Animated Solution for Mathematics - Differential Equations: 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 $y(0)={\log }_e(1/2)$, then the value of $\alpha$ is equal to

Welcome, future engineer! Today, we are going to dismantle a differential equation that, at first glance, might seem like a chaotic mess of exponentials. We are given the equation $dy=e^{\alpha x+y}dx$.

When you see an expression like $e^{\alpha x+y}$, I want you to immediately think of the fundamental law of exponents: $e^{a+b}=e^a\cdot e^b$. This is the key that unlocks the door.

By applying this, our equation transforms into $dy=e^{\alpha x}\cdot e^{y}dx$. Suddenly, the chaos organizes itself. We can now group the $y$ terms on the left and the $x$ terms on the right.

Dividing both sides by $e^y$, we get $e^{-y}dy=e^{\alpha x}dx$. We have successfully separated the variables. This is the moment where the problem stops being a 'differential equation' and starts being a simple calculus exercise.

Now that we have $e^{-y}dy=e^{\alpha x}dx$, we must integrate both sides. This is the heart of the problem.

On the left, the integral of $e^{-y}$ with respect to $y$ is $-e^{-y}$. On the right, the integral of $e^{\alpha x}$ with respect to $x$ is:

And, of course, we must add our constant of integration, $C$. So, our general solution stands tall:

This equation is the foundation upon which we will build our final answer. It connects $x$, $y$, $\alpha$, and $C$ in a beautiful, rigid structure.

Locking the Constants

We have two unknowns, $C$ and $\alpha$, and we have two boundary conditions to help us find them. First, let's look at $y(0)=\ln (\frac{1}{2})$. Using the property of logarithms, we know $\ln (\frac{1}{2})=-\ln 2$.

Substituting $x=0$ and $y=-\ln 2$ into our general solution, we get:

Simplifying this, $-e^{\ln 2}=\frac{1}{\alpha }+C$, which leads us to:

Now, for the second condition: $y(\ln 2)=\ln 2$. Substituting $x=\ln 2$ and $y=\ln 2$ into our general solution, we get:

Since $e^{-\ln 2}=e^{\ln (2^{-1})}=\frac{1}{2}$, the left side becomes $-\frac{1}{2}$. On the right side, $e^{\alpha \ln 2}=e^{\ln (2^{\alpha })}=2^{\alpha }$. Thus, we have:

We have two equations: (1) $-2=\frac{1}{\alpha }+C$ and (2) $-\frac{1}{2}=\frac{2^{\alpha }}{\alpha }+C$. To isolate $\alpha$, we simply subtract Equation 1 from Equation 2.

The constant $C$ vanishes, leaving us with:

This simplifies to:

We are left with a transcendental equation, but remember the constraint: $\alpha \in \mathbb{N}$. We don't need complex numerical methods here. Let's test integer values.

If $\alpha =1$, we get $\frac{2^1-1}{1}=1$, which is not $\frac{3}{2}$. If $\alpha =2$, we get:

It matches perfectly! The elegance of the solution lies in the simplicity of the final step. You have conquered the differential equation, navigated the boundary conditions, and solved for the parameter $\alpha =2$.