Let y=f(x) pass through (-2,2); the slope at (x,f(x)) is given by f(x)+xf'(x)=x^2.

Question

Let $y=f(x)$ pass through $(-2,2)$; the slope at $(x,f(x))$ is given by $f(x)+xf'(x)=x^2$. Then:

Accepted Answer

$$\frac{dy}{dx}+\frac{y}{x}=x.$$

I.F.$=x$. $xy = \dfrac{x^3}{3}+C$. Using $(-2,2)$: $-4=-\dfrac83+C \implies C=-\dfrac43$.

$x^3-3xy-4 = 0.$

Solutions · 1