Question

Solve $(x^3+x^2+x+1)\frac{dy}{dx}=2x^2+x$ if the value of $y=1$ when $x=0$.

Answer 1

Given,

$(x^3+x^2+x+1)\frac{dy}{dx}=2x^2+x$

$\frac{dy}{dx}=\frac{2x^2+x}{x^3+x^2+x+1}$

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

$\frac{2x^2+x}{(x+1)(x^2+1)}=\frac{A}{x+1}+\frac{B}{x^2+1}$

$2x^2+x=A(x^2+1)+(Bx+C)(x+1)$

when $x=-1\to A=\frac{1}{2}$

when $x=0\to C=-\frac{1}{2}$

when $x=1\to B=\frac{3}{2}$

$\frac{2x^2+x}{(x+1)(x^2+1)}=\frac{1}{2(x+1)}+\frac{3x-1}{2(x^2+1)}$

$\therefore \frac{dy}{dx}=\frac{1}{2(x+1)}+\frac{3x-1}{2(x^2+1)}$

Integrating on both sides, we get,

$\int \frac{dy}{dx}=\int \frac{1}{2(x+1)}+\int \frac{3x-1}{2(x^2+1)}$

$y=\frac{1}{2}\log (x+1)+\frac{3}{4}\log (x^2+1)-\frac{1}{2}{\tan }^{-1}x+c$

$x=0,y=1$

$1=\frac{1}{2}\log (0+1)+\frac{3}{4}\log (0^2+1)-\frac{1}{2}{\tan }^{-1}0+c$

$1=c$

$\therefore y=\frac{1}{2}\log (x+1)+\frac{3}{4}\log (x^2+1)-\frac{1}{2}{\tan }^{-1}x+1$