Question

Let $y=y(x)$ be the solution of the differential equation $dy = e^{\alpha x+y}dx; \alpha \in \mathbb{N}$. If $y(\log_e 2) = \log_e 2$ and $y(0)=\log_e(\frac{1}{2})$, then the value of $\alpha$ is equal to ________.

Hint:

After applying boundary conditions, always reduce the equation to a simple integer relation when the parameter is restricted to natural numbers.

Answer 1

Correct Answer: 2

Given differential equation: \[ \frac{dy}{dx}=e^{\alpha x+y} \] Rewrite: \[ \frac{dy}{dx}=e^{\alpha x}e^{y} \] Separating variables, \[ e^{-y}dy=e^{\alpha x}dx \] Integrating both sides, \[ \int e^{-y}dy=\int e^{\alpha x}dx \] \[ -\,e^{-y}=\frac{1}{\alpha}e^{\alpha x}+C \] Using the condition $y(0)=\ln\frac12=-\ln2$: \[ -\,e^{-(-\ln2)}=\frac{1}{\alpha}+C \Rightarrow -2=\frac{1}{\alpha}+C \] \[ C=-2-\frac{1}{\alpha} \] Using the condition $y(\ln2)=\ln2$: \[ -\,e^{-\ln2}=\frac{1}{\alpha}e^{\alpha\ln2}+C \Rightarrow -\frac12=\frac{2^\alpha}{\alpha}+C \] Substitute $C$: \[ -\frac12=\frac{2^\alpha}{\alpha}-2-\frac{1}{\alpha} \] \[ \frac{3}{2}=\frac{2^\alpha-1}{\alpha} \Rightarrow 3\alpha=2(2^\alpha-1) \] Since $\alpha\in\mathbb{N}$, test values: \[ \alpha=2:\quad 3(2)=2(4-1)=6 \quad \text{(satisfies)} \] \[ \boxed{\alpha=2} \]