If y = y(x) is the solution of the equation e^(sin y) cos y (dy/dx) + e^(sin y) cos x = cos x, y(0)

Q: Wait, why do we reject the other solutions for $\sin y = 0$? Couldn't $y$ be $n\pi$?

That is a fantastic observation! Mathematically, $\sin y = 0$ implies $y = n\pi$. However, we are given the initial condition $y(0) = 0$. Since the differential equation is continuous and the initial condition pins the solution at $0$, the only continuous function satisfying this is the constant function $y(x) = 0$. If $y$ were to jump to $\pi$ or $-\pi$, it would violate the continuity required for the derivative to exist everywhere.

Q: Why can't we just integrate the original equation directly? It looks like it might be separable.

It is tempting to try separation of variables, but look closely at the term $e^{\sin y} \cos x$. The variables $x$ and $y$ are 'entangled' in a way that prevents simple separation. The substitution $t = e^{\sin y}$ is the 'key' that untangles this knot, turning a non-linear mess into a linear, solvable structure.

Q: Is it always this simple? The final answer is just 1?

In JEE Advanced, the complexity often lies in the 'journey' rather than the final number. The problem tests your ability to recognize patterns and maintain composure. The fact that the answer is 1 is a reward for your rigorous algebraic manipulation. Trust the process, and the math will reveal its elegance.

Question

If $y = y(x)$ is the solution of the equation $e^{\sin y} \cos y \frac{dy}{dx} + e^{\sin y} \cos x = \cos x, y(0) = 0$; then $1+y(\frac{\pi}{2}) + \sqrt{\frac{\pi}{2}} y'(\frac{\pi}{2}) + \frac{1}{\sqrt{2}} y''(\frac{\pi}{2})$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The given differential equation is:

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

In the heat of the JEE Advanced exam, your first instinct might be panic. You see exponentials, trigonometric functions, and a derivative all tangled together. However, the secret to mastering differential equations is to stop looking at the equation as a whole and start looking at the relationships between its parts.

### The Spark of Substitution

Look at the term $e^{\sin y}$ and the term $\cos y \frac{dy}{dx}$. The latter is the derivative of $\sin y$. This is the 'Spark' of the problem.

We define a new variable $t = e^{\sin y}$. When we differentiate this with respect to $x$, we apply the chain rule:

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

By substituting this back into our original equation, we transform a non-linear nightmare into a linear dream:

$$\frac{dt}{dx} + t \cos x = \cos x$$

This is a standard linear differential equation of the form $\frac{dt}{dx} + P(x)t = Q(x)$, where $P(x) = \cos x$ and $Q(x) = \cos x$.

### The Machinery of Integration

Now that we have a linear form, we calculate the Integrating Factor (I.F.):

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

We multiply the entire equation by this factor:

$$e^{\sin x} \frac{dt}{dx} + t \cos x e^{\sin x} = \cos x e^{\sin x}$$

Notice that the left side is the derivative of the product $t e^{\sin x}$. Thus, we have:

$$\frac{d}{dx} (t e^{\sin x}) = \cos x e^{\sin x}$$

Integrating both sides, we get:

$$t e^{\sin x} = \int \cos x e^{\sin x} dx$$

The integral on the right is solved by substituting $u = \sin x$, which gives $du = \cos x dx$. The integral becomes $\int e^u du = e^u + C = e^{\sin x} + C$. Thus:

$$t e^{\sin x} = e^{\sin x} + C$$

### The Final Revelation

Substituting back $t = e^{\sin y}$, we obtain:

$$e^{\sin y} e^{\sin x} = e^{\sin x} + C$$

Applying the initial condition $y(0) = 0$:

$$e^{\sin 0} e^{\sin 0} = e^{\sin 0} + C \Rightarrow 1 \cdot 1 = 1 + C \Rightarrow C = 0$$

The equation simplifies to $e^{\sin y} e^{\sin x} = e^{\sin x}$. Since $e^{\sin x}$ is never zero, we divide by it to get $e^{\sin y} = 1$, which implies $\sin y = 0$. Given the initial condition, **$y(x) = 0$** is the only continuous solution.

The final expression $1 + y(\frac{\pi}{2}) + \sqrt{\frac{\pi}{2}} y'(\frac{\pi}{2}) + \frac{1}{\sqrt{2}} y''(\frac{\pi}{2})$ evaluates to:

$$1 + 0 + 0 + 0 = \mathbf{1}$$

Animated Solution for Mathematics - Differential Equations: If $y = y (x)$ is the solution of the equation $e^{s i n y} cos y \frac{d y}{d x} + e^{s i n y} cos x = cos x, y (0) = 0$ ; then $1 + y (\frac{π}{2}) + \frac{π}{2} y^{'} (\frac{π}{2}) + \frac{1}{2} y^{''} (\frac{π}{2})$ is equal to

Enter Numerical Value:

Visualized Solution

The Given Differential Equation

Substitution $t = e^{s i n y}$

Transforming to Linear Form

Calculating the Integrating Factor

General Solution Setup & Integration

Substituting back $t$

Applying Initial Condition $y (0) = 0$

Simplifying the Solution

Deducing $y (x) = 0$

Finding Derivatives at $x = \frac{π}{2}$

Final Evaluation

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Spark of Substitution

The Machinery of Integration

The Final Revelation

Similar Questions

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Let $y^{'} (x) + y (x) g^{'} (x) = g (x) g^{'} (x)$ , $y (0) = 0$ , $x \in R$ , where $f^{'} (x)$ denotes $\frac{df ( x )}{d x}$ and $g (x)$ is a given non-constant differentiable function on $R$ with $g (0) = g (2) = 0$ . Then the value of $y (2)$ is

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Enter Numerical Value:

Visualized Solution

The Given Differential Equation

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

Transforming to Linear Form

Calculating the Integrating Factor

General Solution Setup & Integration

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

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

Simplifying the Solution

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

Finding Derivatives at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2π​