Differential Equations Question 7
Question 7 - 2024 (27 Jan Shift 2)
If $y=y(x)$ is the solution curve of the differential equation $\left(x^{2}-4\right) d y-\left(y^{2}-3 y\right) d x=0$, $x>2, y(4)=\frac{3}{2}$ and the slope of the curve is never zero, then the value of $y(10)$ equals :
(1) $\frac{3}{1+(8)^{1 / 4}}$
(2) $\frac{3}{1+2 \sqrt{2}}$
(3) $\frac{3}{1-2 \sqrt{2}}$
(4) $\frac{3}{1-(8)^{1 / 4}}$
Show Answer
Answer (1)
Solution
$$ \begin{aligned} & \left(x^{2}-4\right) d y-\left(y^{2}-3 y\right) d x=0 \\ \Rightarrow & \int \frac{d y}{y^{2}-3 y}=\int \frac{d x}{x^{2}-4} \\ \Rightarrow & \frac{1}{3} \int \frac{y-(y-3)}{y(y-3)} d y=\int \frac{d x}{x^{2}-4} \\ \Rightarrow & \frac{1}{3}(\ln |y-3|-\ln |y|)=\frac{1}{4} \ln \left|\frac{x-2}{x+2}\right|+C \\ \Rightarrow & \frac{1}{3} \ln \left|\frac{y-3}{y}\right|=\frac{1}{4} \ln \left|\frac{x-2}{x+2}\right|+C \end{aligned} $$
At $x=4, y=\frac{3}{2}$
$\therefore C=\frac{1}{4} \ln 3$
$\therefore \frac{1}{3} \ln \left|\frac{y-3}{y}\right|=\frac{1}{4} \ln \left|\frac{x-2}{x+2}\right|+\frac{1}{4} \ln (3)$
At $x=10$
$\frac{1}{3} \ln \left|\frac{y-3}{y}\right|=\frac{1}{4} \ln \left|\frac{2}{3}\right|+\frac{1}{4} \ln (3)$
$\ln \left|\frac{y-3}{y}\right|=\ln 2^{3 / 4}, \forall x>2, \frac{dy}{dx}<0$
as $y(4)=\frac{3}{2} \Rightarrow y \in(0,3)$
$-y+3=8^{1 / 4} \cdot y$
$y=\frac{m}{1+8^{1 / 4}}$
