Let y=y(x) be the solution of the differential equation [x^4,dy + (4x^3y+2 x),dx=0, x0, y…

Question

Let $y=y(x)$ be the solution of the differential equation
$$x^4\,dy + (4x^3y+2\sin x)\,dx=0,\quad x>0,\qquad y\!\left(\frac{\pi}{2}\right)=0.$$
Then $\pi^4\,y\!\left(\frac{\pi}{3}\right)$ is equal to:

Accepted Answer

$$\frac{dy}{dx}+\frac{4}{x}y = \frac{-2\sin x}{x^4}.$$

I.F. $=e^{\int 4/x\,dx}=x^4$.
$$x^4\frac{dy}{dx}+4x^3y=-2\sin x \implies \frac{d}{dx}(x^4y) = -2\sin x \implies x^4y = 2\cos x+c.$$

Using $y(\pi/2)=0 \Rightarrow c=0$. So $y=\dfrac{2\cos x}{x^4}$.

$$y\!\left(\frac{\pi}{3}\right) = \frac{2\cdot\frac12}{(\pi/3)^4} = \frac{1}{(\pi/3)^4} = \frac{81}{\pi^4} \implies \pi^4 y\!\left(\frac{\pi}{3}\right) = 81.$$

Solutions · 1