Q.Show that the differential equation $(x - y)\frac{dy}{dx} = x + 2y$ is homogeneous and solve it.

This is a homogeneous differential equation; substituting $y=vx$ and separating variables gives $x\tan(y/2x) = C$.

$x\frac{dy}{dx} - y + x\sin\!\left(\frac{y}{x}\right) = 0$

Step 1 — rewrite in the form $\dfrac{dy}{dx} =$ function of $y/x$.

$x\frac{dy}{dx} = y - x\sin\!\left(\frac{y}{x}\right) \implies \frac{dy}{dx} = \frac{y}{x} - \sin\!\left(\frac{y}{x}\right)$

This is homogeneous (the right side depends only on $y/x$).

Step 2 — substitute $y = vx$, so $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$:

$v + x\frac{dv}{dx} = v - \sin v \implies x\frac{dv}{dx} = -\sin v$

Step 3 — separate variables.

$\frac{dv}{\sin v} = -\frac{dx}{x} \implies \int \operatorname{cosec} v\,dv = -\int \frac{dx}{x}$

Step 4 — integrate both sides (using $\displaystyle\int \operatorname{cosec} v\,dv = \ln\left|\tan\frac{v}{2}\right| + C_1$):

$\ln\left|\tan\frac{v}{2}\right| = -\ln|x| + C_1$

$\ln\left|\tan\frac{v}{2}\right| + \ln|x| = C_1 \implies \ln\left|x\tan\frac{v}{2}\right| = C_1$

$x\tan\frac{v}{2} = C \quad (\text{where } C = \pm e^{C_1})$

Step 5 — back-substitute $v = y/x$.

$x\tan\left(\frac{y}{2x}\right) = C$

This is a homogeneous first-order ODE; substitute $y=vx$ to separate variables.

$(x^2-y^2)dx+2xy\,dy=0 \implies \dfrac{dy}{dx} = \dfrac{y^2-x^2}{2xy}$

Let $y=vx$, so $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$:

$v+x\dfrac{dv}{dx} = \dfrac{v^2x^2-x^2}{2x\cdot vx} = \dfrac{v^2-1}{2v}$

$x\dfrac{dv}{dx} = \dfrac{v^2-1}{2v}-v = \dfrac{v^2-1-2v^2}{2v} = \dfrac{-(v^2+1)}{2v}$

Separate: $\dfrac{2v}{v^2+1}dv = -\dfrac{dx}{x}$

Integrate: $\ln(v^2+1) = -\ln x+C \implies \ln\big[x(v^2+1)\big]=C \implies x(v^2+1)=C_1$

Substitute back $v=y/x$: $x\Big(\dfrac{y^2}{x^2}+1\Big)=C_1 \implies \dfrac{y^2+x^2}{x}=C_1 \implies x^2+y^2=C_1x$

Both coefficients depend only on $x/y$, so the equation is homogeneous; the substitution $x=vy$ separates the variables and integrates to $x+ye^{x/y}=C$.

Given $\left(1+e^{x/y}\right)dx + e^{x/y}\!\left(1-\dfrac{x}{y}\right)dy=0$.

Homogeneity: writing $\dfrac{dy}{dx}=-\dfrac{1+e^{x/y}}{e^{x/y}(1-x/y)}$, the right side depends on $x$ and $y$ only through the ratio $x/y$ (replacing $x\to\lambda x,\ y\to\lambda y$ leaves it unchanged) — so this is a homogeneous differential equation of degree $0$.

Solving: put $x=vy$ (so $v=x/y$), giving $dx=v\,dy+y\,dv$. Substitute:

$(1+e^{v})(v\,dy+y\,dv) + e^{v}(1-v)\,dy = 0$

$\big[(1+e^v)v + e^v(1-v)\big]dy + (1+e^v)y\,dv = 0$

The bracket simplifies: $v+ve^v+e^v-ve^v = v+e^v$.

$(v+e^v)\,dy + (1+e^v)y\,dv = 0 \;\Longrightarrow\; \dfrac{dy}{y} = -\dfrac{1+e^v}{v+e^v}\,dv$

Integrating, and noting $\dfrac{d}{dv}(v+e^v)=1+e^v$:

$\ln|y| = -\ln|v+e^v| + C_0 \;\Longrightarrow\; \ln\big|y(v+e^v)\big| = C_0 \;\Longrightarrow\; y(v+e^v)=C$

Substitute back $v=x/y$:

$y\!\left(\dfrac xy+e^{x/y}\right)=C \;\Longrightarrow\; x+y\,e^{x/y}=C$