Question

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

• A. $y.\frac{x^2-1}{x}=\frac{x^2}{2}\log x-\frac{x^2}{4}+c.$
• B. $y.\frac{x^2+1}{x}=\frac{x^2}{2}\log x+\frac{x^2}{4}+c.$
• C. $y.\frac{x^2-1}{x}=\frac{x^2}{2}\log x+\frac{x^2}{4}+c.$
• D. $y.\frac{x^2+1}{x}=\frac{x^2}{2}\log x-\frac{x^2}{4}+c.$

Answer 1

The correct option is D $y.\frac{x^2+1}{x}=\frac{x^2}{2}\log x-\frac{x^2}{4}+c.$

Given, $x(x^2+1)\frac{dy}{dx}=y(1-x^2)+x^3\log x$

$\Rightarrow \frac{dy}{dx}+\frac{x^2-1}{x(x^2+1)}y=\frac{x^2\log x}{x^2+1}$

Here $P=\frac{x^2-1}{x(x^2+1)}\Rightarrow \int Pdx=\int \frac{x^2-1}{x(x^2+1)}dx$

Put $x^2=t\Rightarrow 2xdx=dt$

$\therefore \int Pdx=\frac{1}{2}\int \frac{t-1}{t(t+1)}dt=\int (\frac{1}{t+1}-\frac{1}{2t})dt$

$=\log (t+1)-\frac{1}{2}\log t=\log \frac{t+1}{\sqrt{t}}=\log \frac{x^2+1}{x}$

$\therefore I.F.=\frac{x^2+1}{x}$

Multiplying (1) by $I.F.$ we get

$\frac{x^2+1}{x}\frac{dy}{dx}+\frac{x^2-1}{x^2}y=x\log x$

Integrating both sides, we get

$y.\frac{x^2+1}{x}=\int \frac{x^2\log x}{x^2+1}.\frac{x^2+1}{x}dt=\int x\log xdx$

$\Rightarrow y.\frac{x^2+1}{x}=\frac{x^2}{2}\log x-\int \frac{x^2}{2}.\frac{1}{x}dx$

$\Rightarrow y.\frac{x^2+1}{x}=\frac{x^2}{2}\log x-\frac{x^2}{4}+c$