Question

The solution of the differential equation $\frac{dy}{dx}+\frac{y}{x}=x$, with the condition that y = 1 at x = 1, is

• A. y = $\frac{2}{3x^2}+\frac{x}{3}$
• B. $y=\frac{x}{2}+\frac{1}{2x}$
• C. $y=\frac{2}{3}+\frac{x}{3}$
• D. $y=\frac{2}{3x}+\frac{x^2}{3}$

Answer 1

The correct option is D $y=\frac{2}{3x}+\frac{x^2}{3}$

$\frac{dy}{dx}+\frac{y}{x}=x,y\left(1\right)=1...\left(i\right)$

it is a linear differential equation of first order.

I.F = $e^{\int \left(\frac{1}{x}\right)dx}=e^{lnx}=x$

$\therefore$ Solution of (i) is

y(I.F) = $\int x.(I.F)dx+C$

yx = $\int x.xdx+C=\frac{x^3}{3}+C$

Given that y(1) = 1

$\Rightarrow 1\times 1=\frac{1}{3}+C$

$\Rightarrow C=\frac{2}{3}(\because x=1,y=1)$

y = $\frac{x^2}{3}+\frac{2}{3x}$