Question

Solve for differntial equation: $(x^3-x)\frac{dy}{dx}-(3x^2-1)y=x^5-2x^3+x$

• A. $y.\frac{1}{x(x^2+1)}=\log x+c.$
• B. $y.\frac{1}{x(x^2-1)}=-\log x+c.$
• C. $y.\frac{1}{x(x^2-1)}=\log x+c.$
• D. None of these.

Answer 1

Answer: (C)

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

Given, $(x^3-x)\frac{dy}{dx}-({3x}^2-1)y=x^5-{2x}^3+x$

$\Rightarrow \frac{dy}{dx}-\frac{{3x}^2-1}{x^3-x}y=\frac{x{(x^2-1)}^2}{x(x^2-1)}=x^2-1$ ...(1)

Here $P=\frac{{3x}^2-1}{x^3-x}\Rightarrow \int Pdx=\int \frac{{3x}^2-1}{x^3-x}dx=-\log (x^3-x)$

$\therefore I.F=e^{\int Pdx}=\frac{1}{x^3-x}=\frac{1}{x(x^2-1)}$

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

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

Integrating both sides, we get

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