JEE PYQ: Differential Equation Question 39
Question 39 - 2019 (12 Apr Shift 1)
Consider the differential equation $y^2,dx + \left(x - \frac{1}{y}\right),dy = 0$. If the value of $y$ is 1 when $x = 1$, then the value of $x$ for which $y = 2$ is:
(1) $\frac{5}{2} + \frac{1}{\sqrt{e}}$
(2) $\frac{3}{2} - \frac{1}{\sqrt{e}}$
(3) $\frac{1}{2} + \frac{1}{\sqrt{e}}$
(4) $\frac{3}{2} - \sqrt{e}$
Show Answer
Answer: (2)
Solution
$\frac{dx}{dy} + \frac{x}{y^{-2}} = \frac{1}{y^3}$… Actually: $\frac{dx}{dy} \cdot y^2 = \frac{1}{y} - x$, so $\frac{dx}{dy} + \frac{x}{y^2} = \frac{1}{y^3}$. I.F. $= e^{-1/y}$. $xe^{-1/y} = \int \frac{e^{-1/y}}{y^3},dy$. Put $-\frac{1}{y} = u$: $= -\int ue^u,du = -(u-1)e^u + C = \left(\frac{1}{y}+1\right)e^{-1/y} + C$. At $y = 1$, $x = 1$: $e^{-1} = 2e^{-1} + C$, $C = -e^{-1}$. At $y = 2$: $xe^{-1/2} = \frac{3}{2}e^{-1/2} - e^{-1}$, $x = \frac{3}{2} - e^{-1/2} = \frac{3}{2} - \frac{1}{\sqrt{e}}$.
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