Let y = y(x) be the solution of the differential equation x⁴ dy + (4x³y + 2 sin x)dx = 0, x 0, y(π

Q: Wait, how was I supposed to know to group $x^4 dy$ and $4x^3y dx$ together? Is there a trick?

That is a fantastic question! In JEE Advanced, whenever you see a differential equation with terms like $x^n dy$ and $n x^{n-1} y dx$, it is almost always a signal to look for an exact differential. The pattern $x^n dy + n x^{n-1} y dx$ is the signature of $d(x^n y)$. Train your eyes to spot these patterns; they are designed to reward students who practice recognizing the structure of derivatives.

Q: Can I solve this using the standard linear differential equation method, $\frac{dy}{dx} + P(x)y = Q(x)$?

Yes, you absolutely can! If you divide the entire equation by $x^4 dx$, you get $\frac{dy}{dx} + \frac{4}{x}y = -\frac{2 \sin x}{x^4}$. This is a linear differential equation where the integrating factor is $e^{\int \frac{4}{x} dx} = e^{4 \ln x} = x^4$. Multiplying by $x^4$ brings you back to the same exact differential we identified. While the linear method is robust, recognizing the exact differential is much faster and less prone to calculation errors.

Q: Why did the constant $C$ turn out to be zero? Did I make a mistake?

Not at all! You are doing great. The constant $C$ is determined entirely by the initial condition. When we substituted $x = \frac{\pi}{2}$ and $y = 0$ into $x^4 y = 2 \cos x + C$, we got $0 = 2 \cos(\frac{\pi}{2}) + C$. Since $\cos(\frac{\pi}{2}) = 0$, it forces $C$ to be $0$. It is perfectly normal for constants to vanish in these problems.

Question

Let $y = y(x)$ be the solution of the differential equation $x^4 dy + (4x^3y + 2 \sin x)dx = 0, x > 0, y\left(\frac{\pi}{2}\right) = 0$. Then $\pi^4 y\left(\frac{\pi}{3}\right)$ is equal to :

Options:

(A) 72

(B) 92

(C) 64

(D) 81

Sigma Percentile · Accepted Answer

### The Art of Recognizing Patterns

Welcome, fellow traveler on the road to JEE excellence. Today, we are going to dissect a differential equation that, at first glance, might seem like a daunting wall of variables.

Our equation is $x^4 dy + (4x^3y + 2 \sin x)dx = 0$, with the boundary condition $y(\frac{\pi}{2}) = 0$. Our mission is to find the value of $\pi^4 y(\frac{\pi}{3})$.

### The Detective Phase

Let us start by expanding the terms. When we distribute the $dx$, we get:

$$x^4 dy + 4x^3y dx + 2 \sin x dx = 0$$

Now, look at those first two terms: $x^4 dy + 4x^3y dx$. If you have spent time mastering the product rule, $d(uv) = u dv + v du$, you might feel a spark of recognition.

If we set $u = x^4$ and $v = y$, then:

$$d(x^4 y) = x^4 dy + y d(x^4) = x^4 dy + 4x^3y dx$$

This is the "Aha!" moment. The first two terms are not just random; they are the exact differential of the product $x^4 y$. Recognizing this is the key that unlocks the entire problem.

### The Integration Journey

With this insight, our equation simplifies dramatically. We can rewrite it as:

$$d(x^4 y) + 2 \sin x dx = 0$$

To solve this, we move the trigonometric term to the other side:

$$d(x^4 y) = -2 \sin x dx$$

Now, we integrate both sides. The integral of a differential $d(z)$ is simply $z$. On the right side, the integral of $-2 \sin x$ is $2 \cos x$.

Don't forget the constant of integration, $C$. We now have our general solution:

$$x^4 y = 2 \cos x + C$$

### The Anchor Point

We need to find the specific curve that satisfies our initial condition, $y(\frac{\pi}{2}) = 0$. By substituting $x = \frac{\pi}{2}$ and $y = 0$ into our general solution, we get:

$$\left( \frac{\pi}{2} ight)^4 (0) = 2 \cos\left( \frac{\pi}{2} ight) + C$$

Since $\cos(\frac{\pi}{2}) = 0$, the equation simplifies to $0 = 0 + C$, which means $C = 0$. Our particular solution is simply:

$$x^4 y = 2 \cos x$$

### The Final Stretch

Now, we reach the final act. We need to find the value of $\pi^4 y(\frac{\pi}{3})$. We substitute $x = \frac{\pi}{3}$ into our particular solution:

$$\left( \frac{\pi}{3} ight)^4 y\left( \frac{\pi}{3} ight) = 2 \cos\left( \frac{\pi}{3} ight)$$

We know that $\cos(\frac{\pi}{3}) = \frac{1}{2}$. So, the right side becomes $2 	imes \frac{1}{2} = 1$.

On the left side, we have:

$$\frac{\pi^4}{81} y\left( \frac{\pi}{3} ight) = 1$$

Multiplying both sides by $81$, we arrive at our destination:

$$\pi^4 y\left( \frac{\pi}{3} ight) = \mathbf{81}$$

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $x^{4} d y + (4 x^{3} y + 2 sin x) d x = 0, x > 0, y (\frac{π}{2}) = 0$ . Then $π^{4} y (\frac{π}{3})$ is equal to :

Select Answer:

Visualized Solution

Analyze the Differential Equation

Rearranging the Terms

Identify the Exact Differential $d (x^{4} y)$

Rewrite the Equation

Integrate Both Sides

Apply the Initial Condition $y (\frac{π}{2}) = 0$

Find the Constant of Integration $C$

Set Up for the Target Value $x = \frac{π}{3}$

Evaluate the Target Expression

Final Calculation for $π^{4} y (\frac{π}{3})$

The Sigma Insight: Linear Differential Equations

The Art of Recognizing Patterns

The Detective Phase

The Integration Journey

The Anchor Point

The Final Stretch

Similar Questions

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Select Answer:

Visualized Solution

Analyze the Differential Equation

Rearranging the Terms

Identify the Exact Differential .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d(x4y)

Rewrite the Equation

Integrate Both Sides

Apply the Initial Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(2π​)=0

Find the Constant of Integration .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Set Up for the Target Value .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=3π​

Evaluate the Target Expression

Final Calculation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } π4y(3π​)

The Sigma Insight: Linear Differential Equations

The Art of Recognizing Patterns

The Detective Phase

The Integration Journey

The Anchor Point

The Final Stretch

Similar Questions

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