Question

Solve the differential equation
$\frac{dy}{dx}=\frac{3y-7x+7}{3x-7y-3}$

Answer 1

$\frac{dy}{dx}=\frac{-7x+3y+7}{3x-7y-3}$

The point of intersection of $-7x+3y+7=0$ and $3x-7y-3=0$ is $(1,0)$

Let $X=x-1$ and $Y=y$

we get $\frac{dY}{dX}=\frac{-7X+3Y}{3X-7Y}$

Let $Y=kX$ , we get $k+X\frac{dk}{dX}=\frac{3k-7}{3-7k}$

$\Rightarrow X\frac{dk}{dX}=\frac{7(k^2-1)}{3-7k}$

$\Rightarrow 7\frac{dX}{X}=(\frac{3}{k^2-1}-7\frac{k}{k^2-1})dk$

By applying integral we get $7\ln X=\frac{1}{2}\ln \left(\frac{k-1}{k+1}\right)-\frac{7}{2}\ln (k^2-1)=\frac{1}{2}\ln \left(\frac{Y-X}{Y+X}\right)-\frac{7}{2}\ln \left(\frac{Y^2-X^2}{X^2}\right)$

Now put $Y=y$ and $X=x-1$ , we get $7\ln (x-1)=\frac{1}{2}\ln \left(\frac{y-x+1}{y+x-1}\right)-\frac{7}{2}\ln \left(\frac{y^2-{(x-1)}^2}{{(x-1)}^2}\right)$