Let y=y(x) solve (x-x^3),dy=(y+yx^2-3x^4),dx, x2.

Question

Let $y=y(x)$ solve $(x-x^3)\,dy=(y+yx^2-3x^4)\,dx$, $x>2$. If $y(3)=3$, then $y(4)$ is:

Accepted Answer

$$\frac{xdy-ydx}{x^2} = (ydx+xdy)-3x^2dx \implies d\left(\frac{y}{x}\right)=d(xy)-3x^2dx.$$

Integrating: $\dfrac{y}{x}=xy-x^3+c$. Using $(3,3)$: $1=9-27+c \implies c=19$.

At $x=4$: $\dfrac y4=4y-64+19 \implies 15y=4\times45 \implies y=12$.

Solutions · 1