If x^3,dy+xy,dx=x^2,dy+2y,dx, y(2)=e, x1, then y(4) equals:

Question

If $x^3\,dy+xy\,dx=x^2\,dy+2y\,dx$, $y(2)=e$, $x>1$, then $y(4)$ equals:

Accepted Answer

$$\frac{dy}{y} = \frac{2-x}{x^2(x-1)}\,dx = \left(-\frac1x-\frac{2}{x^2}+\frac{1}{x-1}\right)dx.$$

Integrating: $\ln y = -\ln x+\dfrac2x+\ln|x-1|+C$. Using $y(2)=e$: $C=\ln2$.

At $x=4$: $\ln y(4) = -\ln4+\dfrac12+\ln3+\ln2 = \ln\dfrac32+\dfrac12 \implies y(4)=\dfrac32\sqrt e$.

Solutions · 1