JEE PYQ: Differential Equation Question 46
Question 46 - 2019 (11 Jan Shift 1)
If $y(x)$ is the solution of the differential equation $\frac{dy}{dx} + \left(\frac{2x+1}{x}\right)y = e^{-2x}$, $x > 0$, where $y(1) = \frac{1}{2}e^{-2}$, then:
(1) $y(\log_e 2) = \log_e 4$
(2) $y(\log_e 2) = \frac{\log_e 2}{4}$
(3) $y(x)$ is decreasing in $\left(\frac{1}{2}, 1\right)$
(4) $y(x)$ is decreasing in $(0, 1)$
Show Answer
Answer: (3)
Solution
I.F. $= e^{2x+\ln x} = xe^{2x}$. Solution: $yxe^{2x} = \int x,dx = \frac{x^2}{2} + c$. $y(1) = \frac{1}{2}e^{-2}$: $\frac{1}{2} = \frac{1}{2} + c$, $c = 0$. So $y = \frac{x}{2}e^{-2x}$. $y’(x) = \frac{e^{-2x}}{2}(1-2x) < 0$ for $x \in (\frac{1}{2}, 1)$. So $y(x)$ is decreasing in $(\frac{1}{2}, 1)$.
BETA
