Homogeneous Differential Equations PYQs

Let $f(x,y) = \frac{x-y}{x+y}$. Replacing $x$ with $\lambda x$ and $y$ with $\lambda y$ gives back the same function. It is homogeneous.

Put $y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}$.

$v + x\frac{dv}{dx} = \frac{x-vx}{x+vx} = \frac{1-v}{1+v}$.

$x\frac{dv}{dx} = \frac{1-v}{1+v} - v = \frac{1-v-v-v^2}{1+v} = \frac{1-2v-v^2}{1+v}$.

$\frac{1+v}{1-2v-v^2} \, dv = \frac{dx}{x}$.

$-\frac{1}{2}\log|1-2v-v^2| = \log|x| + \log C$.

Substitute $v = y/x$: $x^2(1 - 2y/x - y^2/x^2) = C' \Rightarrow x^2 - 2xy - y^2 = C'$.

Put $y = vx$. Then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.

$v + x\frac{dv}{dx} = \frac{2x(vx)}{x^2 + (vx)^2} = \frac{2v}{1+v^2}$.

$x\frac{dv}{dx} = \frac{2v}{1+v^2} - v = \frac{2v-v-v^3}{1+v^2} = \frac{v(1-v^2)}{1+v^2}$.

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

Using partial fractions: $(\frac{1}{v} + \frac{1}{1-v} - \frac{1}{1+v}) \, dv = \frac{dx}{x}$.

$\log|v| - \log|1-v| - \log|1+v| = \log|x| + \log C \Rightarrow \frac{v}{1-v^2} = Cx$.

Put $y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}$.

$v + x\frac{dv}{dx} = \frac{x+vx}{x-vx} = \frac{1+v}{1-v}$.

$x\frac{dv}{dx} = \frac{1+v}{1-v} - v = \frac{1+v-v+v^2}{1-v} = \frac{1+v^2}{1-v}$.

$\frac{1-v}{1+v^2} \, dv = \frac{dx}{x}$.

$\int \frac{1}{1+v^2} \, dv - \int \frac{v}{1+v^2} \, dv = \int \frac{dx}{x}$.

$\tan^{-1} v - \frac{1}{2}\log(1+v^2) = \log|x| + C$.

$\tan^{-1}(y/x) = \log|x| + \log\sqrt{1+y^2/x^2} + C$.

$\tan^{-1}(y/x) = \log\sqrt{x^2+y^2} + C$.

Put $x = vy$ (easier here) or $y=vx$. Let's use $y=vx$.

$v + x\frac{dv}{dx} = \frac{1}{1+v} \Rightarrow x\frac{dv}{dx} = \frac{1}{1+v} - v = \frac{1-v-v^2}{1+v}$.

$\frac{1+v}{1-v-v^2} \, dv = \frac{dx}{x}$.

Solve using standard integral forms.

Put $y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}$.

$v + x\frac{dv}{dx} = \frac{1}{1+v} \Rightarrow x\frac{dv}{dx} = \frac{1-v-v^2}{1+v}$.

$\int \frac{1+v}{1-v-v^2} \, dv = \int \frac{dx}{x}$.

Let $x = X+h, y = Y+k$. Select $h, k$ such that $h-k+1=0$ and $h+k+1=0$.

Solving gives $h=-1, k=0$.

Equation becomes $\frac{dY}{dX} = \frac{X-Y}{X+Y}$.

Put $Y = vX$.

Final Answer: $(x+1)^2 - 2(x+1)y - y^2 = C$.

Put $y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}$.

$v + x\frac{dv}{dx} = \frac{1-v}{1+v} \Rightarrow x\frac{dv}{dx} = \frac{1-2v-v^2}{1+v}$.

$\frac{1+v}{1-2v-v^2} \, dv = \frac{dx}{x}$.

Final Answer: $x^2 - 2xy - y^2 = C$.