If the solution curve of (2x-10y^3),dy+y,dx=0 passes through (0,1) and (2,), then is a roo…

Question

If the solution curve of $(2x-10y^3)\,dy+y\,dx=0$ passes through $(0,1)$ and $(2,\beta)$, then $\beta$ is a root of:

Accepted Answer

$$\frac{dx}{dy}+\frac{2x}{y}=10y^2.$$

I.F.$=y^2$. $xy^2 = 2y^5+C$. Using $(0,1)$: $C=-2$.

At $(2,\beta)$: $2\beta^2=2\beta^5-2 \implies \beta^5-\beta^2-1=0$.

Solutions · 1