Let y = y(x) be solution of the following differential equation eʸ dy/dx - 2eʸ sin x + sin x cos² x

Q: Why can't we just separate the variables $x$ and $y$?

Separation of variables is a powerful tool, but it only works if you can isolate all $y$ terms on one side and all $x$ terms on the other. Here, the term $-2e^y \sin x$ couples the variables in a way that makes simple separation impossible. That is why we look for linear forms.

Q: How do I know when to use the $e^y = t$ substitution?

Whenever you see a function $f(y)$ and its derivative $f'(y)$ appearing in the same equation, it is a massive red flag that a substitution is required. Here, $f(y) = e^y$ and $f'(y) = e^y \frac{dy}{dx}$. Recognizing this pattern is the hallmark of a seasoned problem solver.

Q: Is there a trick to solving $\int u^2 e^{2u} du$ without doing integration by parts twice?

Yes! You can use the tabular method or the generalized formula $\int u^n e^{au} du = \frac{e^{au}}{a} \sum_{k=0}^{n} (-1)^k \frac{n!}{(n-k)!} \frac{u^{n-k}}{a^k}$. However, for $n=2$, standard integration by parts is perfectly safe and less prone to memorization errors.

Question

Let $y = y(x)$ be solution of the following differential equation $e^y \frac{dy}{dx} - 2e^y \sin x + \sin x \cos^2 x = 0, y\left(\frac{\pi}{2}\right) = 0$. If $y(0) = \log_e(\alpha + \beta e^{-2})$, then $4(\alpha + \beta)$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow problem solvers. Today, we stand before a differential equation that might seem daunting at first glance:

$$e^y \frac{dy}{dx} - 2e^y \sin x + \sin x \cos^2 x = 0$$

In the arena of JEE Advanced, the first step is rarely calculation; it is observation. We have $e^y$ and its derivative $e^y \frac{dy}{dx}$ present in the equation.

Whenever you see a function and its derivative in the same equation, your mathematical intuition should scream **substitution**. By setting $t = e^y$, we transform this non-linear looking beast into a familiar, friendly linear differential equation.

### The Transformation: Entering the Linear Realm

Let us execute the substitution $t = e^y$. Differentiating with respect to $x$, we get:

$$\frac{dt}{dx} = e^y \frac{dy}{dx}$$

Substituting this into our original equation, we obtain:

$$\frac{dt}{dx} - 2t \sin x + \sin x \cos^2 x = 0$$

Rearranging this into the standard linear form $\frac{dt}{dx} + P(x)t = Q(x)$, we find:

$$\frac{dt}{dx} - (2\sin x)t = -\sin x \cos^2 x$$

Here, $P(x) = -2\sin x$ and $Q(x) = -\sin x \cos^2 x$. We have successfully tamed the equation and are ready to apply the standard solution method.

### The Engine: Integrating Factor

The heart of any linear differential equation is the **Integrating Factor ($I.F.$)**. It is the bridge that allows us to integrate the equation.

We calculate it as:

$$I.F. = e^{\int P(x) dx} = e^{\int -2\sin x dx} = e^{2\cos x}$$

This is the key that unlocks the solution. We multiply our entire equation by this factor, leading to the general solution template:

$$t \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + c$$

Substituting our values, we get:

$$t e^{2\cos x} = \int (-\sin x \cos^2 x) e^{2\cos x} dx + c$$

### The Final Stretch: Integration and Boundary Conditions

Now, we evaluate the integral $\int (-\sin x \cos^2 x) e^{2\cos x} dx$. Let $u = \cos x$, then $du = -\sin x dx$.

The integral transforms into $\int u^2 e^{2u} du$. Using integration by parts, we solve this to get:

$$\frac{e^{2u}}{4} (2u^2 - 2u + 1)$$

Substituting back $u = \cos x$ and $t = e^y$, we arrive at the general solution:

$$e^y e^{2\cos x} = \frac{e^{2\cos x}}{4}(2\cos^2 x - 2\cos x + 1) + c$$

Finally, we use the initial condition $y(\frac{\pi}{2}) = 0$ to find $c$. At $x = \frac{\pi}{2}$, $\cos x = 0$, so:

$$1 \cdot 1 = \frac{1}{4}(0 - 0 + 1) + c \Rightarrow c = \frac{3}{4}$$

With $c$ in hand, finding $y(0)$ is a matter of arithmetic. Substituting $x=0$ and $c=\frac{3}{4}$ leads us to:

$$e^{y(0)} = \frac{1}{4} + \frac{3}{4} e^{-2}$$

Comparing this to the given form $\log_e(\alpha + \beta e^{-2})$, we identify $\alpha = \frac{1}{4}$ and $\beta = \frac{3}{4}$. The final result, **$4(\alpha + \beta)$**, is simply **$4$**.

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be solution of the following differential equation $e^{y} \frac{d y}{d x} - 2 e^{y} sin x + sin x cos^{2} x = 0, y (\frac{π}{2}) = 0$ . If $y (0) = lo g_{e} (α + β e^{- 2})$ , then $4 (α + β)$ is equal to

Enter Numerical Value:

Visualized Solution

Analyze the Differential Equation

Substitution: $e^{y} = t$

Standard Linear Form

Calculating Integrating Factor (I.F.)

General Solution Setup

Solving the Integral

Back Substitution for $t$

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

Finding $y (0)$

Comparing and Final Answer

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Transformation

The Engine

The Final Stretch

Similar Questions

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Enter Numerical Value:

Visualized Solution

Analyze the Differential Equation

Substitution: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ey=t

Standard Linear Form

Calculating Integrating Factor (I.F.)

General Solution Setup

Solving the Integral

Back Substitution for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

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

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