The curve y=y(x) solves 2(x^2+x^5/4),dy-y(x+x^1/4),dx=2x^9/4,dx, x0, passing through (1,1…

Question

The curve $y=y(x)$ solves $2(x^2+x^{5/4})\,dy-y(x+x^{1/4})\,dx=2x^{9/4}\,dx$, $x>0$, passing through $\left(1,1-\dfrac43\log_e2\right)$. Find $y(16)$:

Accepted Answer

$$\frac{dy}{dx}-\frac{y}{2x} = \frac{x^{9/4}}{x^{5/4}(x^{3/4}+1)}.$$

I.F. $=x^{-1/2}$. With $x=z^4$, the RHS integral evaluates to $\dfrac{4z^3}{3}-\dfrac43\log|z^3+1|+C$.

Applying the given point fixes $C=-1/3$; then
$$y = \frac43x^{5/4}-\frac43\sqrt x\,\log(x^{3/4}+1)-\frac{\sqrt x}{3}.$$

At $x=16$: $y(16) = 4\left(\dfrac{31}{3}-\dfrac83\log_e3\right)$.

Solutions · 1