JEE PYQ: Differential Equation Question 18
Question 18 - 2020 (02 Sep Shift 2)
If a curve $y = f(x)$, passing through the point $(1, 2)$, is the solution of the differential equation $2x^2,dy = (2xy + y^2),dx$, then $f\left(\frac{1}{2}\right)$ is equal to:
(1) $\frac{1}{1+\log_e 2}$
(2) $\frac{1}{1-\log_e 2}$
(3) $1+\log_e 2$
(4) $\frac{-1}{1+\log_e 2}$
Show Answer
Answer: (1)
Solution
$\frac{dy}{dx} = \frac{y}{x} + \frac{y^2}{2x^2}$. Homogeneous: put $y = vx$. $v + x\frac{dv}{dx} = v + \frac{v^2}{2}$. So $\frac{2dv}{v^2} = \frac{dx}{x}$. $\frac{-2}{v} = \ln x + c$, i.e. $\frac{-2x}{y} = \ln x + c$. At $(1,2)$: $c = -1$. So $\frac{-2x}{y} = \ln x - 1$. At $x = \frac{1}{2}$: $\frac{-1}{y} = -\ln 2 - 1$, $y = \frac{1}{1+\log_e 2}$.
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