Animated Solution for Mathematics - Differential Equations: Let $y=y(x)$ be the solution of the differential equation, $x{y}^{\prime }-y=x^2(x\cos x+\sin x)$, $x>0$. If $y(\pi )=\pi$, then $y^{\prime \prime }\left(\frac{\pi }{2}\right)+y\left(\frac{\pi }{2}\right)$ is equal to :

Imagine you are standing at the edge of a vast, uncharted mathematical landscape. Before you lies the differential equation $x{y}^{\prime }-y=x^2(x\cos x+\sin x)$.

At first glance, it might seem like a chaotic jumble of variables, a tangled mess of $x$ and $y$ that refuses to yield. But do not be intimidated; in the world of JEE Advanced, every complex problem is just a puzzle waiting for the right key.

The first step in our journey is to bring order to the chaos. We are looking at a first-order differential equation, and our goal is to mold it into the standard linear form:

To isolate $y^{\prime }$, we must divide the entire equation by $x$. Since the problem guarantees $x>0$, we can do this without fear of dividing by zero.

Dividing every term by $x$, we get:

Now, the chaos has vanished, replaced by a clean, elegant structure where $P(x)=-\frac{1}{x}$ and $Q(x)=x(x\cos x+\sin x)$.

Now that we have our $P(x)$, we need the magic wand of differential equations: the Integrating Factor ($I.F.$). The formula is defined as:

Substituting our $P(x)$, we have $e^{\int -\frac{1}{x}dx}$. The integral of $\frac{1}{x}$ is $\ln x$, so we get $e^{-\ln x}$.

Using the properties of logarithms, this simplifies beautifully to $e^{\ln (x^{-1})}$, which is simply $\frac{1}{x}$. This is our key.

When we multiply our standardized equation by $\frac{1}{x}$, the left side undergoes a miraculous transformation:

If you look closely, the left side is the exact derivative of the product $\frac{y}{x}$. We have successfully collapsed the equation into:

Now, we integrate both sides with respect to $x$. On the left, the integral of a derivative is just the function itself: $\frac{y}{x}$.

On the right, we face $\int (x\cos x+\sin x)dx$. Recognizing that $x\cos x+\sin x$ is the derivative of $x\sin x$ saves us precious time and effort.

So, we have:

Multiplying by $x$, we arrive at our general solution:

We have a family of curves, but we need the specific one that satisfies $y(\pi )=\pi$. Substituting $x=\pi$ and $y=\pi$ into our general solution, we get:

Since $\sin (\pi )=0$, the equation simplifies to $\pi =C\pi$, which means $C=1$. Our particular solution is:

The question asks for $y^{\prime \prime }\left(\frac{\pi }{2}\right)+y\left(\frac{\pi }{2}\right)$. First, let's find $y\left(\frac{\pi }{2}\right)$:

Next, we differentiate $y$ to find $y^{\prime }$. Using the product rule:

Differentiating again for $y^{\prime \prime }$, we get:

Evaluating $y^{\prime \prime }$ at $x=\frac{\pi }{2}$:

Finally, we add them:

The $\frac{{\pi }^2}{4}$ terms cancel out, leaving us with the final result: $2+\frac{\pi }{2}$.