If the solution y(x) of the given differential equation (e^y + 1) cos x dx + e^y sin x dy = 0 passes

Q: Why can't we just integrate the terms as they are without dividing?

Because the variables $x$ and $y$ are multiplied together in the differential terms. In a differential equation, you must isolate the variables so that all $x$ terms are with $dx$ and all $y$ terms are with $dy$. Only then can you perform the integration independently.

Q: Why did we use $\ln C$ instead of just $C$ for the constant of integration?

It is a strategic choice. Since both sides of our equation resulted in natural logarithms, using $\ln C$ allows us to combine the terms using logarithmic properties, making the final algebra much cleaner and avoiding messy exponential constants.

Q: What if the point $(\frac{\pi}{2}, 0)$ was not given?

Without the initial condition, you would have a family of curves defined by $(e^y + 1) \sin x = C$. The point allows us to 'pin down' the specific curve that satisfies the physical reality of the problem, turning a general solution into a particular one.

Question

If the solution $y(x)$ of the given differential equation $(e^y + 1) \cos x dx + e^y \sin x dy = 0$ passes through the point $(\frac{\pi}{2}, 0)$, then the value of $e^{y(\frac{\pi}{6})}$ is equal to_________

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given the differential equation:
$$(e^y + 1) \cos x \, dx + e^y \sin x \, dy = 0$$

At first glance, it looks like a tangled mess of variables. However, in the world of JEE Advanced, complexity is just an invitation to simplify.

### The Art of Separation

We cannot integrate while $x$ and $y$ are dancing together in the same term. We need to isolate them.

By dividing the entire equation by $(e^y + 1) \sin x$, we perform a surgical separation. The equation transforms into:

$$\frac{\cos x}{\sin x} \, dx + \frac{e^y}{e^y + 1} \, dy = 0$$

Suddenly, the chaos vanishes. We have the $x$-world on the left and the $y$-world on the right.

### The Phase of Integration

Now, we integrate both sides:

$$\int \cot x \, dx + \int \frac{e^y}{e^y + 1} \, dy = \int 0$$

The first integral is a standard result: $\int \cot x \, dx = \ln |\sin x|$.

The second integral is where the beauty of calculus shines. Notice that the numerator, $e^y$, is the derivative of the denominator, $e^y + 1$. This is the classic $\frac{f'(y)}{f(y)}$ form, which integrates to $\ln (e^y + 1)$.

We combine these using the property $\ln A + \ln B = \ln (AB)$, giving us:

$$\ln |(e^y + 1) \sin x| = \ln C$$

By exponentiating both sides, we arrive at the general solution:

$$(e^y + 1) \sin x = C$$

### Applying Boundary Conditions

We are given the point $(\frac{\pi}{2}, 0)$. Substituting $x = \frac{\pi}{2}$ and $y = 0$, we get:

$$(e^0 + 1) \sin \frac{\pi}{2} = C$$

Since $e^0 = 1$ and $\sin \frac{\pi}{2} = 1$, we find that $C = 2$. Our particular solution is:

$$(e^y + 1) \sin x = 2$$

### Final Calculation

Finally, we seek the value of $e^{y(\frac{\pi}{6})}$. Substituting $x = \frac{\pi}{6}$, we have:

$$(e^{y(\frac{\pi}{6})} + 1) \sin \frac{\pi}{6} = 2$$

Since $\sin \frac{\pi}{6} = \frac{1}{2}$, the equation becomes:

$$(e^{y(\frac{\pi}{6})} + 1) \cdot \frac{1}{2} = 2$$

Multiplying by 2, we get $e^{y(\frac{\pi}{6})} + 1 = 4$. Subtracting 1, we arrive at the elegant conclusion:

**$e^{y(\frac{\pi}{6})} = 3$**

Animated Solution for Mathematics - Differential Equations: If the solution $y (x)$ of the given differential equation $(e^{y} + 1) cos x d x + e^{y} sin x d y = 0$ passes through the point $(\frac{π}{2}, 0)$ , then the value of $e^{y (\frac{π}{6})}$ is equal to_________

Enter Numerical Value:

Visualized Solution

Given Differential Equation

Separating Variables

Integration Setup

Executing Integration

Logarithmic Properties

General Solution

Applying Initial Condition

Finding $C$

Particular Solution

Substituting $x = \frac{π}{6}$

Final Answer

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Art of Separation

The Phase of Integration

Applying Boundary Conditions

Final Calculation

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