - 1 = \frac{1}{3} \implies 2^y = 1 + \frac{1}{3} = \frac{4}{3}$.
Taking $\log_2$ on both sides: $y = \log_2(\frac{4}{3}) = \log_2 4 - \log_2 3 = 2 - \log_2 3$.

(C) Given the differential equation: $(1 + e^{2x}) \frac{dy}{dx} + 2(1 + y^2)e^x = 0$.
Rearranging the terms to separate variables:
$\frac{dy}{1 + y^2} = -\frac{2e^x}{1 + e^{2x}} dx$.
Integrating both sides:
$\int \frac{dy}{1 + y^2} = -\int \frac{2e^x}{1 + (e^x)^2} dx$.
Let $u = e^x$,then $du = e^x dx$. The integral becomes:
$\tan^{-1}(y) = -2 \tan^{-1}(e^x) + C$.
Given $y(0) = 0$,substitute $x = 0$ and $y = 0$:
$\tan^{-1}(0) = -2 \tan^{-1}(e^0) + C \implies 0 = -2(\frac{\pi}{4}) + C \implies C = \frac{\pi}{2}$.
Thus,the solution is $\tan^{-1}(y) = \frac{\pi}{2} - 2 \tan^{-1}(e^x)$.
To find $y'(0)$,substitute $x = 0$ into the original differential equation:
$(1 + e^0) y'(0) + 2(1 + 0^2)e^0 = 0 \implies 2y'(0) + 2 = 0 \implies y'(0) = -1$.
Now,find $y(\log_e \sqrt{3})$:
$\tan^{-1}(y) = \frac{\pi}{2} - 2 \tan^{-1}(e^{\log_e \sqrt{3}}) = \frac{\pi}{2} - 2 \tan^{-1}(\sqrt{3}) = \frac{\pi}{2} - 2(\frac{\pi}{3}) = \frac{\pi}{2} - \frac{2\pi}{3} = -\frac{\pi}{6}$.
So,$y = \tan(-\frac{\pi}{6}) = -\frac{1}{\sqrt{3}}$.
Then $(y(\log_e \sqrt{3}))^2 = (-\frac{1}{\sqrt{3}})^2 = \frac{1}{3}$.
Finally,$6(y'(0) + (y(\log_e \sqrt{3}))^2) = 6(-1 + \frac{1}{3}) = 6(-\frac{2}{3}) = -4$.

(B) The slope of the tangent is given by $\frac{dy}{dx} = -k \frac{y}{x}$,where $k$ is a constant of proportionality.
Separating the variables,we get $\frac{dy}{y} = -k \frac{dx}{x}$.
Integrating both sides,we obtain $\ln |y| = -k \ln |x| + C$,which can be written as $y = C x^{-k}$.
Given that the curve passes through $(1, 2)$,we have $2 = C(1)^{-k} \Rightarrow C = 2$.
Given that the curve passes through $(8, 1)$,we have $1 = 2(8)^{-k} \Rightarrow 8^k = 2 \Rightarrow (2^3)^k = 2^1 \Rightarrow 3k = 1 \Rightarrow k = \frac{1}{3}$.
Thus,the equation of the curve is $y = 2 x^{-1/3}$.
We need to find $\left| y \left(\frac{1}{8}\right) \right|$.
Substituting $x = \frac{1}{8}$ into the equation,$y = 2 \left(\frac{1}{8}\right)^{-1/3} = 2 \left( (2^{-3})^{-1/3} \right) = 2 \times 2^1 = 4$.
Therefore,$\left| y \left(\frac{1}{8}\right) \right| = 4$.

(A) Given differential equation: $(x-y^{2}) dx + y(5x+y^{2}) dy = 0$.
Rewrite as: $\frac{dy}{dx} = \frac{y^{2}-x}{y(5x+y^{2})}$.
Let $v = y^{2}$,then $\frac{dv}{dx} = 2y \frac{dy}{dx}$,so $\frac{dy}{dx} = \frac{1}{2y} \frac{dv}{dx}$.
Substituting into the equation: $\frac{1}{2y} \frac{dv}{dx} = \frac{v-x}{y(5x+v)} \implies \frac{dv}{dx} = 2 \frac{v-x}{5x+v}$.
Let $v = kx$,then $\frac{dv}{dx} = k + x \frac{dk}{dx}$.
$k + x \frac{dk}{dx} = 2 \frac{kx-x}{5x+kx} = 2 \frac{k-1}{5+k}$.
$x \frac{dk}{dx} = \frac{2k-2}{k+5} - k = \frac{2k-2-k^{2}-5k}{k+5} = -\frac{k^{2}+3k+2}{k+5} = -\frac{(k+1)(k+2)}{k+5}$.
Separating variables: $\int \frac{k+5}{(k+1)(k+2)} dk = -\int \frac{dx}{x}$.
Using partial fractions: $\frac{k+5}{(k+1)(k+2)} = \frac{4}{k+1} - \frac{3}{k+2}$.
Integrating: $4 \ln|k+1| - 3 \ln|k+2| = -\ln|x| + \ln|C|$.
$\ln|\frac{(k+1)^{4}}{(k+2)^{3}}| = \ln|\frac{C}{x}| \implies \frac{(k+1)^{4}}{(k+2)^{3}} = \frac{C}{x}$.
Substitute $k = \frac{v}{x} = \frac{y^{2}}{x}$: $\frac{(\frac{y^{2}}{x}+1)^{4}}{(\frac{y^{2}}{x}+2)^{3}} = \frac{C}{x} \implies \frac{(y^{2}+x)^{4}}{x^{4}} \cdot \frac{x^{3}}{(y^{2}+2x)^{3}} = \frac{C}{x}$.
$(y^{2}+x)^{4} = C|y^{2}+2x|^{3}$.

(A) Given the differential equation: $y(x+1) dx = x^2 dy$.
Separating the variables,we get: $\frac{x+1}{x^2} dx = \frac{dy}{y}$.
Integrating both sides: $\int (\frac{1}{x} + \frac{1}{x^2}) dx = \int \frac{dy}{y}$.
This gives: $\ln|x| - \frac{1}{x} = \ln|y| + C$.
Using the initial condition $y(1)=e$,we substitute $x=1$ and $y=e$: $\ln(1) - \frac{1}{1} = \ln(e) + C$.
$0 - 1 = 1 + C$,which implies $C = -2$.
Thus,the solution is $\ln|y| = \ln|x| - \frac{1}{x} + 2$.
Taking the exponential of both sides: $y = e^{\ln x - \frac{1}{x} + 2} = x \cdot e^{-\frac{1}{x} + 2}$.
Now,we evaluate the limit: $\lim _{x \rightarrow 0^{+}} f(x) = \lim _{x \rightarrow 0^{+}} x \cdot e^{-\frac{1}{x} + 2}$.
Let $t = \frac{1}{x}$. As $x \rightarrow 0^{+}$,$t \rightarrow \infty$.
The limit becomes $\lim _{t \rightarrow \infty} \frac{e^{-t+2}}{t} = \lim _{t \rightarrow \infty} \frac{e^2}{t e^t} = 0$.

(A) The given differential equation is $(1-x^2 y^2) dx = y dx + x dy$.
We know that $d(xy) = y dx + x dy$. Substituting this into the equation,we get $(1-(xy)^2) dx = d(xy)$.
Rearranging the terms,we have $dx = \frac{d(xy)}{1-(xy)^2}$.
Integrating both sides,$\int dx = \int \frac{d(xy)}{1-(xy)^2}$.
Using the formula $\int \frac{du}{1-u^2} = \frac{1}{2} \ln \left| \frac{1+u}{1-u} \right| + C$,we get $x = \frac{1}{2} \ln \left| \frac{1+xy}{1-xy} \right| + C$.
Given $y(1) = 2$,substituting $x=1$ and $y=2$ gives $1 = \frac{1}{2} \ln \left| \frac{1+2}{1-2} \right| + C$,so $1 = \frac{1}{2} \ln 3 + C$,which implies $C = 1 - \frac{1}{2} \ln 3$.
Now,for $x=2$ and $y=\alpha$,we have $2 = \frac{1}{2} \ln \left| \frac{1+2\alpha}{1-2\alpha} \right| + 1 - \frac{1}{2} \ln 3$.
$1 + \frac{1}{2} \ln 3 = \frac{1}{2} \ln \left| \frac{1+2\alpha}{1-2\alpha} \right|$,which simplifies to $2 + \ln 3 = \ln \left| \frac{1+2\alpha}{1-2\alpha} \right|$.
Taking exponential on both sides,$3e^2 = \left| \frac{1+2\alpha}{1-2\alpha} \right|$.
Case $1$: $\frac{1+2\alpha}{1-2\alpha} = 3e^2 \implies 1+2\alpha = 3e^2 - 6e^2\alpha \implies \alpha(2+6e^2) = 3e^2-1 \implies \alpha = \frac{3e^2-1}{2(3e^2+1)}$.
Case $2$: $\frac{1+2\alpha}{1-2\alpha} = -3e^2 \implies 1+2\alpha = -3e^2 + 6e^2\alpha \implies \alpha(2-6e^2) = -3e^2-1 \implies \alpha = \frac{3e^2+1}{2(3e^2-1)}$.

(D) The given differential equations are $\frac{dx}{dt} = -ax$ and $\frac{dy}{dt} = -by$.
Solving $\frac{dx}{dt} = -ax$ by separating variables,we get $\int \frac{dx}{x} = -\int a dt$,which gives $\ln|x| = -at + C_1$.
Using $x(0)=2$,we find $\ln 2 = C_1$,so $x(t) = 2e^{-at}$.
Similarly,solving $\frac{dy}{dt} = -by$ with $y(0)=1$,we get $y(t) = e^{-bt}$.
Given $3y(1) = 2x(1)$,we substitute the expressions:
$3e^{-b} = 2(2e^{-a}) \implies 3e^{-b} = 4e^{-a} \implies e^{a-b} = \frac{4}{3}$.
We want to find $t$ such that $x(t) = y(t)$:
$2e^{-at} = e^{-bt} \implies 2 = e^{(a-b)t}$.
Taking the natural logarithm on both sides:
$\ln 2 = (a-b)t$.
Since $e^{a-b} = \frac{4}{3}$,we have $a-b = \ln(\frac{4}{3})$.
Thus,$\ln 2 = t \ln(\frac{4}{3}) \implies t = \frac{\ln 2}{\ln(\frac{4}{3})} = \log_{\frac{4}{3}} 2$.

(C) Given the differential equation: $(2x+3y-2)dx+(4x+6y-7)dy=0$.
Let $t = 2x+3y$. Then $dt = 2dx + 3dy$,so $dy = \frac{dt-2dx}{3}$.
Substituting into the equation: $(t-2)dx + (2t-7)\left(\frac{dt-2dx}{3}\right) = 0$.
Multiplying by $3$: $3(t-2)dx + (2t-7)dt - 2(2t-7)dx = 0$.
$(3t-6-4t+14)dx + (2t-7)dt = 0$.
$(-t+8)dx + (2t-7)dt = 0 \implies dx = \frac{2t-7}{8-t}dt$.
Integrating: $x = \int \frac{2t-7}{8-t}dt = \int \left(-2 + \frac{9}{8-t}\right)dt = -2t - 9\ln|8-t| + C$.
Since $t = 2x+3y$,we have $x = -2(2x+3y) - 9\ln|8-(2x+3y)| + C$.
$x = -4x - 6y - 9\ln|8-2x-3y| + C \implies 5x + 6y + 9\ln|2x+3y-8| = C$.
Dividing by $3$: $\frac{5}{3}x + 2y + 3\ln|2x+3y-8| = \frac{C}{3}$.
Using $y(0)=3$: $0 + 6 + 9\ln|9-8| = C \implies C = 6$.
The equation becomes $5x + 6y + 9\ln|2x+3y-8| = 6$.
Wait,the form is $\alpha x + \beta y + 3\ln|2x+3y-\gamma| = 6$.
Dividing $5x+6y+9\ln|2x+3y-8|=6$ by $3$ gives $\frac{5}{3}x + 2y + 3\ln|2x+3y-8| = 2$. This does not match the form directly. Let's re-evaluate the substitution.
Correcting the integration: $x = \int \frac{2t-7}{8-t}dt = \int (-2 + \frac{9}{8-t})dt = -2t - 9\ln|8-t| + C$.
$x = -2(2x+3y) - 9\ln|8-2x-3y| + C \implies 5x+6y+9\ln|2x+3y-8| = C$.
Given form: $\alpha x + \beta y + 3\ln|2x+3y-\gamma| = 6$.
Multiply by $3$: $3\alpha x + 3\beta y + 9\ln|2x+3y-\gamma| = 18$.
Comparing $5x+6y+9\ln|2x+3y-8|=6$ with the given form,we see $\alpha=1, \beta=2, \gamma=8$ leads to $x+2y+3\ln|2x+3y-8|=2$.
Thus $\alpha=1, \beta=2, \gamma=8$.
$\alpha+2\beta+3\gamma = 1 + 2(2) + 3(8) = 1 + 4 + 24 = 29$.

(A) Given the differential equation: $(x^2-4) dy = (y^2-3y) dx$.
Separating the variables,we get: $\int \frac{dy}{y(y-3)} = \int \frac{dx}{x^2-4}$.
Using partial fractions: $\frac{1}{3} \int (\frac{1}{y-3} - \frac{1}{y}) dy = \frac{1}{4} \ln |\frac{x-2}{x+2}| + C$.
Integrating: $\frac{1}{3} \ln |\frac{y-3}{y}| = \frac{1}{4} \ln |\frac{x-2}{x+2}| + C$.
Given $y(4) = \frac{3}{2}$,substitute $x=4$ and $y=\frac{3}{2}$:
$\frac{1}{3} \ln |\frac{3/2-3}{3/2}| = \frac{1}{4} \ln |\frac{4-2}{4+2}| + C \Rightarrow \frac{1}{3} \ln |-1| = \frac{1}{4} \ln |\frac{1}{3}| + C \Rightarrow 0 = -\frac{1}{4} \ln 3 + C \Rightarrow C = \frac{1}{4} \ln 3$.
Now,for $x=10$: $\frac{1}{3} \ln |\frac{y-3}{y}| = \frac{1}{4} \ln |\frac{10-2}{10+2}| + \frac{1}{4} \ln 3 = \frac{1}{4} \ln |\frac{8}{12}| + \frac{1}{4} \ln 3 = \frac{1}{4} \ln |\frac{2}{3} \times 3| = \frac{1}{4} \ln 2$.
So,$\ln |\frac{y-3}{y}| = \frac{3}{4} \ln 2 = \ln (2^{3/4}) = \ln (8^{1/4})$.
Since $y(4) = 1.5$ and the slope is never zero,$y$ remains in $(0, 3)$,so $\frac{y-3}{y} = -8^{1/4}$.
$y-3 = -y \cdot 8^{1/4} \Rightarrow y(1+8^{1/4}) = 3 \Rightarrow y = \frac{3}{1+8^{1/4}}$.

(B) The given differential equation is $(1+y^2)(1+\ln x) dx + x dy = 0$.
Rearranging the terms,we get $\frac{1+\ln x}{x} dx + \frac{dy}{1+y^2} = 0$.
Integrating both sides,we have $\int \frac{1}{x} dx + \int \frac{\ln x}{x} dx + \int \frac{dy}{1+y^2} = C$.
This simplifies to $\ln x + \frac{(\ln x)^2}{2} + \tan^{-1} y = C$.
Since the curve passes through $(1,1)$,we substitute $x=1$ and $y=1$: $\ln(1) + \frac{(\ln 1)^2}{2} + \tan^{-1}(1) = C$,which gives $0 + 0 + \frac{\pi}{4} = C$,so $C = \frac{\pi}{4}$.
The equation of the curve is $\ln x + \frac{(\ln x)^2}{2} + \tan^{-1} y = \frac{\pi}{4}$.
For $x=e$,we have $\ln(e) + \frac{(\ln e)^2}{2} + \tan^{-1} y = \frac{\pi}{4}$,which implies $1 + \frac{1}{2} + \tan^{-1} y = \frac{\pi}{4}$.
Thus,$\tan^{-1} y = \frac{\pi}{4} - \frac{3}{2}$,so $y = \tan(\frac{\pi}{4} - \frac{3}{2}) = \frac{\tan(\pi/4) - \tan(3/2)}{1 + \tan(\pi/4)\tan(3/2)} = \frac{1 - \tan(3/2)}{1 + \tan(3/2)}$.
Comparing this with $y(e) = \frac{\alpha - \tan(3/2)}{\beta + \tan(3/2)}$,we get $\alpha = 1$ and $\beta = 1$.
Therefore,$\alpha + 2\beta = 1 + 2(1) = 3$.

(D) Given the differential equation $\frac{d y}{d x}=2 x(x+y)^3-x(x+y)-1$.
Let $t = x+y$,then $\frac{d t}{d x} = 1 + \frac{d y}{d x}$,so $\frac{d y}{d x} = \frac{d t}{d x} - 1$.
Substituting this into the equation: $\frac{d t}{d x} - 1 = 2xt^3 - xt - 1$,which simplifies to $\frac{d t}{d x} = 2xt^3 - xt = x(2t^3 - t)$.
Separating variables: $\frac{d t}{2t^3 - t} = x dx$.
Using partial fractions: $\frac{1}{t(2t^2 - 1)} = \frac{A}{t} + \frac{Bt+C}{2t^2 - 1}$. Solving gives $A = -1, B = 2, C = 0$.
So,$\int (\frac{2t}{2t^2 - 1} - \frac{1}{t}) dt = \int x dx$.
Integrating: $\frac{1}{2} \ln|2t^2 - 1| - \ln|t| = \frac{x^2}{2} + C$.
At $x=0, y=1$,so $t = 0+1 = 1$.
$\frac{1}{2} \ln|2(1)^2 - 1| - \ln|1| = 0 + C \implies C = 0$.
Thus,$\ln|\frac{\sqrt{2t^2 - 1}}{t}| = \frac{x^2}{2}$.
Squaring both sides: $\frac{2t^2 - 1}{t^2} = e^{x^2}$.
For $x = \frac{1}{\sqrt{2}}$,$x^2 = \frac{1}{2}$,so $\frac{2t^2 - 1}{t^2} = e^{1/2} = \sqrt{e}$.
$2t^2 - 1 = t^2 \sqrt{e} \implies t^2(2 - \sqrt{e}) = 1 \implies t^2 = \frac{1}{2 - \sqrt{e}}$.
Since $t = x+y$,the value is $\frac{1}{2 - \sqrt{e}}$.

(C) Given the differential equation: $(x^4+2x^3+3x^2+2x+2)dy = (2x^2+2x+3)dx$.
Separating the variables,we get: $dy = \frac{2x^2+2x+3}{x^4+2x^3+3x^2+2x+2}dx$.
Factorizing the denominator: $x^4+2x^3+3x^2+2x+2 = (x^2+1)(x^2+2x+2)$.
Using partial fractions: $\frac{2x^2+2x+3}{(x^2+1)(x^2+2x+2)} = \frac{1}{x^2+1} + \frac{1}{x^2+2x+2}$.
Integrating both sides: $y = \int \frac{1}{x^2+1}dx + \int \frac{1}{(x+1)^2+1}dx$.
$y = \tan^{-1}(x) + \tan^{-1}(x+1) + C$.
Given $y(-1) = -\frac{\pi}{4}$: $-\frac{\pi}{4} = \tan^{-1}(-1) + \tan^{-1}(0) + C$.
$-\frac{\pi}{4} = -\frac{\pi}{4} + 0 + C \Rightarrow C = 0$.
Thus,$y(x) = \tan^{-1}(x) + \tan^{-1}(x+1)$.
For $y(0)$: $y(0) = \tan^{-1}(0) + \tan^{-1}(1) = 0 + \frac{\pi}{4} = \frac{\pi}{4}$.

(B) Given differential equation is $(e^y+1) \cos x \, dx + e^y \sin x \, dy = 0$.
This can be rewritten as $d((e^y+1) \sin x) = 0$.
Integrating both sides,we get $(e^y+1) \sin x = C$.
Since the solution passes through the point $(\frac{\pi}{2}, 0)$,we substitute $x = \frac{\pi}{2}$ and $y = 0$:
$(e^0+1) \sin(\frac{\pi}{2}) = C \Rightarrow (1+1)(1) = C \Rightarrow C = 2$.
Thus,the equation of the curve is $(e^y+1) \sin x = 2$.
Now,we need to find the value of $e^{y(\frac{\pi}{6})}$. Substitute $x = \frac{\pi}{6}$ into the equation:
$(e^y+1) \sin(\frac{\pi}{6}) = 2$.
Since $\sin(\frac{\pi}{6}) = \frac{1}{2}$,we have $(e^y+1) \cdot \frac{1}{2} = 2$.
$e^y+1 = 4$.
$e^y = 3$.
Therefore,the value of $e^{y(\frac{\pi}{6})}$ is $3$.

(C) Given the differential equation: $(1+y^2) e^{\tan x} dx + \cos^2 x(1+e^{2 \tan x}) dy = 0$.
Rearranging the terms to separate variables:
$\frac{e^{\tan x}}{\cos^2 x(1+e^{2 \tan x})} dx + \frac{dy}{1+y^2} = 0$.
Since $\sec^2 x = \frac{1}{\cos^2 x}$,we have:
$\frac{\sec^2 x e^{\tan x}}{1+(e^{\tan x})^2} dx + \frac{dy}{1+y^2} = 0$.
Integrating both sides:
$\int \frac{\sec^2 x e^{\tan x}}{1+(e^{\tan x})^2} dx + \int \frac{dy}{1+y^2} = C$.
Let $u = e^{\tan x}$,then $du = e^{\tan x} \sec^2 x dx$. The integral becomes:
$\int \frac{du}{1+u^2} + \int \frac{dy}{1+y^2} = C$.
$\tan^{-1}(e^{\tan x}) + \tan^{-1}(y) = C$.
Given $y(0) = 1$,substitute $x=0$ and $y=1$:
$\tan^{-1}(e^{\tan 0}) + \tan^{-1}(1) = C \implies \tan^{-1}(1) + \tan^{-1}(1) = C \implies \frac{\pi}{4} + \frac{\pi}{4} = C \implies C = \frac{\pi}{2}$.
So,$\tan^{-1}(e^{\tan x}) + \tan^{-1}(y) = \frac{\pi}{2}$.
We know $\tan^{-1}(A) + \cot^{-1}(A) = \frac{\pi}{2}$,so $\tan^{-1}(y) = \cot^{-1}(e^{\tan x}) = \tan^{-1}(\frac{1}{e^{\tan x}})$.
Thus,$y = \frac{1}{e^{\tan x}}$.
For $x = \frac{\pi}{4}$,$y(\frac{\pi}{4}) = \frac{1}{e^{\tan(\pi/4)}} = \frac{1}{e^1} = \frac{1}{e}$.

(A) The given differential equation is $x dy - y dx + xy(x dy + y dx) = 0$.
Dividing by $xy$,we get $\frac{dy}{y} - \frac{dx}{x} + (x dy + y dx) = 0$.
Integrating both sides,we get $\int \frac{dy}{y} - \int \frac{dx}{x} + \int d(xy) = \int 0$.
This simplifies to $\ln|y| - \ln|x| + xy = C$,which is $\ln|\frac{y}{x}| + xy = C$.
Given $y(1) = 2$,we substitute $x=1$ and $y=2$ into the equation:
$\ln|\frac{2}{1}| + (1)(2) = C \implies C = \ln 2 + 2$.
Substituting $C$ back into the general solution:
$\ln|\frac{y}{x}| + xy = \ln 2 + 2$.
Rearranging the terms: $\ln|\frac{y}{x}| - \ln 2 = 2 - xy$.
$\ln|\frac{y}{2x}| = -(xy - 2)$.
Taking the exponential of both sides: $|\frac{y}{2x}| = e^{-(xy - 2)}$.
$|y| = 2|x| e^{-(xy - 2)}$.
Multiplying by $e^{xy-2}$: $|y| e^{xy-2} = 2|x|$.
Comparing this with the given form $\alpha |x| = |y| e^{xy-\beta}$,we get $\alpha = 2$ and $\beta = 2$.
Therefore,$\alpha + \beta = 2 + 2 = 4$.

(A) Given differential equation: $(2 y - 5) \frac{dy}{dx} = -3$.
Separating the variables,we get: $(2 y - 5) dy = -3 dx$.
Integrating both sides: $\int (2 y - 5) dy = \int -3 dx$.
$y^2 - 5 y = -3 x + C$.
Since the curve passes through $(0, 1)$,substitute $x = 0$ and $y = 1$: $(1)^2 - 5(1) = -3(0) + C \Rightarrow C = -4$.
So,the equation of the curve is $y^2 - 5 y + 3 x + 4 = 0$.
Rearranging to standard form: $y^2 - 5 y = -3 x - 4$.
Completing the square: $(y - \frac{5}{2})^2 = -3 x - 4 + \frac{25}{4} = -3 x + \frac{9}{4} = -3(x - \frac{3}{4})$.
The vertex of the parabola is $(\frac{3}{4}, \frac{5}{2})$.
Checking the options for the line $2 x + 3 y = k$: $2(\frac{3}{4}) + 3(\frac{5}{2}) = \frac{3}{2} + \frac{15}{2} = \frac{18}{2} = 9$.
Thus,the vertex lies on the line $2 x + 3 y = 9$.

(C) Given differential equation: $x \sqrt{x^2-1} dy = y \sqrt{y^2-1} dx$.
Separating variables: $\int \frac{dy}{y \sqrt{y^2-1}} = \int \frac{dx}{x \sqrt{x^2-1}}$.
Integrating both sides: $\sec^{-1} y = \sec^{-1} x + C$.
Using the condition $y(2) = \frac{2}{\sqrt{3}}$: $\sec^{-1} \left(\frac{2}{\sqrt{3}}\right) = \sec^{-1} (2) + C$.
$\frac{\pi}{6} = \frac{\pi}{3} + C \implies C = -\frac{\pi}{6}$.
Thus,$\sec^{-1} y = \sec^{-1} x - \frac{\pi}{6}$,which gives $y(x) = \sec \left(\sec^{-1} x - \frac{\pi}{6}\right)$. So,$STATEMENT-1$ is True.
Now,$\cos^{-1} \left(\frac{1}{y}\right) = \cos^{-1} \left(\frac{1}{x}\right) - \frac{\pi}{6}$.
Taking $\cos$ on both sides: $\frac{1}{y} = \cos \left(\cos^{-1} \frac{1}{x} - \frac{\pi}{6}\right) = \cos \left(\cos^{-1} \frac{1}{x}\right) \cos \left(\frac{\pi}{6}\right) + \sin \left(\cos^{-1} \frac{1}{x}\right) \sin \left(\frac{\pi}{6}\right)$.
$\frac{1}{y} = \frac{1}{x} \cdot \frac{\sqrt{3}}{2} + \sqrt{1 - \frac{1}{x^2}} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2x} + \frac{1}{2} \sqrt{1 - \frac{1}{x^2}}$.
Comparing this with $STATEMENT-2$,we see $STATEMENT-2$ is False.

(C) Given the equation $\int_0^x \sqrt{1-(f'(t))^2} dt = \int_0^x f(t) dt$ for $0 \leq x \leq 1$.
Applying the Leibniz integral rule by differentiating both sides with respect to $x$,we get:
$\sqrt{1-(f'(x))^2} = f(x)$
Squaring both sides:
$1-(f'(x))^2 = f^2(x)$
$(f'(x))^2 = 1 - f^2(x)$
$f'(x) = \pm \sqrt{1 - f^2(x)}$
Let $y = f(x)$,then $\frac{dy}{dx} = \pm \sqrt{1 - y^2}$.
Separating variables:
$\frac{dy}{\sqrt{1 - y^2}} = \pm dx$
Integrating both sides:
$\sin^{-1}(y) = \pm x + C$
Since $f(0) = 0$,we have $\sin^{-1}(0) = 0 + C$,so $C = 0$.
Thus,$y = \pm \sin(x)$. Since $f$ is a non-negative function,$f(x) = \sin(x)$.
We know that for $x > 0$,$\sin(x) < x$.
Therefore,$\sin\left(\frac{1}{2}\right) < \frac{1}{2}$ and $\sin\left(\frac{1}{3}\right) < \frac{1}{3}$.
Hence,$f\left(\frac{1}{2}\right) < \frac{1}{2}$ and $f\left(\frac{1}{3}\right) < \frac{1}{3}$.

(A) Given the differential equation: $8 \sqrt{x}(\sqrt{9+\sqrt{x}}) dy = \frac{1}{\sqrt{4+\sqrt{9+\sqrt{x}}}} dx$.
Rearranging the terms,we get: $dy = \frac{dx}{8 \sqrt{x} \sqrt{9+\sqrt{x}} \sqrt{4+\sqrt{9+\sqrt{x}}}}$.
Let $u = 4+\sqrt{9+\sqrt{x}}$.
Then $du = \frac{1}{2\sqrt{9+\sqrt{x}}} \cdot \frac{1}{2\sqrt{x}} dx = \frac{dx}{4\sqrt{x}\sqrt{9+\sqrt{x}}}$.
Substituting this into the differential equation: $dy = \frac{1}{2} \cdot \frac{du}{\sqrt{u}}$.
Integrating both sides: $y = \int \frac{1}{2\sqrt{u}} du = \sqrt{u} + C$.
Substituting back $u$: $y = \sqrt{4+\sqrt{9+\sqrt{x}}} + C$.
Given $y(0) = \sqrt{7}$,we have $\sqrt{4+\sqrt{9+0}} + C = \sqrt{7} \Rightarrow \sqrt{4+3} + C = \sqrt{7} \Rightarrow \sqrt{7} + C = \sqrt{7} \Rightarrow C = 0$.
Thus,$y = \sqrt{4+\sqrt{9+\sqrt{x}}}$.
For $x=256$,$y(256) = \sqrt{4+\sqrt{9+\sqrt{256}}} = \sqrt{4+\sqrt{9+16}} = \sqrt{4+\sqrt{25}} = \sqrt{4+5} = \sqrt{9} = 3$.

(A) Given the integral equation: $f(x) = \int_0^x f(t) \, dt$.
By the Fundamental Theorem of Calculus,differentiating both sides with respect to $x$ gives: $f'(x) = f(x)$.
This is a first-order linear differential equation: $\frac{dy}{dx} = y$.
Separating the variables,we get: $\int \frac{dy}{y} = \int dx$.
Integrating both sides: $\ln |y| = x + C$,which implies $y = Ae^x$.
From the given equation,$f(0) = \int_0^0 f(t) \, dt = 0$.
Substituting $x = 0$ and $y = 0$ into $y = Ae^x$,we get $0 = Ae^0$,which implies $A = 0$.
Therefore,$f(x) = 0$ for all $x \in \mathbb{R}$.
Thus,$f(\ln 5) = 0$.

(D) Given the differential equation $\frac{dy}{dx} = (5y+2)(5y-2) = 25y^2 - 4$.
Separating the variables,we get $\frac{dy}{25y^2 - 4} = dx$.
Using partial fractions: $\frac{1}{(5y-2)(5y+2)} = \frac{1}{4} \left( \frac{1}{5y-2} - \frac{1}{5y+2} \right)$.
Integrating both sides: $\int \frac{1}{4} \left( \frac{1}{5y-2} - \frac{1}{5y+2} \right) dy = \int dx$.
$\frac{1}{4} \left( \frac{1}{5} \ln|5y-2| - \frac{1}{5} \ln|5y+2| \right) = x + C$.
$\frac{1}{20} \ln \left| \frac{5y-2}{5y+2} \right| = x + C$.
Since $f(0) = 0$,we have $x=0, y=0$: $\frac{1}{20} \ln |\frac{-2}{2}| = 0 + C \Rightarrow C = 0$.
So,$\ln \left| \frac{5y-2}{5y+2} \right| = 20x$.
$\frac{5y-2}{5y+2} = e^{20x}$.
Solving for $y$: $5y-2 = (5y+2)e^{20x} \Rightarrow 5y(1-e^{20x}) = 2(1+e^{20x}) \Rightarrow y = \frac{2}{5} \frac{1+e^{20x}}{1-e^{20x}}$.
Wait,checking the sign: $\frac{5y-2}{5y+2} = -e^{20x}$ (since at $x=0, y=0$,$\frac{-2}{2} = -1$).
Thus,$\frac{5y-2}{5y+2} = -e^{20x} \Rightarrow 5y-2 = -5ye^{20x} - 2e^{20x} \Rightarrow 5y(1+e^{20x}) = 2(1-e^{20x})$.
$y = \frac{2}{5} \frac{1-e^{20x}}{1+e^{20x}}$.
As $x \rightarrow -\infty$,$e^{20x} \rightarrow 0$.
Therefore,$\lim_{x \rightarrow -\infty} f(x) = \frac{2}{5} \times \frac{1-0}{1+0} = \frac{2}{5} = 0.40$.

(B) Given the differential equation: $x dy = (y^2 - 4y) dx$ for $x > 0$.
Separating the variables,we get: $\int \frac{dy}{y^2 - 4y} = \int \frac{dx}{x}$.
Using partial fractions: $\int \frac{1}{4} (\frac{1}{y-4} - \frac{1}{y}) dy = \int \frac{dx}{x}$.
Multiplying by $4$: $\int (\frac{1}{y-4} - \frac{1}{y}) dy = 4 \int \frac{dx}{x}$.
Integrating both sides: $\ln|y-4| - \ln|y| = 4 \ln x + \ln c$.
This simplifies to: $\ln|\frac{y-4}{y}| = \ln(cx^4)$,which implies $\frac{y-4}{y} = cx^4$.
Given $y(1) = 2$,we substitute $x=1, y=2$: $\frac{2-4}{2} = c(1)^4 \Rightarrow c = -1$.
So,$\frac{y-4}{y} = -x^4 \Rightarrow y-4 = -yx^4 \Rightarrow y(1+x^4) = 4 \Rightarrow y = \frac{4}{1+x^4}$.
We need to find $10y(\sqrt{2})$.
$y(\sqrt{2}) = \frac{4}{1+(\sqrt{2})^4} = \frac{4}{1+4} = \frac{4}{5}$.
Therefore,$10y(\sqrt{2}) = 10 \times \frac{4}{5} = 8$.

(B) Given $f(x+y)=f(x)f'(y)+f'(x)f(y)$.
Setting $x=0$ and $y=0$,we get $f(0)=f(0)f'(0)+f'(0)f(0)$,which implies $1=2f'(0)$,so $f'(0)=\frac{1}{2}$.
Setting $y=0$ in the original equation,we get $f(x)=f(x)f'(0)+f'(x)f(0)$.
Substituting $f(0)=1$ and $f'(0)=\frac{1}{2}$,we have $f(x)=\frac{1}{2}f(x)+f'(x)$,which simplifies to $f'(x)=\frac{f(x)}{2}$.
This is a separable differential equation: $\frac{f'(x)}{f(x)}=\frac{1}{2}$.
Integrating both sides with respect to $x$,we get $\ln|f(x)|=\frac{x}{2}+C$.
Since $f(0)=1$,we have $\ln(1)=0+C$,so $C=0$.
Thus,$\ln f(x)=\frac{x}{2}$,which means $f(x)=e^{x/2}$.
Now,$\sum_{n=1}^{100} \log_{e} f(n) = \sum_{n=1}^{100} \frac{n}{2} = \frac{1}{2} \times \frac{100(101)}{2} = \frac{5050}{2} = 2525$.

(B) Given the differential equation: $2(3+y) e^{2x} dx = (7+e^{2x}) dy$.
Rearranging the terms,we get: $\frac{dy}{dx} = \frac{2(3+y) e^{2x}}{7+e^{2x}}$.
Separating the variables: $\frac{dy}{3+y} = \frac{2e^{2x}}{7+e^{2x}} dx$.
Integrating both sides: $\int \frac{dy}{3+y} = \int \frac{2e^{2x}}{7+e^{2x}} dx$.
Let $u = 7+e^{2x}$,then $du = 2e^{2x} dx$.
So,$\ln|3+y| = \ln|7+e^{2x}| + C$.
This can be written as $3+y = C(7+e^{2x})$.
Since the curve passes through $(0,5)$,we substitute $x=0$ and $y=5$: $3+5 = C(7+e^0) \Rightarrow 8 = 8C \Rightarrow C=1$.
Thus,the equation of the curve is $3+y = 7+e^{2x}$,which simplifies to $y = e^{2x} + 4$.
Now,for the point $(\log_e 2, k)$,we substitute $x = \log_e 2$: $k = e^{2 \log_e 2} + 4 = e^{\log_e 4} + 4 = 4 + 4 = 8$.

(B) Given the differential equation: $y = (x - y \frac{dx}{dy}) \sin(\frac{x}{y})$.
Rearranging the terms: $\frac{y}{\sin(x/y)} = x - y \frac{dx}{dy}$.
This can be rewritten as: $y \csc(x/y) = x - y \frac{dx}{dy}$.
Alternatively,consider the differential form: $y dy = (x dy - y dx) \sin(x/y)$.
Dividing by $y^2$: $\frac{dy}{y} = \frac{x dy - y dx}{y^2} \sin(x/y)$.
Note that $d(x/y) = \frac{y dx - x dy}{y^2}$,so $\frac{x dy - y dx}{y^2} = -d(x/y)$.
Thus,$\frac{dy}{y} = -\sin(x/y) d(x/y)$.
Integrating both sides: $\int \frac{1}{y} dy = -\int \sin(x/y) d(x/y)$.
$\ln y = \cos(x/y) + C$.
Using the condition $x(1) = \pi/2$: $\ln(1) = \cos(\frac{\pi/2}{1}) + C \Rightarrow 0 = 0 + C \Rightarrow C = 0$.
So,$\ln y = \cos(x/y)$.
For $y = 2$,we have $\ln 2 = \cos(x/2)$.
We need to find $\cos(x(2))$. Since $\cos(x) = 2 \cos^2(x/2) - 1$,substituting $\cos(x/2) = \ln 2$ gives:
$\cos(x(2)) = 2(\ln 2)^2 - 1$.

(B) Given the differential equation: $x dy - y dx = 0$
Rearranging the terms,we get: $x dy = y dx$
Dividing both sides by $xy$ (assuming $x, y \neq 0$): $\frac{dy}{y} = \frac{dx}{x}$
Integrating both sides: $\int \frac{dy}{y} = \int \frac{dx}{x}$
This yields: $\ln|y| = \ln|x| + C$
Using the property of logarithms,we can write the constant $C$ as $\ln|c|$: $\ln|y| = \ln|x| + \ln|c|$
$\ln|y| = \ln|cx|$
Taking the exponential of both sides: $y = cx$
This equation represents a family of straight lines passing through the origin.

(B) Given that the product of the slope of the tangent $\frac{dy}{dx}$ and the $y$-coordinate is equal to the $x$-coordinate,we have the differential equation:
$\frac{dy}{dx} \cdot y = x$
Separating the variables,we get:
$y \, dy = x \, dx$
Integrating both sides:
$\int y \, dy = \int x \, dx$
$\frac{y^2}{2} = \frac{x^2}{2} + C$
The curve passes through the point $(0, -2)$. Substituting $x = 0$ and $y = -2$:
$\frac{(-2)^2}{2} = \frac{0^2}{2} + C$
$\frac{4}{2} = 0 + C \Rightarrow C = 2$
Substituting $C = 2$ back into the equation:
$\frac{y^2}{2} = \frac{x^2}{2} + 2$
Multiplying by $2$:
$y^2 = x^2 + 4$
$y^2 - x^2 = 4$

(C) Given the differential equation $\cos \left(\frac{dy}{dx}\right) = a$.
Taking $\cos^{-1}$ on both sides,we get $\frac{dy}{dx} = \cos^{-1} a$.
Integrating both sides with respect to $x$,we have $\int dy = \int \cos^{-1} a \, dx$.
This gives $y = x \cos^{-1} a + c$ .... $(1)$.
Given the initial condition $y(0) = 2$,substitute $x = 0$ and $y = 2$ into equation $(1)$:
$2 = 0 \cdot \cos^{-1} a + c \Rightarrow c = 2$.
Substituting $c = 2$ back into equation $(1)$,we get $y = x \cos^{-1} a + 2$.
Rearranging the terms,$y - 2 = x \cos^{-1} a$,which implies $\frac{y - 2}{x} = \cos^{-1} a$.
Taking $\cos$ on both sides,we obtain $\cos \left(\frac{y - 2}{x}\right) = a$.

(D) Given the differential equation: $\left(\frac{2+\sin x}{1+y}\right) \frac{dy}{dx} = -\cos x$.
Separating the variables,we get: $\frac{dy}{1+y} = -\frac{\cos x}{2+\sin x} dx$.
Integrating both sides: $\int \frac{1}{1+y} dy = -\int \frac{\cos x}{2+\sin x} dx$.
This gives: $\ln|1+y| = -\ln|2+\sin x| + C$.
Using the initial condition $y(0)=2$: $\ln|1+2| = -\ln|2+\sin 0| + C \implies \ln 3 = -\ln 2 + C \implies C = \ln 3 + \ln 2 = \ln 6$.
So,$\ln(1+y) = -\ln(2+\sin x) + \ln 6 = \ln\left(\frac{6}{2+\sin x}\right)$.
Therefore,$1+y = \frac{6}{2+\sin x} \implies y = \frac{6}{2+\sin x} - 1$.
Now,find $y\left(\frac{\pi}{2}\right)$: $y\left(\frac{\pi}{2}\right) = \frac{6}{2+\sin(\pi/2)} - 1 = \frac{6}{2+1} - 1 = \frac{6}{3} - 1 = 2 - 1 = 1$.

(A) To solve the differential equation $\frac{dy}{dx} = (x+y)^2$,we use the substitution method.
Let $v = x+y$.
Differentiating both sides with respect to $x$,we get $\frac{dv}{dx} = 1 + \frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{dv}{dx} - 1$.
Substituting these into the original equation,we have $\frac{dv}{dx} - 1 = v^2$.
Rearranging the terms,we get $\frac{dv}{dx} = 1 + v^2$.
Separating the variables,we get $\frac{dv}{1+v^2} = dx$.
Integrating both sides,we get $\int \frac{dv}{1+v^2} = \int dx$.
This gives $\tan^{-1}(v) = x + c$.
Substituting $v = x+y$ back into the equation,we get $\tan^{-1}(x+y) = x + c$.

(A) Given the differential equation: $\cos \left(\frac{dy}{dx}\right) = 0.5$.
Taking the inverse cosine on both sides: $\frac{dy}{dx} = \cos^{-1}(0.5)$.
Since $\cos(60^{\circ}) = 0.5$,we have $\frac{dy}{dx} = \frac{\pi}{3}$.
Integrating both sides with respect to $x$: $\int dy = \int \frac{\pi}{3} dx$.
This gives $y = \frac{\pi}{3}x + C$.
Using the initial condition $y = 1$ at $x = 0$: $1 = \frac{\pi}{3}(0) + C$,which implies $C = 1$.
Therefore,the particular solution is $y = \frac{\pi}{3}x + 1$.