JEE PYQ: Differential Equation Question 33
Question 33 - 2020 (09 Jan Shift 2)
If $\frac{dy}{dx} = \frac{xy}{x^2 + y^2}$; $y(1) = 1$; then a value of $x$ satisfying $y(x) = e$ is:
(1) $\frac{1}{2}\sqrt{3},e$
(2) $\frac{e}{\sqrt{2}}$
(3) $\sqrt{2},e$
(4) $\sqrt{3},e$
Show Answer
Answer: (4)
Solution
Homogeneous equation. Put $y = vx$: $\frac{1+v^2}{v^3},dv = -\frac{dx}{x}$. Integrating: $-\frac{x^2}{2y^2} + \ln y = -\ln x + c$. At $(1,1)$: $c = -\frac{1}{2}$. So $x^2 = y^2(1 + 2\ln y)$. At $y = e$: $x^2 = e^2 \cdot 3$, $x = \sqrt{3},e$.
BETA
