Let y=y(x) solve dydx=4y^3+2yx^23xy^2+x^3, y(1)=1.

Question

Let $y=y(x)$ solve $\dfrac{dy}{dx}=\dfrac{4y^3+2yx^2}{3xy^2+x^3}$, $y(1)=1$. If $y(2)\in[n-1,n)$ for some $n\in\mathbb{N}$, then $n$ is \_\_\_\_\_\_.

Answer: 3

Accepted Answer

With $y=vx$: $\dfrac{dy}{dx}=\dfrac{y}{x}\cdot\dfrac{4v^2+2}{3v^2+1}$; separating,
$$\int\frac{3v^2+1}{v^3+v}\,dv = \int\frac{dx}{x} \implies \ln|v^3+v|=\ln x+c.$$

I.e. $\dfrac{y^3}{x^3}+\dfrac{y}{x}=Cx$. Using $y(1)=1$: $C=2$.

At $x=2$: $\dfrac{y^3}{8}+\dfrac{y}{2}=4 \implies y(2)\approx 2.xx$, so $[y(2)]=2 \implies n=3$.

Solutions · 1