JEE Main 2021 (Online) 27th July Evening ShiftQuestion 1
Let $y=y(x)$ be the solution of the differential equation $dy=e^{\alpha x+y}dx$; $\alpha\in\mathbf{N}$. If $y(\log_e2)=\log_e2$ and $y(0)=\log_e\left(\frac12\right)$, then the value of $\alpha$ is equal to ________.
Differential Equation
Answer$2$
Solution
Concept. Separate the variables; the two given values remove the constant and give an equation for $\alpha$. Formulas used. (1) $\int e^{-y}\,dy=-e^{-y}$ and $\int e^{\alpha x}\,dx=\dfrac{e^{\alpha x}}{\alpha}$Step 1 - Integrate. $e^{-y}\,dy=e^{\alpha x}\,dx$, so by (1) $$-e^{-y}=\frac{e^{\alpha x}}{\alpha}+C.$$ Step 2 - Two conditions. $y(0)=-\ln2$ gives $e^{-y}=2$: $-2=\frac1\alpha+C$. $y(\ln2)=\ln2$ gives $e^{-y}=\frac12$: $-\frac12=\frac{2^\alpha}{\alpha}+C$. Step 3 - Subtract. $\dfrac{2^\alpha-1}{\alpha}=\dfrac32$. Step 4 - Solve. $\alpha=1,2,3$ give $1,\ \frac32,\ \frac73$, and the left side increases for $\alpha\ge1$. So $\alpha=2$. Answer: 2