Question

x(x-y) dy=y(x +y) dx solve the homogenous eqn​

Answer 1

Here is u r ans  mate ..

Answer 2

∴ The homogeneous equation is $log(\frac{y}{x} )+\frac{y}{x} =-2logx-2c$

Step-by-step explanation:

Given;

$x(x-y)dy=y(x+y)dx$

⇒$\frac{dy}{dx} =\frac{y(x+y)}{x(x-y)}$___equation-1

Let, $y=vx$

differentiate with respect to $x$;

⇒$\frac{dy}{dx}=v+x\frac{dv}{dx}$

Plug $\frac{dy}{dx}$ value in equation-1;

∴$v+x\frac{dv}{dx}=\frac{vx(x+vx)}{x(x-vx)}$

⇒$v+x\frac{dv}{dx}=\frac{v(1+v)}{(1-v)}$

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

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

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

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

Integrating on both sides;

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

By integrating of $\int \frac{1-v}{2v^{2} } dv$ the value will be come $-\frac{1}{2}log(v)-\frac{1}{2}v$

So,

$-\frac{1}{2}log(v)-\frac{1}{2}v=logx+c$  where $c=$constant

⇒$log(v)+v=-2logx-2c$

∴$log(\frac{y}{x} )+\frac{y}{x} =-2logx-2c$    (∵$v=\frac{y}{x}$ )

So the homogeneous equation is $log(\frac{y}{x} )+\frac{y}{x} =-2logx-2c$