Let y=y(x) solve x(1-x^2)dydx+(3x^2y-y-4x^3)=0, x1, y(2)=-2.

Question

Let $y=y(x)$ solve $x(1-x^2)\dfrac{dy}{dx}+(3x^2y-y-4x^3)=0$, $x>1$, $y(2)=-2$. Then $y(3)$ is:

Accepted Answer

$$\frac{dy}{dx}+\frac{y(3x^2-1)}{x(1-x^2)} = \frac{4x^3}{x(1-x^2)}.$$

I.F. $=\dfrac{1}{x^3-x}$.
$$\frac{y}{x^3-x} = \frac{2}{x^2-1}+C.$$

Using $y(2)=-2$: $-\dfrac13 = \dfrac23+C \Rightarrow C=-1$.

At $x=3$: $\dfrac{y}{24}=\dfrac28-1 \Rightarrow y=-18$.

Solutions · 1