JEE PYQ: Differential Equation Question 22
Question 22 - 2020 (04 Sep Shift 2)
The solution of the differential equation $\frac{dy}{dx} - \frac{y+3x}{\log_e(y+3x)} + 3 = 0$ is (where $C$ is a constant of integration):
(1) $x - \frac{1}{2}(\log_e(y+3x))^2 = C$
(2) $x - \log_e(y+3x) = C$
(3) $y + 3x - \frac{1}{2}(\log_e x)^2 = C$
(4) $x - 2\log_e(y+3x) = C$
Show Answer
Answer: (1)
Solution
Let $y + 3x = t$: $\frac{dt}{dx} = \frac{dy}{dx} + 3$. The equation becomes $\frac{dt}{dx} - 3 - \frac{t}{\log_e t} + 3 = 0$, i.e. $\frac{dt}{dx} = \frac{t}{\ln t}$. So $\frac{\ln t}{t},dt = dx$, giving $\frac{(\ln t)^2}{2} = x + C’$, i.e. $x - \frac{1}{2}(\log_e(y+3x))^2 = C$.
BETA
