JEE PYQ: Differential Equation Question 28
Question 28 - 2020 (07 Jan Shift 1)
If $y = y(x)$ is the solution of the differential equation $e^y\left(\frac{dy}{dx} - 1\right) = e^x$ such that $y(0) = 0$, then $y(1)$ is equal to:
(1) $1 + \log_e 2$
(2) $2 + \log_e 2$
(3) $2e$
(4) $\log_e 2$
Show Answer
Answer: (1)
Solution
Let $e^y = t$: $\frac{dt}{dx} - t = e^x$. I.F. $= e^{-x}$. Solution: $te^{-x} = \int dx = x + c$. So $e^{y-x} = x + c$. At $(0,0)$: $c = 1$. At $x = 1$: $e^{y-1} = 2$, $y = 1 + \ln 2$.
BETA
