Animated Solution for Mathematics - Differential Equations: If the solution $y=y(x)$ of the differential equation $(x^4+2x^3+3x^2+2x+2)dy-(2x^2+2x+3)dx=0$ satisfies $y(-1)=-\frac{\pi }{4}$, then $y(0)$ is equal to :

Welcome, fellow traveler of the mathematical landscape. Today, we are going to dismantle a differential equation that, at first glance, looks like a chaotic mess of powers and coefficients.

But as we peel back the layers, you will see that it is actually a beautifully choreographed dance of algebraic symmetry. Let us begin with our given equation:

Our first instinct is to isolate the variables. We want $y$ on one side and $x$ on the other. By rearranging, we get:

Now, look at that denominator. It is a quartic polynomial. If you try to find its roots using the rational root theorem, you might find yourself lost in a sea of trial and error.

But wait—look closer. Do you see the pattern? We can group the terms to reveal a hidden structure:

By factoring out common terms, we get $x^2(x^2+1)+2x(x^2+1)+2(x^2+1)$. Suddenly, the beast is tamed! The denominator is simply $(x^2+1)(x^2+2x+2)$.

Now, we face the fraction:

Most students would immediately write $\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+2x+2}$. But pause. Look at the numerator again.

Note that $2x^2+2x+3$ is exactly $(x^2+1)+(x^2+2x+2)$. This is the 'Aha!' moment. We can split the fraction into two incredibly simple parts:

This is the elegance of JEE-level mathematics—the problem is designed to reward those who look for patterns before they start calculating.

Now, we integrate both sides:

The first term is a standard integral: $\int \frac{1}{x^2+1}dx={\tan }^{-1}(x)$. For the second term, we complete the square: $x^2+2x+2=(x+1{)}^2+1$.

This transforms the integral into $\int \frac{1}{(x+1{)}^2+1}dx$, which is simply ${\tan }^{-1}(x+1)$. Thus, our general solution is:

We are given the initial condition $y(-1)=-\frac{\pi }{4}$. Let us plug in $x=-1$:

Since ${\tan }^{-1}(-1)=-\frac{\pi }{4}$ and ${\tan }^{-1}(0)=0$, we find that $-\frac{\pi }{4}=-\frac{\pi }{4}+C$, which means $C=0$. The constant vanishes, leaving us with the clean, elegant solution $y(x)={\tan }^{-1}(x)+{\tan }^{-1}(x+1)$.

Finally, to find $y(0)$, we substitute $x=0$:

You have navigated the complexity and arrived at the truth. Remember, the math is not just about the final number; it is about the journey of discovery. Keep practicing, keep questioning, and keep falling in love with the logic.