Question

A curve satisfying the initial condition, y(1) = 0, satisfies the differential equation, $x\frac{dy}{dx}=y-x^2$. The area bounded by the curve and the x-axis is

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{6}$

Answer 1

The correct option is D $\frac{1}{6}$

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

$\frac{dy}{dx}=\frac{y}{x}-x$

$\frac{y}{x}=m$

$\frac{dy}{dx}=x\frac{dm}{dx}+m$

$x\frac{dm}{dx}+m=m-x$

$=\frac{dm}{dx}=-1$

$m=-x+c$

$y=-x^2+cx$

$y\left(1\right)=0$

$0=-1+c$

$c=1$

$y=-x^2+x$

${\int }_0^1(-x^2+x)dx$

$=-\frac{1}{3}+\frac{1}{2}$

$=\frac{1}{6}$