If y=y(x) is the solution curve of the differential equation (x^2-4)dy-(y^2-3y)dx=0,x2,y(…

Question

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

Accepted Answer

$\begin{gathered} ({x}^{2}-4)dy-({y}^{2}-3y)dx=0 \ \Rightarrow \int \frac{dy}{{y}^{2}-3y}=\int \frac{dx}{{x}^{2}-4} \ \Rightarrow \frac{1}{3}\int \frac{y-(y-3)}{y(y-3)}dy=\int \frac{dx}{{x}^{2}-4} \ \Rightarrow \frac{1}{3}(\ln |y-3|-\ln |y|)=\frac{1}{4}\ln |\frac{x-2}{x+2}|+C \ \Rightarrow \frac{1}{3}\ln |\frac{y-3}{y}|=\frac{1}{4}\ln |\frac{x-2}{x+2}|+C \end{gathered}$

$\begin{gathered} x=4,y=\frac{3}{2} \ 	herefore C=\frac{1}{4}\ln 3 \ 	herefore \frac{1}{3}\ln |\frac{y-3}{y}|=\frac{1}{4}\ln |\frac{x-2}{x+2}|+\frac{1}{4}\ln (3) \ x=10 \ \frac{1}{3}\ln |\frac{y-3}{y}|=\frac{1}{4}\ln |\frac{2}{3}|+\frac{1}{4}\ln (3) \ \ln |\frac{y-3}{y}|=\ln {2}^{3/4},\forallx>2,\frac{dy}{dx}<0 \ y(4)=\frac{3}{2}\Rightarrow y\in (0,3) \ -y+3={8}^{1/4}\cdot y \ y=\frac{3}{1+{8}^{1/4}} \end{gathered}$

Solutions · 1