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 $\mathrm{x}=4, \mathrm{y}=\frac{3}{2}$
$\therefore \mathrm{C}=\frac{1}{4} \ln 3$
$\therefore \frac{1}{3} \ln \left|\frac{\mathrm{y}-3}{\mathrm{y}}\right|=\frac{1}{4} \ln \left|\frac{\mathrm{x}-2}{\mathrm{x}+2}\right|+\frac{1}{4} \ln (3)$
At $\mathrm{x}=10$
$\frac{1}{3} \ln \left|\frac{\mathrm{y}-3}{\mathrm{y}}\right|=\frac{1}{4} \ln \left|\frac{2}{3}\right|+\frac{1}{4} \ln (3)$
$\ln \left|\frac{\mathrm{y}-3}{\mathrm{y}}\right|=\ln 2^{3 / 4}, \forall \mathrm{x}>2, \frac{\mathrm{dy}}{\mathrm{dx}}<0$
as $\mathrm{y}(4)=\frac{3}{2} \Rightarrow \mathrm{y} \in(0,3)$
$-\mathrm{y}+3=8^{1 / 4} \cdot \mathrm{y}$
$\mathrm{y}=\frac{\mathrm{m}}{1+8^{1 / 4}}$
