JEE PYQ: Differential Equation Question 20
Question 20 - 2020 (03 Sep Shift 2)
If $x^3,dy + xy,dx = x^2,dy + 2y,dx$; $y(2) = e$ and $x > 1$, then $y(4)$ is equal to:
(1) $\frac{3}{2}+\sqrt{e}$
(2) $\frac{3}{2}\sqrt{e}$
(3) $\frac{1}{2}+\sqrt{e}$
(4) $\frac{\sqrt{e}}{2}$
Show Answer
Answer: (2)
Solution
$(x^3-x^2),dy = (2-x)y,dx$, so $\frac{dy}{y} = \frac{2-x}{x^2(x-1)},dx$. By partial fractions: $\frac{2-x}{x^2(x-1)} = \frac{-1}{x} - \frac{2}{x^2} + \frac{1}{x-1}$. Integrating: $\ln y = -\ln x + \frac{2}{x} + \ln|x-1| + C$. At $(2,e)$: $1 = -\ln 2 + 1 + 0 + C$, so $C = \ln 2$. At $x = 4$: $\ln y = -\ln 4 + \frac{1}{2} + \ln 3 + \ln 2 = \ln\frac{3}{2} + \frac{1}{2}$. So $y = \frac{3}{2}\sqrt{e}$.
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