Let y = y(x) be the solution of the differential equation sec x dy/dx - 2y = 2 + 3 sin x, x ∈ (-π/2,

Q: Why do we multiply by $\cos x$ at the beginning?

Great question! The standard form for a linear differential equation is $\frac{dy}{dx} + P(x)y = Q(x)$. In our problem, the $\sec x$ term is attached to the $\frac{dy}{dx}$, which prevents us from using the standard integrating factor method directly. Since $\sec x = \frac{1}{\cos x}$, multiplying by $\cos x$ is the most efficient way to 'clean' the derivative term and bring the equation into the standard form.

Q: How do I know when to use Integration by Parts?

Integration by Parts is your best friend when you see a product of two different types of functions that don't easily simplify with basic substitution. Here, we had an algebraic part $(2 + 3t)$ and an exponential part $e^{-2t}$. The ILATE rule (Inverse trig, Logarithmic, Algebraic, Trigonometric, Exponential) suggests choosing the algebraic part as $u$ and the exponential part as $dv$ to reduce the complexity of the integral.

Q: What happens if I forget the constant $C$?

If you forget the constant of integration $C$, you will end up with a general solution rather than the specific, particular solution required by the initial condition $y(0) = -\frac{7}{4}$. Without $C$, you cannot satisfy the initial condition, and your final answer will be incorrect. Always remember that $C$ is the 'memory' of the differential equation's general family of curves!

Question

Let $y = y(x)$ be the solution of the differential equation $\sec x \frac{dy}{dx} - 2y = 2 + 3 \sin x, x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right), y(0) = -\frac{7}{4}$. Then $y\left(\frac{\pi}{6}\right)$ is equal to :

Options:

(A) $-\frac{5}{4}$

(B) $-\frac{5}{2}$

(C) $-3\sqrt{3} - 7$

(D) $-3\sqrt{2} - 7$

Sigma Percentile · Accepted Answer

### The Art of Unmasking: Solving the Differential Equation

Differential equations often feel like riddles wrapped in a layer of complexity. When you first look at the equation $\sec x \frac{dy}{dx} - 2y = 2 + 3 \sin x$, it might seem intimidating.

But in the world of JEE Advanced, complexity is often just a mask. Our goal is to peel back that mask and reveal the elegant structure underneath.

### Phase 1: The Standardization

The first step is to bring order to chaos. We have a $\sec x$ term interfering with our derivative $\frac{dy}{dx}$.

Since $\sec x$ is the reciprocal of $\cos x$, multiplying the entire equation by $\cos x$ is the perfect move. This transforms our equation into:

$$\frac{dy}{dx} - 2 \cos x \cdot y = (2 + 3 \sin x) \cos x$$

Now, we have a clear linear form: $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = -2 \cos x$ and $Q(x) = (2 + 3 \sin x) \cos x$.

Notice how the negative sign in $P(x)$ is crucial; missing it is a classic trap that can derail your entire calculation.

### Phase 2: The Integrating Factor

With our $P(x)$ identified, we need the 'magic key'—the Integrating Factor ($IF$). The formula is $IF = e^{\int P(x) dx}$.

Substituting our $P(x)$, we get $IF = e^{\int -2 \cos x dx}$. Since the integral of $\cos x$ is $\sin x$, our $IF$ becomes:

$$IF = e^{-2 \sin x}$$

This function is the bridge that allows us to collapse the left side of our differential equation into the derivative of a product: $\frac{d}{dx} (y \cdot e^{-2 \sin x})$.

### Phase 3: The Transformation

Now, we set up the general solution:

$$y \cdot e^{-2 \sin x} = \int (2 + 3 \sin x) \cos x \cdot e^{-2 \sin x} dx + C$$

This integral looks daunting, but look closely at the structure. We have $\sin x$ and its derivative $\cos x$ sitting right there. This is a clear signal for substitution.

Let $t = \sin x$, which implies $dt = \cos x dx$. The integral transforms into:

$$\int (2 + 3t) e^{-2t} dt$$

This is much friendlier! We use Integration by Parts, where we let $u = 2 + 3t$ and $dv = e^{-2t} dt$.

Following the formula $\int u dv = uv - \int v du$, we find that the integral evaluates to:

$$e^{-2t} \left( -\frac{7}{4} - \frac{3}{2} t \right)$$

Substituting $\sin x$ back for $t$, we get our integral result.

### Phase 4: The Final Reveal

After dividing by the $IF$, we arrive at the general solution:

$$y(x) = -\frac{7}{4} - \frac{3}{2} \sin x + C e^{2 \sin x}$$

We are given the initial condition $y(0) = -\frac{7}{4}$. Plugging in $x=0$, we find that $\sin(0) = 0$ and $e^0 = 1$, which leads to $-\frac{7}{4} = -\frac{7}{4} + C$.

Thus, $C = 0$. Our particular solution is simply $y(x) = -\frac{7}{4} - \frac{3}{2} \sin x$.

Finally, evaluating at $x = \frac{\pi}{6}$, we use $\sin(\frac{\pi}{6}) = \frac{1}{2}$ to get:

$$y\left(\frac{\pi}{6}\right) = -\frac{7}{4} - \frac{3}{2}\left(\frac{1}{2}\right) = -\frac{7}{4} - \frac{3}{4} = -\frac{10}{4} = -\frac{5}{2}$$

We have successfully navigated the complexity and arrived at the elegant solution of **$-\frac{5}{2}$**.

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $sec x \frac{d y}{d x} - 2 y = 2 + 3 sin x, x \in (- \frac{π}{2}, \frac{π}{2}), y (0) = - \frac{7}{4}$ . Then $y (\frac{π}{6})$ is equal to :

Select Answer:

Visualized Solution

Standardizing the Differential Equation

Identifying $P (x)$ and $Q (x)$

Calculating the Integrating Factor ( $I F$ )

Setting up the General Solution

Substitution for the Integral

Integration by Parts

Completing the Integration

The General Solution Equation

Applying Initial Condition $y (0) = - \frac{7}{4}$

The Particular Solution

Final Calculation for $y (\frac{π}{6})$

The Sigma Insight: Linear Differential Equations

The Art of Unmasking

Phase 1

Phase 2

Phase 3

Phase 4

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Select Answer:

Visualized Solution

Standardizing the Differential Equation

Identifying .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)

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