Let y=y(x) be the solution of the differential equation [x(dy)/(dx) - 2y = x^3(2-x^3)^2 y,…

Question

Let $y=y(x)$ be the solution of the differential equation
$$x\frac{dy}{dx} - \sin 2y = x^3(2-x^3)\cos^2 y,\quad x\ne 0,\qquad y(2)=0.$$
Then $\tan(y(1))$ is equal to:

Accepted Answer

Dividing by $x\cos^2 y$: $\sec^2 y\dfrac{dy}{dx} - \dfrac{2\tan y}{x} = 2x^2-x^5$.

Let $\tan y=t$: $\dfrac{dt}{dx}-\dfrac{2}{x}t = 2x^2-x^5$.

I.F. $= e^{\int(-2/x)dx}=1/x^2$.
$$\frac{t}{x^2} = \int(2-x^3)\,dx = 2x-\frac{x^4}{4}+C \implies \tan y = x^2\left(2x-\frac{x^4}{4}+C\right).$$

Using $y(2)=0$: $0 = 4(4-4+C) \implies C=0$. So $\tan y = 2x^3-\dfrac{x^6}{4}$.

At $x=1$: $\tan(y(1)) = 2-\dfrac14 = \dfrac74$.

Solutions · 1