Animated Solution for Mathematics - Differential Equations: If $x=x(y)$ is the solution of the differential equation $y\frac{dx}{dy}=2x+y^3(y+1)e^y,x(1)=0$; then $x(e)$ is equal to :

Welcome, future engineer. Today, we are going to dissect a problem that, at first glance, might make your heart skip a beat. You see a differential equation like $y\frac{dx}{dy}=2x+y^3(y+1)e^y$, and the instinct is to panic.

You might think, "Wait, is this separable? Is it homogeneous?" But I want you to take a deep breath. In the world of JEE Advanced, the most complex-looking problems often hide a very elegant, very simple structure. The secret is not to solve, but to observe.

The Anatomy of the Equation

Look at the equation again. We have $\frac{dx}{dy}$. This is our first clue. Usually, we are trained to see $\frac{dy}{dx}$, but here, the roles are reversed. We are treating $y$ as our independent variable and $x$ as our dependent variable.

This is a Linear Differential Equation in disguise. To reveal its true form, we must bring it into the standard shape:

By dividing the entire equation by $y$, we get:

Now, move the $x$ term to the left. We get:

Suddenly, the chaos settles. We have identified our $P(y)=-\frac{2}{y}$ and our $Q(y)=y^2(y+1)e^y$. This is the moment where you gain control over the problem.

The Integrating Factor (The Hero)

Now, we need the magic tool: the Integrating Factor ($I.F.$). The formula is $I.F.=e^{\int P(y)dy}$.

Let's calculate this carefully. We have $\int -\frac{2}{y}dy$, which is $-2\ln (y)$. Using the laws of logarithms, this becomes $\ln (y^{-2})$.

When we raise $e$ to this power, we get $e^{\ln (y^{-2})}$, which simplifies beautifully to $y^{-2}$ or $\frac{1}{y^2}$.

This $I.F.$ is the key that unlocks the door. When we multiply our entire differential equation by $\frac{1}{y^2}$, the left side transforms into the derivative of the product $x\cdot (I.F.)$.

It is not just algebra; it is a geometric alignment. We are essentially collapsing the complexity of the equation into a single, manageable derivative.

The Integration (The 'Aha!' Moment)

After multiplying, our equation becomes:

Now, look at that integral. Many students would jump straight into Integration by Parts. But wait! Look closer.

We have $e^y$ multiplied by $(y+1)$. This is the classic special form:

If we set $f(y)=y$, then $f^{\prime }(y)=1$. The expression $(y+1)$ is exactly $f(y)+f^{\prime }(y)$.

The integral evaluates instantly to $y{e}^y+C$. This is the kind of elegance that JEE examiners love. It rewards the student who looks for patterns rather than brute-forcing calculations.

The Final Stretch

We are almost there. We have:

Multiplying by $y^2$, we get the general solution:

Now, we apply the initial condition $x(1)=0$. Substituting $y=1$ and $x=0$, we find:

Our particular solution is $x=y^3{e}^y-e{y}^2$. Finally, the question asks for $x(e)$.

Substituting $y=e$, we get:

Factoring out $e^3$, we arrive at the final answer:

This problem was not a test of your endurance, but a test of your ability to recognize patterns. You started with a messy equation and, through systematic steps, revealed a beautiful, clean answer. Keep this mindset—always look for the structure, always look for the pattern, and you will conquer any problem they throw at you.