Question

The solution of the differential equation $x^2\frac{dy}{dx}+2xy-x+1=0$ given that at x = 1, y = 0 is

• A. $\frac{1}{2}-\frac{1}{x}+\frac{1}{2x^2}$
• B. $\frac{1}{2}-\frac{1}{x}-\frac{1}{2x^2}$
• C. $\frac{1}{2}+\frac{1}{x}+\frac{1}{2x^2}$
• D. $-\frac{1}{2}+\frac{1}{x}+\frac{1}{2x^2}$

Answer 1

The correct option is A $\frac{1}{2}-\frac{1}{x}+\frac{1}{2x^2}$

$x^2\frac{dy}{dx}+2xy-x+1=0$ ...(i)

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

I.F = $e^{\int \frac{2}{x}dx}$

= $e^{2lnx}=x^2$

$\therefore$ Solution of (i) is given by

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

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

$y{x}^2=\frac{x^2}{2}-x+c$

Given at

x = 1

y = 0

0 = $\frac{1}{2}-1+c\Rightarrow c=\frac{1}{2}$

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