Homogeneous differential equations Questions in English

(D) Given the differential equation $\frac{dy}{dx} = \tan \left(\frac{y}{x}\right) + \frac{y}{x}$.
Substitute $\frac{y}{x} = v$,which implies $y = vx$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the original equation: $v + x \frac{dv}{dx} = \tan(v) + v$.
Simplifying,we get $x \frac{dv}{dx} = \tan(v)$.
Separating the variables,we have $\frac{dv}{\tan(v)} = \frac{dx}{x}$,which is $\cot(v) \, dv = \frac{dx}{x}$.
Integrating both sides: $\int \cot(v) \, dv = \int \frac{dx}{x}$.
This gives $\ln |\sin(v)| = \ln |x| + \ln |c|$.
Taking the exponential of both sides,we get $\sin(v) = cx$.
Substituting back $v = \frac{y}{x}$,the general solution is $\sin \left(\frac{y}{x}\right) = cx$.

(C) The given differential equation is $x \frac{dy}{dx} = y - x \tan \left( \frac{y}{x} \right)$.
Dividing by $x$,we get $\frac{dy}{dx} = \frac{y}{x} - \tan \left( \frac{y}{x} \right)$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = v - \tan v$.
Subtracting $v$ from both sides: $x \frac{dv}{dx} = - \tan v$.
Separating the variables: $\frac{dv}{\tan v} = - \frac{dx}{x}$,which is $\cot v \, dv = - \frac{dx}{x}$.
Integrating both sides: $\int \cot v \, dv = - \int \frac{1}{x} \, dx$.
This gives $\ln |\sin v| = - \ln |x| + \ln |c|$.
Using logarithmic properties: $\ln |\sin v| + \ln |x| = \ln |c| \Rightarrow \ln |x \sin v| = \ln |c|$.
Thus,$x \sin v = c$. Substituting $v = \frac{y}{x}$ back,we get $x \sin \left( \frac{y}{x} \right) = c$.

(B) Given the differential equation: $\frac{dy}{dx} = \tan \left(\frac{y}{x}\right) + \frac{y}{x}$.
Let $v = \frac{y}{x}$,so $y = vx$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the original equation:
$v + x \frac{dv}{dx} = \tan v + v$.
Subtracting $v$ from both sides:
$x \frac{dv}{dx} = \tan v$.
Separating the variables:
$\frac{dv}{\tan v} = \frac{dx}{x}$,which is $\cot v \, dv = \frac{dx}{x}$.
Integrating both sides:
$\int \cot v \, dv = \int \frac{1}{x} \, dx$.
$\log |\sin v| = \log |x| + \log |c|$.
$\log |\sin v| = \log |cx|$.
Taking the exponential of both sides:
$\sin v = cx$.
Substituting $v = \frac{y}{x}$ back:
$\sin \left(\frac{y}{x}\right) = cx$.

(D) Given the differential equation: $x \frac{dy}{dx} = y(\log(\frac{y}{x}) + 1)$.
Divide by $x$: $\frac{dy}{dx} = \frac{y}{x}(\log(\frac{y}{x}) + 1)$.
Let $v = \frac{y}{x}$,so $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = v(\log v + 1)$.
$v + x \frac{dv}{dx} = v \log v + v$.
$x \frac{dv}{dx} = v \log v$.
Separating variables: $\frac{dv}{v \log v} = \frac{dx}{x}$.
Integrating both sides: $\int \frac{1}{v \log v} dv = \int \frac{1}{x} dx$.
Let $u = \log v$,then $du = \frac{1}{v} dv$. The integral becomes $\int \frac{1}{u} du = \int \frac{1}{x} dx$.
$\log|u| = \log|x| + \log|c|$.
$\log|\log v| = \log|cx|$.
$\log v = cx$.
Substituting $v = \frac{y}{x}$ back: $\log(\frac{y}{x}) = cx$.

(A) Given the differential equation: $(y^2 - x^2) dx = xy dy$.
This can be rewritten as: $\frac{dy}{dx} = \frac{y^2 - x^2}{xy}$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{(vx)^2 - x^2}{x(vx)} = \frac{v^2 - 1}{v}$.
$x \frac{dv}{dx} = \frac{v^2 - 1}{v} - v = \frac{v^2 - 1 - v^2}{v} = -\frac{1}{v}$.
Separating the variables: $v dv = -\frac{1}{x} dx$.
Integrating both sides: $\int v dv = -\int \frac{1}{x} dx$.
$\frac{v^2}{2} = -\log |x| + c$.
Since $v = \frac{y}{x}$,we have $\frac{y^2}{2x^2} = -\log |x| + c$.
Multiplying by $2x^2$: $y^2 = -2x^2 \log |x| + 2cx^2$.
Rearranging gives: $2x^2 \log |x| + y^2 - 2cx^2 = 0$.
Replacing $-c$ with a new constant $C$,we get $2x^2 \log |x| + y^2 + 2Cx^2 = 0$.

(A) The slope of the tangent is given by $\frac{dy}{dx} = \frac{x^2+y^2}{2xy}$.
This is a homogeneous differential equation.
Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting into the equation: $v + x \frac{dv}{dx} = \frac{x^2 + v^2x^2}{2x(vx)} = \frac{1+v^2}{2v}$.
$x \frac{dv}{dx} = \frac{1+v^2}{2v} - v = \frac{1+v^2-2v^2}{2v} = \frac{1-v^2}{2v}$.
Separating variables: $\frac{2v}{1-v^2} dv = \frac{dx}{x}$.
Integrating both sides: $-\ln|1-v^2| = \ln|x| + C$.
$\ln|1-v^2|^{-1} = \ln|x| + C \implies \frac{1}{1-v^2} = Cx$.
Substituting $v = \frac{y}{x}$: $\frac{1}{1-(y/x)^2} = Cx \implies \frac{x^2}{x^2-y^2} = Cx \implies x^2 = C x (x^2-y^2) \implies x = C(x^2-y^2)$.
Since the curve passes through $(2, 1)$,$2 = C(4-1) = 3C \implies C = \frac{2}{3}$.
So,$x = \frac{2}{3}(x^2-y^2) \implies 3x = 2x^2 - 2y^2$ (This does not match options,re-evaluating).
Correcting the integration: $\int \frac{2v}{1-v^2} dv = \int \frac{dx}{x} \implies -\ln|1-v^2| = \ln|x| + \ln|k| \implies \ln|\frac{1}{1-v^2}| = \ln|kx| \implies \frac{1}{1-v^2} = kx$.
Using $(2, 1)$: $\frac{1}{1-(1/4)} = k(2) \implies \frac{1}{3/4} = 2k \implies \frac{4}{3} = 2k \implies k = \frac{2}{3}$.
$\frac{x^2}{x^2-y^2} = \frac{2}{3}x \implies 3x = 2(x^2-y^2) \implies 2y^2 = 2x^2 - 3x$. The provided options were incorrect; the correct curve is $2y^2 = 2x^2 - 3x$.

(A) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} + \sec \left(\frac{y}{x}\right)$.
This is a homogeneous differential equation. Let $v = \frac{y}{x}$,so $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation:
$v + x \frac{dv}{dx} = v + \sec v$
$x \frac{dv}{dx} = \sec v$
$\cos v \, dv = \frac{1}{x} \, dx$.
Integrating both sides:
$\int \cos v \, dv = \int \frac{1}{x} \, dx$
$\sin v = \log x + C$.
Substituting $v = \frac{y}{x}$ back:
$\sin \left(\frac{y}{x}\right) = \log x + C$.
The curve passes through $\left(1, \frac{\pi}{6}\right)$,so:
$\sin \left(\frac{\pi/6}{1}\right) = \log(1) + C$
$\sin \left(\frac{\pi}{6}\right) = 0 + C \Rightarrow C = \frac{1}{2}$.
Thus,the equation of the curve is $\sin \left(\frac{y}{x}\right) = \log x + \frac{1}{2}$.

(A) Given the differential equation $\frac{d y}{d x}=\frac{3 x+y}{x-y}$.
Since it is a homogeneous differential equation,let $y=vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Substituting these into the equation:
$v + x\frac{dv}{dx} = \frac{3x+vx}{x-vx} = \frac{3+v}{1-v}$.
$x\frac{dv}{dx} = \frac{3+v}{1-v} - v = \frac{3+v-v+v^2}{1-v} = \frac{3+v^2}{1-v}$.
Separating the variables:
$\int \frac{1-v}{3+v^2} dv = \int \frac{dx}{x}$.
$\int \frac{1}{3+v^2} dv - \int \frac{v}{3+v^2} dv = \int \frac{dx}{x}$.
$\frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{v}{\sqrt{3}}\right) - \frac{1}{2} \log(3+v^2) = \log|x| + C$.
Substituting $v = \frac{y}{x}$ back:
$\frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{y}{x\sqrt{3}}\right) - \frac{1}{2} \log\left(3+\frac{y^2}{x^2}\right) = \log|x| + C$.
$\frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{y}{x\sqrt{3}}\right) - \frac{1}{2} \log\left(\frac{3x^2+y^2}{x^2}\right) = \log|x| + C$.
$\frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{y}{x\sqrt{3}}\right) - \log\left(\frac{y^2+3x^2}{x^2}\right)^{\frac{1}{2}} = \log|x| + C$.

(C) Given differential equation: $x^2+y^2-2xy \frac{dy}{dx}=0$
Rearranging the terms: $2xy \frac{dy}{dx} = x^2+y^2$
$\frac{dy}{dx} = \frac{x^2+y^2}{2xy}$
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{x^2 + (vx)^2}{2x(vx)} = \frac{x^2(1+v^2)}{2x^2v} = \frac{1+v^2}{2v}$
$x \frac{dv}{dx} = \frac{1+v^2}{2v} - v = \frac{1+v^2-2v^2}{2v} = \frac{1-v^2}{2v}$
Separating the variables: $\frac{2v}{1-v^2} dv = \frac{dx}{x}$
Integrating both sides: $\int \frac{2v}{1-v^2} dv = \int \frac{dx}{x}$
Let $1-v^2 = t$,then $-2v dv = dt$,so $2v dv = -dt$.
$-\int \frac{dt}{t} = \int \frac{dx}{x} \Rightarrow -\ln|t| = \ln|x| + \ln|C_1|$
$-\ln|1-v^2| = \ln|x| + \ln|C_1| = \ln|C_1 x|$
$\ln|1-v^2|^{-1} = \ln|C_1 x| \Rightarrow \frac{1}{1-v^2} = C_1 x$
Substituting $v = y/x$: $\frac{1}{1-(y^2/x^2)} = C_1 x \Rightarrow \frac{x^2}{x^2-y^2} = C_1 x$
$\frac{x}{x^2-y^2} = C_1 \Rightarrow x = C_1(x^2-y^2)$
Rearranging gives $x^2-y^2 = Cx$ (where $C = 1/C_1$ is an arbitrary constant).

(C) Given the differential equation $\frac{dy}{dx} = \frac{y}{x + \sqrt{xy}}$.
Since it is a homogeneous differential equation,let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Substituting these into the equation: $v + x\frac{dv}{dx} = \frac{vx}{x + \sqrt{x^2v}} = \frac{vx}{x(1 + \sqrt{v})} = \frac{v}{1 + \sqrt{v}}$.
$x\frac{dv}{dx} = \frac{v}{1 + \sqrt{v}} - v = \frac{v - v - v\sqrt{v}}{1 + \sqrt{v}} = \frac{-v\sqrt{v}}{1 + \sqrt{v}}$.
Separating variables: $-\frac{1 + \sqrt{v}}{v\sqrt{v}} dv = \frac{dx}{x} \Rightarrow -(\frac{1}{v^{3/2}} + \frac{1}{v}) dv = \frac{dx}{x}$.
Integrating both sides: $-\int (v^{-3/2} + v^{-1}) dv = \int \frac{1}{x} dx$.
$-(-2v^{-1/2} - \ln|v|) = \ln|x| + C$.
$2\frac{1}{\sqrt{v}} + \ln|v| = \ln|x| + C$.
Since $v = y/x$,we have $2\sqrt{x/y} + \ln(y/x) = \ln|x| + C$.
$2\sqrt{x/y} + \ln|y| - \ln|x| = \ln|x| + C$.
$2\sqrt{x/y} = \ln|x| - \ln|y| + \ln|x| + C = 2\ln|x| - \ln|y| + C$.

(D) Given differential equation is $(3xy+y^2) dx + (x^2+xy) dy = 0$.
Rearranging,we get $\frac{dy}{dx} = -\frac{3xy+y^2}{x^2+xy}$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Substituting these into the equation: $v + x\frac{dv}{dx} = -\frac{3v x^2 + v^2 x^2}{x^2 + vx^2} = -\frac{3v+v^2}{1+v}$.
$x\frac{dv}{dx} = -\frac{3v+v^2}{1+v} - v = -\frac{3v+v^2+v+v^2}{1+v} = -\frac{2v^2+4v}{1+v}$.
Separating variables: $\int \frac{1+v}{2v^2+4v} dv = -\int \frac{1}{x} dx$.
Multiply by $2$: $\int \frac{1+v}{v^2+2v} dv = -2\int \frac{1}{x} dx$.
Let $u = v^2+2v$,then $du = (2v+2) dv = 2(v+1) dv$.
So,$\frac{1}{2} \int \frac{du}{u} = -2\ln|x| + C'$.
$\frac{1}{2} \ln|v^2+2v| = -2\ln|x| + C'$.
$\ln|v^2+2v| = -4\ln|x| + C''$.
$\ln|v^2+2v| + \ln|x^4| = C''$.
$\ln|x^4(v^2+2v)| = C''$.
$x^4(\frac{y^2}{x^2} + \frac{2y}{x}) = c$.
$x^2(y^2+2xy) = c$.

(C) differential equation $\frac{dy}{dx} = f(x, y)$ is homogeneous if $f(x, y)$ is a homogeneous function of degree $0$.
For $I: f(x, y) = \frac{y+x}{x} = \frac{y}{x} + 1$. This is a homogeneous function of degree $0$. Thus,$I$ is a homogeneous differential equation.
For $II: f(x, y) = \frac{x^2+y}{x^3} = \frac{1}{x} + \frac{y}{x^3}$. This is not a homogeneous function because the degrees of the terms are not the same. Thus,$II$ is not homogeneous.
For $III: f(x, y) = \frac{2xy}{y^2-x^2}$. Dividing numerator and denominator by $x^2$,we get $f(x, y) = \frac{2(y/x)}{(y/x)^2-1}$. This is a homogeneous function of degree $0$. Thus,$III$ is a homogeneous differential equation.
Since $I$ and $III$ are homogeneous,statement $S3$ is valid.

(D) Given the differential equation $\frac{dy}{dx} = \frac{x^2 + y^2}{xy} = \frac{x}{y} + \frac{y}{x}$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{1}{v} + v$.
Subtracting $v$ from both sides: $x \frac{dv}{dx} = \frac{1}{v}$.
Separating variables: $v \, dv = \frac{1}{x} \, dx$.
Integrating both sides: $\int v \, dv = \int \frac{1}{x} \, dx \Rightarrow \frac{v^2}{2} = \log |x| + C$.
Substituting $v = \frac{y}{x}$: $\frac{y^2}{2x^2} = \log |x| + C \Rightarrow y^2 = 2x^2 \log |x| + 2x^2 C$.
Since $\log |x| = \frac{1}{2} \log x^2$,we have $y^2 = x^2 \log x^2 + 2x^2 C$.
Using the condition $y(1) = -2$: $(-2)^2 = (1)^2 \log(1)^2 + 2(1)^2 C \Rightarrow 4 = 0 + 2C \Rightarrow C = 2$.
Substituting $C = 2$ into the general solution: $y^2 = x^2 \log x^2 + 4x^2$.

(C) Given the differential equation: $\frac{dy}{dx} = \frac{x+y}{x-y}$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{x+vx}{x-vx} = \frac{1+v}{1-v}$.
$x \frac{dv}{dx} = \frac{1+v}{1-v} - v = \frac{1+v-v+v^2}{1-v} = \frac{1+v^2}{1-v}$.
Separating the variables: $\int \frac{1-v}{1+v^2} dv = \int \frac{dx}{x}$.
$\int \frac{1}{1+v^2} dv - \frac{1}{2} \int \frac{2v}{1+v^2} dv = \int \frac{dx}{x}$.
Integrating both sides: $\tan^{-1}(v) - \frac{1}{2} \log |1+v^2| = \log |x| + c$.
Substituting $v = \frac{y}{x}$: $\tan^{-1}(\frac{y}{x}) - \frac{1}{2} \log |1 + \frac{y^2}{x^2}| = \log |x| + c$.
$\tan^{-1}(\frac{y}{x}) - \frac{1}{2} \log |\frac{x^2+y^2}{x^2}| = \log |x| + c$.
$\tan^{-1}(\frac{y}{x}) - \frac{1}{2} (\log |x^2+y^2| - \log |x^2|) = \log |x| + c$.
$\tan^{-1}(\frac{y}{x}) - \frac{1}{2} \log |x^2+y^2| + \log |x| = \log |x| + c$.
Thus,$\tan^{-1}(\frac{y}{x}) - \frac{1}{2} \log |x^2+y^2| = c$.

(A) Given the differential equation $x^{2} \frac{dy}{dx} = y^{2} + xy$.
Dividing by $x^{2}$,we get $\frac{dy}{dx} = \frac{y^{2} + xy}{x^{2}} = \left(\frac{y}{x}\right)^{2} + \frac{y}{x}$.
This is a homogeneous differential equation. Let $y = ux$,then $\frac{dy}{dx} = u + x \frac{du}{dx}$.
Substituting these into the equation: $u + x \frac{du}{dx} = u^{2} + u$.
Subtracting $u$ from both sides: $x \frac{du}{dx} = u^{2}$.
Separating the variables: $\frac{du}{u^{2}} = \frac{dx}{x}$.
Integrating both sides: $\int u^{-2} du = \int \frac{1}{x} dx$.
$-u^{-1} = \log |x| + c$.
Since $u = \frac{y}{x}$,we have $-\frac{x}{y} = \log |x| + c$.
Rearranging the terms: $\frac{x}{y} + \log |x| = -c$,which can be written as $\frac{x}{y} + \log |x| = C$ (where $C = -c$).

(D) Given the slope of the tangent is $\frac{dy}{dx} = 1 + \frac{y}{x} + \left(\frac{y}{x}\right)^2$.
This is a homogeneous differential equation.
Let $y = ux$,then $\frac{dy}{dx} = u + x\frac{du}{dx}$.
Substituting this into the equation:
$u + x\frac{du}{dx} = 1 + u + u^2$.
Subtracting $u$ from both sides:
$x\frac{du}{dx} = 1 + u^2$.
Separating the variables:
$\int \frac{du}{1 + u^2} = \int \frac{dx}{x}$.
Integrating both sides:
$\tan^{-1}(u) = \log|x| + C$.
Substituting $u = \frac{y}{x}$:
$\tan^{-1}\left(\frac{y}{x}\right) = \log|x| + C$.
Since the curve passes through $(1,0)$,we have $\tan^{-1}(0) = \log|1| + C$,which gives $0 = 0 + C$,so $C = 0$.
Thus,the equation is $\tan^{-1}\left(\frac{y}{x}\right) = \log|x|$.

(C) Given the differential equation: $x \sin \left(\frac{y}{x}\right) dy = \left[y \sin \left(\frac{y}{x}\right) - x\right] dx$
Dividing both sides by $dx \cdot x \sin \left(\frac{y}{x}\right)$,we get:
$\frac{dy}{dx} = \frac{y \sin \left(\frac{y}{x}\right) - x}{x \sin \left(\frac{y}{x}\right)} = \frac{y}{x} - \frac{1}{\sin \left(\frac{y}{x}\right)}$
Let $v = \frac{y}{x}$,so $y = vx$. Then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation:
$v + x \frac{dv}{dx} = v - \frac{1}{\sin v}$
$x \frac{dv}{dx} = -\frac{1}{\sin v} = -\operatorname{cosec} v$
Separating the variables:
$\sin v \, dv = -\frac{1}{x} dx$
Integrating both sides:
$\int \sin v \, dv = -\int \frac{1}{x} dx$
$-\cos v = -\log |x| + c$
$\cos v = \log |x| + c$
Substituting $v = \frac{y}{x}$ back:
$\cos \left(\frac{y}{x}\right) = \log |x| + c$

(A) The given differential equation is $(3xy + y^2)dx + (x^2 + xy)dy = 0$.
Rearranging the terms,we get $\frac{dy}{dx} = -\frac{3xy + y^2}{x^2 + xy}$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Substituting these into the equation: $v + x\frac{dv}{dx} = -\frac{3x^2v + x^2v^2}{x^2 + x^2v} = -\frac{3v + v^2}{1 + v}$.
$x\frac{dv}{dx} = -\frac{3v + v^2}{1 + v} - v = -\frac{3v + v^2 + v + v^2}{1 + v} = -\frac{2v^2 + 4v}{1 + v} = -\frac{2v(v + 2)}{v + 1}$.
Separating the variables: $\frac{v + 1}{v(v + 2)} dv = -\frac{2}{x} dx$.
Using partial fractions: $\frac{v + 1}{v(v + 2)} = \frac{A}{v} + \frac{B}{v + 2} \Rightarrow v + 1 = A(v + 2) + Bv$.
For $v = 0$,$1 = 2A \Rightarrow A = 1/2$. For $v = -2$,$-1 = -2B \Rightarrow B = 1/2$.
So,$\int (\frac{1}{2v} + \frac{1}{2(v + 2)}) dv = -\int \frac{2}{x} dx$.
$\frac{1}{2} \ln|v| + \frac{1}{2} \ln|v + 2| = -2 \ln|x| + C'$.
$\ln|v(v + 2)| = -4 \ln|x| + 2C' \Rightarrow \ln|v(v + 2)x^4| = C''$.
$v(v + 2)x^4 = c^2$.
Substituting $v = y/x$: $\frac{y}{x}(\frac{y}{x} + 2)x^4 = c^2 \Rightarrow y(y + 2x)x^2 = c^2 \Rightarrow x^2(y^2 + 2xy) = c^2$.

(A) Given the differential equation: $(x^2+y^2) dy = xy dx$.
Rearranging,we get: $\frac{dy}{dx} = \frac{xy}{x^2+y^2}$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting into the equation: $v + x \frac{dv}{dx} = \frac{x(vx)}{x^2+(vx)^2} = \frac{vx^2}{x^2(1+v^2)} = \frac{v}{1+v^2}$.
$x \frac{dv}{dx} = \frac{v}{1+v^2} - v = \frac{v - v - v^3}{1+v^2} = -\frac{v^3}{1+v^2}$.
Separating variables: $\int \frac{1+v^2}{v^3} dv = -\int \frac{dx}{x}$.
$\int (v^{-3} + v^{-1}) dv = -\ln|x| + C$.
$-\frac{1}{2v^2} + \ln|v| = -\ln|x| + C$.
$-\frac{x^2}{2y^2} + \ln|y/x| = -\ln|x| + C$.
$-\frac{x^2}{2y^2} + \ln|y| - \ln|x| = -\ln|x| + C$.
$-\frac{x^2}{2y^2} + \ln|y| = C$.
Using $y(1) = 1$: $-\frac{1^2}{2(1)^2} + \ln(1) = C \Rightarrow C = -\frac{1}{2}$.
So,$-\frac{x^2}{2y^2} + \ln|y| = -\frac{1}{2}$.
For $y(x_0) = e$: $-\frac{x_0^2}{2e^2} + \ln(e) = -\frac{1}{2}$.
$-\frac{x_0^2}{2e^2} + 1 = -\frac{1}{2} \Rightarrow \frac{x_0^2}{2e^2} = \frac{3}{2}$.
$x_0^2 = 3e^2 \Rightarrow x_0 = \sqrt{3}e$.

(D) function $f(x, y)$ is said to be a homogeneous function of degree $n$ if $f(\lambda x, \lambda y) = \lambda^n f(x, y)$.
$1$. For $f(x, y) = y^2+2xy$,$f(\lambda x, \lambda y) = (\lambda y)^2 + 2(\lambda x)(\lambda y) = \lambda^2(y^2+2xy) = \lambda^2 f(x, y)$. This is homogeneous of degree $2$.
$2$. For $f(x, y) = 2x-3y$,$f(\lambda x, \lambda y) = 2(\lambda x) - 3(\lambda y) = \lambda(2x-3y) = \lambda^1 f(x, y)$. This is homogeneous of degree $1$.
$3$. For $f(x, y) = \sin\left(\frac{y}{x}\right)$,$f(\lambda x, \lambda y) = \sin\left(\frac{\lambda y}{\lambda x}\right) = \sin\left(\frac{y}{x}\right) = \lambda^0 f(x, y)$. This is homogeneous of degree $0$.
$4$. For $f(x, y) = \cos x + \sin y$,$f(\lambda x, \lambda y) = \cos(\lambda x) + \sin(\lambda y)$. This cannot be expressed in the form $\lambda^n f(x, y)$ for any $n$.
Therefore,$\cos x + \sin y$ is not a homogeneous function.

(B) Let $S$ be the sample space of $50$ tickets: $S = \{00, 01, 02, \ldots, 49\}$,so $n(S) = 50$.
Let $E_1$ be the event that the sum of the digits is $8$: $E_1 = \{08, 17, 26, 35, 44\}$,so $n(E_1) = 5$.
Let $E_2$ be the event that the product of the digits is zero. This occurs if at least one digit is $0$: $E_2 = \{00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40\}$,so $n(E_2) = 14$.
The intersection $E_1 \cap E_2$ is the set of tickets where the sum is $8$ and the product is $0$: $E_1 \cap E_2 = \{08\}$,so $n(E_1 \cap E_2) = 1$.
The conditional probability is $P(E_1 | E_2) = \frac{n(E_1 \cap E_2)}{n(E_2)} = \frac{1}{14}$.

(D) The prime numbers between $1$ and $19$ are $2, 3, 5, 7, 11, 13, 17, 19$. There are $8$ such numbers.
Since the first guest is given a room with a prime number,there are $8 - 1 = 7$ prime-numbered rooms remaining.
The total number of rooms remaining for the second guest is $19 - 1 = 18$.
Therefore,the probability that the second guest is given a room with a prime number is $\frac{7}{18}$.

(B) Given $P(A') = 0.6$,we have $P(A) = 1 - 0.6 = 0.4$.
We are given $P(B) = 0.8$ and $P(B/A) = 0.3$.
Using the multiplication theorem,$P(A \cap B) = P(A) \cdot P(B/A) = 0.4 \times 0.3 = 0.12$.
We know that $P(A/B) = \frac{P(A \cap B)}{P(B)}$.
Substituting the values,$P(A/B) = \frac{0.12}{0.8} = \frac{12}{80} = \frac{3}{20}$.

(A) The total number of balls in the bag is $5 + 3 = 8$.
We need to find the probability of selecting one red ball and one green ball in two draws without replacement.
This can happen in two ways: (Red then Green) or (Green then Red).
$\text{Required probability} = P(R_1 \cap G_2) + P(G_1 \cap R_2)$
$= P(R_1) \cdot P(G_2 | R_1) + P(G_1) \cdot P(R_2 | G_1)$
$= \left(\frac{5}{8} \cdot \frac{3}{7}\right) + \left(\frac{3}{8} \cdot \frac{5}{7}\right)$
$= \frac{15}{56} + \frac{15}{56}$
$= \frac{30}{56} = \frac{15}{28}$

(C) The given differential equation is $\left(1+e^{\frac{x}{y}}\right) dx + e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) dy = 0$.
Rearranging the terms,we get $\frac{dx}{dy} = -\frac{e^{\frac{x}{y}}\left(1-\frac{x}{y}\right)}{1+e^{\frac{x}{y}}}$.
Since the equation involves the term $\frac{x}{y}$,it is a homogeneous differential equation of the form $\frac{dx}{dy} = F\left(\frac{x}{y}\right)$.
To solve such an equation,we substitute $x = vy$,where $v$ is a function of $y$.
Therefore,the correct substitution is $x = vy$.

(B) By the definition of conditional probability,$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.
Given $P(A \mid B) = 1$,we have $\frac{P(A \cap B)}{P(B)} = 1$,which implies $P(A \cap B) = P(B)$.
Since $A \cap B$ is a subset of $B$ $(A \cap B \subseteq B)$,and their probabilities are equal,it follows that all outcomes in $B$ must be contained within $A$.
Therefore,$B \subseteq A$ (or $B \subset A$ in the context of the given options).

(B) Given: $3 P(A) = \frac{5}{13} \implies P(A) = \frac{5}{39}$.
$P(B) = \frac{5}{13} = \frac{15}{39}$.
$P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{3}{5}$.
Therefore,$P(A \cap B) = P(A \mid B) \times P(B) = \frac{3}{5} \times \frac{5}{13} = \frac{3}{13} = \frac{9}{39}$.
We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = \frac{5}{39} + \frac{15}{39} - \frac{9}{39} = \frac{11}{39}$.

(A) Given that $2 P(A) = \frac{5}{13} \implies P(A) = \frac{5}{26}$.
Given that $P(B) = \frac{5}{13}$.
Given that $P(A \mid B) = \frac{2}{5}$.
We know that $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.
Therefore,$P(A \cap B) = P(A \mid B) \times P(B) = \frac{2}{5} \times \frac{5}{13} = \frac{2}{13}$.
We use the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the values: $P(A \cup B) = \frac{5}{26} + \frac{5}{13} - \frac{2}{13}$.
$P(A \cup B) = \frac{5}{26} + \frac{10}{26} - \frac{4}{26} = \frac{11}{26}$.

(A) Given the differential equation: $\frac{dt}{dx} = \frac{t}{x + t e^{-2x/t}}$.
Taking the reciprocal of both sides,we get:
$\frac{dx}{dt} = \frac{x + t e^{-2x/t}}{t}$.
Dividing each term in the numerator by $t$,we obtain:
$\frac{dx}{dt} = \frac{x}{t} + \frac{t e^{-2x/t}}{t}$.
Simplifying the expression,we get:
$\frac{dx}{dt} = \frac{x}{t} + e^{-2(x/t)}$.
This is in the form $\frac{dx}{dt} = \phi\left(\frac{x}{t}\right)$,where $\phi(v) = v + e^{-2v}$.
Thus,the correct option is $A$.

(A) Given the differential equation $x dy - y dx = \sqrt{x^2 + y^2} dx$.
Rearranging the terms,we get $\frac{dy}{dx} = \frac{y + \sqrt{x^2 + y^2}}{x}$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{vx + \sqrt{x^2 + v^2x^2}}{x} = v + \sqrt{1 + v^2}$.
Simplifying,we have $x \frac{dv}{dx} = \sqrt{1 + v^2}$.
Separating the variables: $\int \frac{dv}{\sqrt{1 + v^2}} = \int \frac{dx}{x}$.
Integrating both sides: $\ln(v + \sqrt{v^2 + 1}) = \ln|x| + C = \ln|cx|$.
Thus,$v + \sqrt{v^2 + 1} = cx$.
Substituting $v = \frac{y}{x}$: $\frac{y}{x} + \sqrt{\frac{y^2}{x^2} + 1} = cx \Rightarrow y + \sqrt{x^2 + y^2} = cx^2$.
Given $y(\sqrt{3}) = 1$,we have $1 + \sqrt{3 + 1} = c(\sqrt{3})^2 \Rightarrow 1 + 2 = 3c \Rightarrow c = 1$.
Therefore,the solution is $y + \sqrt{x^2 + y^2} = x^2$.

(C) Given the differential equation: $(x+y) y dx + (y-x) x dy = 0$.
Rearranging the terms: $(y-x) x dy = -(x+y) y dx$,which gives $\frac{dy}{dx} = \frac{(x+y) y}{(x-y) x}$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{(x+vx) vx}{(x-vx) x} = \frac{v(1+v)}{1-v} = \frac{v+v^2}{1-v}$.
$x \frac{dv}{dx} = \frac{v+v^2}{1-v} - v = \frac{v+v^2-v+v^2}{1-v} = \frac{2v^2}{1-v}$.
Separating the variables: $\int \frac{1-v}{v^2} dv = \int \frac{2}{x} dx$.
Integrating both sides: $\int (v^{-2} - v^{-1}) dv = 2 \int \frac{1}{x} dx$.
$-v^{-1} - \ln|v| = 2 \ln|x| + C$.
Substituting $v = \frac{y}{x}$: $-\frac{x}{y} - \ln(\frac{y}{x}) = 2 \ln|x| + C$.
$-\frac{x}{y} - \ln|y| + \ln|x| = 2 \ln|x| + C$.
$-\frac{x}{y} = \ln|y| + \ln|x| + C = \ln|xy| + C$.
Multiplying by $-y$: $x = -y \ln|xy| - yC$.
$x + y(\ln|xy| + C) = 0$.
$x + y \ln|cxy| = 0$,where $C = \ln|c|$.

(B) Given differential equation: $(x \sin \frac{y}{x}) dy = (y \sin \frac{y}{x} - x) dx$
Rearranging the terms: $\frac{dy}{dx} = \frac{y \sin \frac{y}{x} - x}{x \sin \frac{y}{x}} = \frac{y}{x} - \frac{1}{\sin \frac{y}{x}} = \frac{y}{x} - \operatorname{cosec} \frac{y}{x}$
Let $\frac{y}{x} = v$,then $y = vx$,so $\frac{dy}{dx} = v + x \frac{dv}{dx}$
Substituting these into the equation: $v + x \frac{dv}{dx} = v - \operatorname{cosec} v$
$x \frac{dv}{dx} = -\operatorname{cosec} v$
Separating the variables: $\frac{dv}{\operatorname{cosec} v} = -\frac{dx}{x} \Rightarrow \sin v dv = -\frac{dx}{x}$
Integrating both sides: $\int \sin v dv = -\int \frac{1}{x} dx$
$-\cos v = -\log_e |x| + C$
$\cos v = \log_e |x| + C'$
Substituting $v = \frac{y}{x}$ back: $\cos \frac{y}{x} = \log_e x + C$

(B) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} + \tan(\frac{y}{x})$.
Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation:
$v + x \frac{dv}{dx} = v + \tan(v)$.
Subtracting $v$ from both sides:
$x \frac{dv}{dx} = \tan(v)$.
Separating the variables:
$\int \cot(v) dv = \int \frac{1}{x} dx$.
Integrating both sides:
$\ln|\sin(v)| = \ln|x| + \ln|C|$.
$\ln|\sin(v)| = \ln|Cx|$.
Taking the exponential of both sides:
$\sin(v) = Cx$.
Substituting $v = \frac{y}{x}$ back:
$\sin(\frac{y}{x}) = Cx$.

(A) Given differential equation is $x dy - y dx = \sqrt{x^2 + y^2} dx$.
Dividing both sides by $x dx$ (assuming $x \neq 0$),we get:
$\frac{dy}{dx} = \frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}}$.
Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation:
$v + x \frac{dv}{dx} = v + \sqrt{1 + v^2}$.
Subtracting $v$ from both sides:
$x \frac{dv}{dx} = \sqrt{1 + v^2}$.
Separating the variables:
$\frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x}$.
Integrating both sides:
$\int \frac{dv}{\sqrt{1 + v^2}} = \int \frac{dx}{x}$.
$\log |v + \sqrt{1 + v^2}| = \log |x| + \log |c|$.
$v + \sqrt{1 + v^2} = cx$.
Substituting $v = \frac{y}{x}$ back:
$\frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}} = cx$.
$\frac{y + \sqrt{x^2 + y^2}}{x} = cx$.
$y + \sqrt{x^2 + y^2} = cx^2$.

(B) Given differential equation: $(xy + y^2) dx - (x^2 - 2xy) dy = 0$
Rearranging the terms: $\frac{dy}{dx} = \frac{xy + y^2}{x^2 - 2xy}$
Divide numerator and denominator by $x^2$: $\frac{dy}{dx} = \frac{\frac{y}{x} + (\frac{y}{x})^2}{1 - 2(\frac{y}{x})}$
Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Substituting into the equation: $v + x\frac{dv}{dx} = \frac{v + v^2}{1 - 2v}$
$x\frac{dv}{dx} = \frac{v + v^2}{1 - 2v} - v = \frac{v + v^2 - v + 2v^2}{1 - 2v} = \frac{3v^2}{1 - 2v}$
Separating variables: $(\frac{1 - 2v}{3v^2}) dv = \frac{dx}{x} \Rightarrow (\frac{1}{3v^2} - \frac{2}{3v}) dv = \frac{dx}{x}$
Integrating both sides: $\int (\frac{1}{3}v^{-2} - \frac{2}{3v}) dv = \int \frac{1}{x} dx$
$-\frac{1}{3v} - \frac{2}{3} \ln|v| = \ln|x| + C_1$
Substitute $v = \frac{y}{x}$: $-\frac{x}{3y} - \frac{2}{3} \ln(\frac{y}{x}) = \ln|x| + C_1$
Multiply by $3$: $-\frac{x}{y} - 2(\ln y - \ln x) = 3\ln x + 3C_1$
$-\frac{x}{y} = 3\ln x + 2\ln y - 2\ln x + C_2 = \ln x + 2\ln y + C_2 = \ln(xy^2) + C_2$
$-\frac{x}{y} = \ln(Cxy^2) \Rightarrow e^{-\frac{x}{y}} = Cxy^2$
Therefore,$Cxy^2 e^{\frac{x}{y}} = 1$.

(C) differential equation of the form $\frac{dy}{dx} = f(x, y)$ is said to be a homogeneous differential equation if the function $f(x, y)$ is a homogeneous function of degree zero.
By definition,a function $f(x, y)$ is homogeneous of degree zero if $f(\lambda x, \lambda y) = \lambda^0 f(x, y) = f(x, y)$.
Such a function can always be expressed in the form $f(x, y) = \phi\left(\frac{y}{x}\right)$ or $f(x, y) = \psi\left(\frac{x}{y}\right)$.
Therefore,the general form of $f(x, y)$ for a homogeneous differential equation $\frac{dy}{dx} = f(x, y)$ is $\phi\left(\frac{y}{x}\right)$.

(B) Given the differential equation: $\left(x \sin \frac{y}{x}\right) dy = \left(y \sin \frac{y}{x} - x\right) dx$.
Rearranging the terms,we get: $\frac{dy}{dx} = \frac{y \sin(y/x) - x}{x \sin(y/x)}$.
Since this is a homogeneous differential equation,let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{vx \sin v - x}{x \sin v} = \frac{v \sin v - 1}{\sin v} = v - \frac{1}{\sin v}$.
Subtracting $v$ from both sides: $x \frac{dv}{dx} = -\frac{1}{\sin v}$.
Separating the variables: $\sin v \, dv = -\frac{1}{x} dx$.
Integrating both sides: $\int \sin v \, dv = -\int \frac{1}{x} dx$.
$-\cos v = -\log |x| + C$.
Rearranging gives: $\log |x| - \cos v = C$.
Substituting $v = \frac{y}{x}$ back,we get: $\log |x| - \cos \frac{y}{x} = C$.

(A) Given the differential equation $\frac{dy}{dx} = -\frac{x+y+1}{x-3y+5}$.
Let $x = X+h$ and $y = Y+k$. We choose $h, k$ such that $h+k+1 = 0$ and $h-3k+5 = 0$.
Solving these,we get $h = -2$ and $k = 1$. So,$x = X-2$ and $y = Y+1$.
The equation becomes $\frac{dY}{dX} = -\frac{X+Y}{X-3Y}$.
This is a homogeneous equation. Let $Y = vX$,then $\frac{dY}{dX} = v + X\frac{dv}{dX}$.
$v + X\frac{dv}{dX} = -\frac{1+v}{1-3v} = \frac{1+v}{3v-1}$.
$X\frac{dv}{dX} = \frac{1+v}{3v-1} - v = \frac{1+v-3v^2+v}{3v-1} = \frac{-3v^2+2v+1}{3v-1}$.
Separating variables: $\int \frac{3v-1}{-3v^2+2v+1} dv = \int \frac{1}{X} dX$.
Integrating,we get $-\frac{1}{2} \ln| -3v^2+2v+1 | = \ln|X| + C$.
This simplifies to $-3v^2+2v+1 = \frac{c}{X^2}$.
Substituting $v = \frac{Y}{X} = \frac{y-1}{x+2}$:
$-3(\frac{y-1}{x+2})^2 + 2(\frac{y-1}{x+2}) + 1 = \frac{c}{(x+2)^2}$.
$-3(y-1)^2 + 2(y-1)(x+2) + (x+2)^2 = c$.
Rearranging gives $3(y-1)^2 - 2(x+2)(y-1) - (x+2)^2 = c$.

(B) The given differential equation is $\frac{dy}{dx} = \frac{x+y}{x-y}$. This is a homogeneous differential equation.
Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Substituting this into the equation: $v + x\frac{dv}{dx} = \frac{x+vx}{x-vx} = \frac{1+v}{1-v}$.
$x\frac{dv}{dx} = \frac{1+v}{1-v} - v = \frac{1+v-v+v^2}{1-v} = \frac{1+v^2}{1-v}$.
Separating variables: $\frac{1-v}{1+v^2} dv = \frac{dx}{x}$.
Integrating both sides: $\int \frac{1}{1+v^2} dv - \int \frac{v}{1+v^2} dv = \int \frac{dx}{x}$.
$\tan^{-1}(v) - \frac{1}{2} \log(1+v^2) = \log|x| + C$.
Substituting $v = \frac{y}{x}$: $\tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2} \log\left(1+\frac{y^2}{x^2}\right) = \log|x| + C$.
$\tan^{-1}\left(\frac{y}{x}\right) = \frac{1}{2} \log\left(\frac{x^2+y^2}{x^2}\right) + \log|x| + C = \frac{1}{2} \log(x^2+y^2) - \log|x| + \log|x| + C$.
$\tan^{-1}\left(\frac{y}{x}\right) = \log\left(\sqrt{x^2+y^2}\right) + C$.
Thus,$\tan^{-1}\left(\frac{y}{x}\right) = \log\left(c\sqrt{x^2+y^2}\right)$.

(A) Given the differential equation $\frac{dy}{dx} = \frac{2x+y-3}{2y-x+3}$.
Let $x = X+h$ and $y = Y+k$. Then $\frac{dy}{dx} = \frac{dY}{dX}$.
We solve the system $2h+k-3=0$ and $-h+2k+3=0$.
Multiplying the second by $2$: $-2h+4k+6=0$. Adding to the first: $5k+3=0 \implies k = -3/5$. Then $2h = 3 - (-3/5) = 18/5 \implies h = 9/5$.
The equation becomes $\frac{dY}{dX} = \frac{2X+Y}{2Y-X}$.
Let $Y = vX$,then $\frac{dY}{dX} = v + X\frac{dv}{dX}$.
$v + X\frac{dv}{dX} = \frac{2+v}{2v-1} \implies X\frac{dv}{dX} = \frac{2+v-2v^2+v}{2v-1} = \frac{-2v^2+2v+2}{2v-1}$.
Separating variables: $\int \frac{2v-1}{-2v^2+2v+2} dv = \int \frac{dX}{X}$.
Let $u = -2v^2+2v+2$,then $du = (-4v+2) dv = -2(2v-1) dv$.
So,$-\frac{1}{2} \int \frac{du}{u} = \ln|X| + C \implies -\frac{1}{2} \ln|-2v^2+2v+2| = \ln|X| + C$.
$\ln|-2(Y/X)^2+2(Y/X)+2| = -2\ln|X| + C' \implies -2Y^2+2YX+2X^2 = C''$.
Substituting $X=x-9/5$ and $Y=y+3/5$,we simplify to $x^2-xy-y^2+3x+3y+c=0$.

(A) Given the differential equation: $(x \sin \frac{y}{x}) dy = (y \sin \frac{y}{x} - x) dx$.
Rearranging,we get: $\frac{dy}{dx} = \frac{y \sin (y/x) - x}{x \sin (y/x)}$.
Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting into the equation: $v + x \frac{dv}{dx} = \frac{vx \sin v - x}{x \sin v} = \frac{v \sin v - 1}{\sin v} = v - \frac{1}{\sin v}$.
Subtracting $v$ from both sides: $x \frac{dv}{dx} = -\frac{1}{\sin v}$.
Separating variables: $\sin v \, dv = -\frac{1}{x} \, dx$.
Integrating both sides: $\int \sin v \, dv = -\int \frac{1}{x} \, dx$.
$-\cos v = -\log |x| + C_1$,which simplifies to $\cos v = \log |x| + c$.
Substituting $v = y/x$ back,we get: $\cos (\frac{y}{x}) = \log |x| + c$.

(D) The substitution $x=vy$ is used to solve homogeneous differential equations of the form $\frac{dx}{dy} = f(\frac{x}{y})$ or $\frac{dy}{dx} = g(\frac{x}{y})$.
For the substitution $x=vy$ (which implies $\frac{x}{y}=v$),the differential equation must be homogeneous in $x$ and $y$ such that $\frac{dy}{dx}$ can be expressed as a function of $\frac{x}{y}$.
Checking option $(d)$:
$(1+2e^{\frac{x}{y}}) + 2e^{\frac{x}{y}}(1-\frac{x}{y})\frac{dy}{dx} = 0$
$\frac{dy}{dx} = -\frac{1+2e^{\frac{x}{y}}}{2e^{\frac{x}{y}}(1-\frac{x}{y})}$
Since the right-hand side is a function of $\frac{x}{y}$,it is a homogeneous differential equation of degree zero,which becomes separable after substituting $x=vy$.

(A) The given differential equation is $x^\alpha \frac{dy}{dx} = y^\beta(\gamma \log x + \delta \log y + 1)$.
For the equation to be homogeneous,the degree of the numerator and denominator must be the same,or the function must be expressible as a function of $\frac{y}{x}$.
Rewriting the equation: $\frac{dy}{dx} = \frac{y^\beta}{x^\alpha} (\gamma \log x + \delta \log y + 1)$.
For homogeneity,the powers of $x$ and $y$ must balance such that the expression depends only on the ratio $\frac{y}{x}$.
This requires $\alpha = \beta$.
Substituting $\alpha = \beta$,we get $\frac{dy}{dx} = (\frac{y}{x})^\alpha (\gamma \log x + \delta \log y + 1) = (\frac{y}{x})^\alpha (\gamma \log x + \delta \log y + \delta \log x - \delta \log x + 1)$.
For the term to be a function of $\frac{y}{x}$,the $\log x$ terms must cancel out,which happens when $\gamma + \delta = 0$,i.e.,$\gamma = -\delta$.

(C) differential equation of the form $\frac{dy}{dx} = F(x, y)$ is homogeneous if $F(x, y)$ is a homogeneous function of degree $0$. Equivalently,for an equation $M(x, y)dx + N(x, y)dy = 0$,it is homogeneous if both $M(x, y)$ and $N(x, y)$ are homogeneous functions of the same degree.
In option $C$,we have $(x^2 + y^2)dx = 2xy dy$.
Here,$M(x, y) = x^2 + y^2$,which is a homogeneous function of degree $2$.
And $N(x, y) = 2xy$,which is also a homogeneous function of degree $2$.
Since both $M$ and $N$ are homogeneous functions of the same degree,the differential equation is homogeneous.

(B) Given the differential equation: $\frac{dy}{dx} = \frac{2x + 3y - 7}{3x + 2y - 8}$ ... $(i)$
To transform this into a homogeneous differential equation,we substitute $x = X - h$ and $y = Y - k$.
Substituting these into $(i)$,we get: $\frac{dY}{dX} = \frac{2(X - h) + 3(Y - k) - 7}{3(X - h) + 2(Y - k) - 8} = \frac{2X + 3Y - (2h + 3k + 7)}{3X + 2Y - (3h + 2k + 8)}$.
For the equation to be homogeneous,the constant terms in the numerator and denominator must be zero:
$2h + 3k + 7 = 0$ ... (ii)
$3h + 2k + 8 = 0$ ... (iii)
Multiplying (ii) by $3$ and (iii) by $2$:
$6h + 9k + 21 = 0$
$6h + 4k + 16 = 0$
Subtracting the equations: $(9k - 4k) + (21 - 16) = 0 \Rightarrow 5k + 5 = 0 \Rightarrow k = -1$.
Wait,re-evaluating the signs: The original equation is $\frac{2x+3y-7}{3x+2y-8}$.
Setting $2h+3k-7=0$ and $3h+2k-8=0$:
$4h+6k-14=0$
$9h+6k-24=0$
Subtracting: $5h - 10 = 0 \Rightarrow h = 2$.
Substituting $h=2$ into $2(2)+3k-7=0 \Rightarrow 4+3k-7=0 \Rightarrow 3k=3 \Rightarrow k=1$.
Thus,$(h, k) = (2, 1)$.

(B) Given the differential equation: $(x \sin \frac{y}{x}) dy = (y \sin \frac{y}{x} - x) dx$
Rearranging the terms to find $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{y \sin(y/x) - x}{x \sin(y/x)} = \frac{y}{x} - \frac{1}{\sin(y/x)}$
Let $v = \frac{y}{x}$,then $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$
Substituting these into the equation:
$v + x \frac{dv}{dx} = v - \frac{1}{\sin v}$
$x \frac{dv}{dx} = -\frac{1}{\sin v}$
Separating the variables:
$-\sin v \, dv = \frac{1}{x} \, dx$
Integrating both sides:
$-\int \sin v \, dv = \int \frac{1}{x} \, dx$
$\cos v = \log x + C$
Substituting $v = \frac{y}{x}$ back:
$\cos \frac{y}{x} = \log x + C$
Comparing this with the given solution $\cos \frac{y}{x} = A \log x + C$,we find $A = 1$.

(D) Given the differential equation $\frac{dy}{dx} = \frac{y^2}{xy - y^2 - x^2}$.
Substitute $y = mx$,then $\frac{dy}{dx} = m + x \frac{dm}{dx}$.
$m + x \frac{dm}{dx} = \frac{m^2 x^2}{x(mx) - m^2 x^2 - x^2} = \frac{m^2}{m - m^2 - 1}$.
$x \frac{dm}{dx} = \frac{m^2}{m - m^2 - 1} - m = \frac{m^2 - m^2 + m^3 + m}{m - m^2 - 1} = \frac{m^3 + m}{m - m^2 - 1}$.
Separating variables: $\int \frac{m - m^2 - 1}{m^3 + m} dm = \int \frac{dx}{x}$.
$int \left( \frac{m}{m(m^2 + 1)} - \frac{m^2 + 1}{m(m^2 + 1)} \right) dm = \ln|x| + C$.
$int \left( \frac{1}{m^2 + 1} - \frac{1}{m} \right) dm = \ln|x| + C$.
$\tan^{-1}(m) - \ln|m| = \ln|x| + C$.
Substitute $m = \frac{y}{x}$: $\tan^{-1}\left(\frac{y}{x}\right) - \ln\left|\frac{y}{x}\right| = \ln|x| + C$.
$\tan^{-1}\left(\frac{y}{x}\right) - (\ln|y| - \ln|x|) = \ln|x| + C$.
$\tan^{-1}\left(\frac{y}{x}\right) = \ln|y| + C$.
Comparing with $\tan^{-1}\left(\frac{y}{x}\right) = f(y) + C$,we get $f(y) = \ln|y|$.
Therefore,$f(e^3) = \ln|e^3| = 3$.

(D) Given the differential equation $\frac{dy}{dx} = \frac{f(x, y)}{g(x, y)}$.
Since $f$ and $g$ are homogeneous functions of the same order,we can write $\frac{dy}{dx} = \frac{f(Vy, y)}{g(Vy, y)} = \frac{y^n f(V, 1)}{y^n g(V, 1)} = \frac{f(V, 1)}{g(V, 1)}$.
We are given the substitution $x = Vy$. Differentiating with respect to $y$,we get $\frac{dx}{dy} = V + y\frac{dV}{dy}$.
Thus,$\frac{dV}{dy} = \frac{1}{y} \left( \frac{dx}{dy} - V \right)$.
Since $\frac{dy}{dx} = \frac{f(x, y)}{g(x, y)}$,we have $\frac{dx}{dy} = \frac{g(x, y)}{f(x, y)} = \frac{g(Vy, y)}{f(Vy, y)} = \frac{g(V, 1)}{f(V, 1)}$.
Substituting this into the expression for $\frac{dV}{dy}$,we get $\frac{dV}{dy} = \frac{1}{y} \left( \frac{g(V, 1)}{f(V, 1)} - V \right)$.
Comparing this with $\frac{dV}{dy} = \frac{1}{y}(F(V))$,we find $F(V) = \left( \frac{g(V, 1)}{f(V, 1)} - V \right)$.

(C) Given the differential equation: $\{x \cos (y/x) + y \sin (y/x)\} y dx = \{y \sin (y/x) - x \cos (y/x)\} x dy$.
Rearranging the terms,we get: $\frac{dy}{dx} = \frac{y \{x \cos (y/x) + y \sin (y/x)\}}{x \{y \sin (y/x) - x \cos (y/x)\}} = \frac{(y/x) \cos (y/x) + (y/x)^2 \sin (y/x)}{(y/x) \sin (y/x) - \cos (y/x)}$.
Let $v = y/x$,so $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{v \cos v + v^2 \sin v}{v \sin v - \cos v}$.
$x \frac{dv}{dx} = \frac{v \cos v + v^2 \sin v - v(v \sin v - \cos v)}{v \sin v - \cos v} = \frac{v \cos v + v^2 \sin v - v^2 \sin v + v \cos v}{v \sin v - \cos v} = \frac{2v \cos v}{v \sin v - \cos v}$.
Separating the variables: $\frac{v \sin v - \cos v}{v \cos v} dv = 2 \frac{dx}{x}$.
Integrating both sides: $\int (\tan v - \frac{1}{v}) dv = 2 \int \frac{1}{x} dx$.
$-\ln |\cos v| - \ln |v| = 2 \ln |x| + C$.
$-\ln |v \cos v| = \ln |x^2| + C$.
$-\ln |(y/x) \cos (y/x)| = \ln |x^2| + C$.
$\ln |(y/x) \cos (y/x)| + \ln |x^2| = -C$.
$\ln |xy \cos (y/x)| = -C$.
$xy \cos (y/x) = \pm e^{-C}$.