Question

Find the particular solution of the differential equation $x(x^2-1)\frac{dy}{dx}=1,y=0$ when $x=2.$

Answer 1

$x(x^2-1)\frac{dy}{dx}=1$

$\Rightarrow x(x^2-1)dy=dx$

$\therefore \frac{dx}{x(x^2-1)}=dy$

Let $t=x^2,dt=2xdx$

$\therefore \frac{dt}{2t(t-1)}=dy$

$\therefore \frac{dt}{t-1}-\frac{dt}{t}=2dy$

Integrating both sides, we get

$\ln \left(t-1\right)-\ln t=2y+c$

$\therefore \ln \frac{t-1}{t}=2y+c$

Substituting $t=x^2,$ we can write the function as

$\ln \left(\frac{x^2-1}{x^2}\right)=2y+c$

Also given that $y=0$ when $x=2$, substituting those values, we get

$\ln \left(\frac{4-1}{4}\right)=c$

$\therefore c=\ln \left(\frac{3}{4}\right)$

The function thus becomes $\ln \left(\frac{4(x^2-1)}{3x^2}\right)=2y$