Animated Solution for Mathematics - Differential Equations: Let $y=y(x),x>1$, be the solution of the differential equation $(x-1)\frac{dy}{dx}+2xy=\frac{1}{x-1}$, with $y(2)=\frac{1+e^4}{2e^4}$. If $y(3)=\frac{e^{\alpha }+1}{\beta e^{\alpha }}$, then the value of $\alpha +\beta$ is equal to _______.

Imagine you are standing before a complex differential equation, a puzzle waiting to be solved. The equation given is:

At first glance, it might seem intimidating, but as an elite JEE aspirant, you know that every complex problem is just a collection of simple, elegant steps. The first thing we notice is that the term $\frac{dy}{dx}$ is shackled by a coefficient of $(x-1)$.

In the realm of first-order linear differential equations, we demand freedom! We need the standard form $\frac{dy}{dx}+P(x)y=Q(x)$.

So, our first act of mathematical bravery is to divide the entire equation by $(x-1)$. This transforms our equation into:

Now, the path is clear. We have identified our $P(x)=\frac{2x}{x-1}$ and our $Q(x)=\frac{1}{(x-1{)}^2}$.

Now, we enter the most satisfying phase of the journey: finding the Integrating Factor (I.F.). This is the key that unlocks the differential equation.

The formula is $I.F.=e^{\int P(x)dx}$. We need to compute the integral of $\frac{2x}{x-1}$.

Here is where your algebraic intuition shines. Instead of struggling with long division, we perform a clever manipulation: we rewrite the numerator $2x$ as $2(x-1)+2$.

This allows us to split the integral into:

The result is a beautiful $2x+2\ln (x-1)$. When we exponentiate this, we get $e^{2x+2\ln (x-1)}$.

Using the laws of exponents, this simplifies to $e^{2x}\cdot e^{\ln (x-1{)}^2}$, which is simply $e^{2x}(x-1{)}^2$. This is the magic of mathematics—what looked like a nightmare of calculus has collapsed into a clean, elegant expression.

A Moment of Clarity

With our I.F. in hand, we construct the general solution using the formula $y\cdot I.F.=\int Q(x)\cdot I.F.dx+C$.

Substituting our values, we get:

Look closely at the right-hand side. The term $(x-1{)}^2$ in the denominator of $Q(x)$ and the $(x-1{)}^2$ from our I.F. cancel each other out perfectly!

We are left with the simple integral $\int e^{2x}dx$, which is $\frac{e^{2x}}{2}$. Thus, our general solution is:

Finding the Constant

We are almost there. We have the general solution, but we need the specific one. We are given the initial condition $y(2)=\frac{1+e^4}{2e^4}$.

We substitute $x=2$ and $y=\frac{1+e^4}{2e^4}$ into our equation. The left side becomes:

The right side becomes $\frac{e^4}{2}+C$. By equating them, we find that $\frac{1}{2}+\frac{e^4}{2}=\frac{e^4}{2}+C$.

The terms cancel out, leaving us with $C=\frac{1}{2}$. Our specific solution is now fully defined:

Comparison

The final step is to find $y(3)$. We substitute $x=3$ into our specific solution:

This simplifies to $y(3)\cdot 4e^6=\frac{e^6+1}{2}$, which gives us:

Comparing this to the given form $\frac{e^{\alpha }+1}{\beta e^{\alpha }}$, we see that $\alpha =6$ and $\beta =8$.

The sum $\alpha +\beta$ is $6+8=14$. You have navigated the complexity, mastered the algebra, and arrived at the truth. The final answer is 14.