If y=y(x) is the solution curve of (x^2-4),dy-(y^2-3y),dx=0, x2, y(4)=32, and the slope o…

Question

If $y=y(x)$ is the solution curve of $(x^2-4)\,dy-(y^2-3y)\,dx=0$, $x>2$, $y(4)=\dfrac32$, and the slope of the curve is never zero, then $y(10)$ equals:

Accepted Answer

Separating variables:
$$\frac{dy}{y^2-3y} = \frac{dx}{x^2-4} \implies \frac13\ln\left|\frac{y-3}{y}\right| = \frac14\ln\left|\frac{x-2}{x+2}\right|+C.$$

Using $y(4)=3/2$ to fix $C$, and evaluating at $x=10$ gives $y(10) = \dfrac{3}{1+8^{1/4}}$.

Solutions · 1