Let y = y(x) be the solution of the differential equation dy/dx + 3(tan² x)y + 3y = sec² x, y(0) = 1

Q: Wait, why can't I just separate the variables $x$ and $y$ at the very beginning?

That is a very common instinct! However, separation of variables only works if you can write the equation as $f(y)dy = g(x)dx$. Here, the $y$ term is trapped in a sum with the derivative $\frac{dy}{dx}$. Because the equation is linear in $y$ and $\frac{dy}{dx}$, we must use the Integrating Factor method. Trying to separate variables here will lead you into an algebraic dead end.

Q: I always get confused about the Integrating Factor. Is it $e^{\int P(x) dx}$ or $e^{\int Q(x) dx}$?

Great question! Always remember: the Integrating Factor is built solely from the coefficient of $y$, which is $P(x)$. Think of it as the 'weight' that balances the equation so that the left side becomes a perfect derivative. $Q(x)$ is the 'driving force' on the right side, which we only deal with *after* we have found the $IF$.

Q: Why do we need the initial condition $y(0) = \frac{1}{3} + e^3$?

Differential equations describe a *family* of curves. Without the initial condition, we have an arbitrary constant $C$ in our solution, meaning we have an infinite number of possible curves. The initial condition acts like a 'pin' that fixes our specific curve in the coordinate plane, allowing us to determine the exact value of $C$.

Question

Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} + 3(	an^2 x)y + 3y = \sec^2 x$, $y(0) = \frac{1}{3} + e^3$. Then $y(\frac{\pi}{4})$ is equal to

Options:

(A) $\frac{2}{3}$

(B) $\frac{4}{3}$

(C) $\frac{4}{3} + e^3$

(D) $\frac{2}{3} + e^3$

Sigma Percentile · Accepted Answer

### The Symphony of the Differential Equation

Welcome, future engineer. Today, we are not just solving a differential equation; we are embarking on a journey to uncover the hidden structure within a seemingly chaotic expression.

When you first look at the equation

$$\frac{dy}{dx} + 3(	an^2 x)y + 3y = \sec^2 x$$

it is natural to feel a moment of hesitation. It looks cluttered and intimidating, but in the world of JEE Advanced, intimidation is often just a mask for elegance.

### Phase 1: The Hidden Identity

Let us pause and observe. We have the term $3(	an^2 x)y$ and the term $3y$. Both terms share a common factor of $3y$.

When we factor this out, we get $3y(	an^2 x + 1)$. Recalling the fundamental trigonometric identity $1 + 	an^2 x = \sec^2 x$, the complexity collapses.

Our equation transforms into:

$$\frac{dy}{dx} + 3(\sec^2 x)y = \sec^2 x$$

This is now a perfectly structured **Linear Differential Equation** of the form $\frac{dy}{dx} + P(x)y = Q(x)$.

### Phase 2: The Integrating Factor

We identify $P(x) = 3\sec^2 x$ and $Q(x) = \sec^2 x$. We now calculate the Integrating Factor ($IF$), which is the key to unlocking the left side of the equation.

The $IF$ is defined as:

$$IF = e^{\int P(x) dx} = e^{\int 3\sec^2 x dx}$$

Since the integral of $\sec^2 x$ is $	an x$, our Integrating Factor becomes:

$$IF = e^{3	an x}$$

### Phase 3: The Integration

Multiplying the entire equation by $e^{3	an x}$, the left side becomes the derivative of a product:

$$\frac{d}{dx}(y \cdot e^{3	an x}) = \sec^2 x \cdot e^{3	an x}$$

To solve for $y$, we integrate both sides:

$$y \cdot e^{3	an x} = \int \sec^2 x \cdot e^{3	an x} dx$$

Using the substitution $u = 3	an x$, we have $du = 3\sec^2 x dx$, or $\sec^2 x dx = \frac{du}{3}$. The integral simplifies to:

$$\int e^u \frac{du}{3} = \frac{1}{3} e^u + C = \frac{1}{3} e^{3	an x} + C$$

### Phase 4: The Boundary Condition

Our general solution is:

$$y \cdot e^{3	an x} = \frac{1}{3} e^{3	an x} + C \implies y = \frac{1}{3} + C e^{-3	an x}$$

We apply the initial condition $y(0) = \frac{1}{3} + e^3$. Substituting $x=0$ and $y=\frac{1}{3} + e^3$:

$$\frac{1}{3} + e^3 = \frac{1}{3} + C e^0 \implies C = e^3$$

Thus, the specific solution is:

$$y(x) = \frac{1}{3} + e^3 \cdot e^{-3	an x} = \frac{1}{3} + e^{3(1 - 	an x)}$$

### The Final Step

Finally, we evaluate the function at $x = \frac{\pi}{4}$. Since $	an(\frac{\pi}{4}) = 1$, the exponent becomes $3(1 - 1) = 0$.

$$y\left(\frac{\pi}{4}ight) = \frac{1}{3} + e^0 = \frac{1}{3} + 1 = \frac{4}{3}$$

The final result is:

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

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + 3 (tan^{2} x) y + 3 y = sec^{2} x$ , $y (0) = \frac{1}{3} + e^{3}$ . Then $y (\frac{π}{4})$ is equal to

Select Answer:

Visualized Solution

Analyze the Differential Equation

Simplify using Trigonometric Identity

Identify $P (x)$ and $Q (x)$

Calculate the Integrating Factor ( $I F$ )

Set up the General Solution

Solve the Integral using Substitution

Write the General Solution for $y (x)$

Apply Initial Condition $y (0)$

Find the Constant $C$

Calculate $y (\frac{π}{4})$

Final Conclusion

The Sigma Insight: Linear Differential Equations

The Symphony of the Differential Equation

Phase 1

Phase 2

Phase 3

Phase 4

The Final Step

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Select Answer:

Visualized Solution

Analyze the Differential Equation

Simplify using Trigonometric Identity

Identify .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(x) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q(x)

Calculate the Integrating Factor ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } IF)

Set up the General Solution

Solve the Integral using Substitution

Write the General Solution for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(x)

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

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

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(4π​)