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 $\left(\frac{\pi}{2}, 0\right)$, then the value of $e^{y\left(\frac{\pi}{6}\right)}$ is equal to ________.

Answer 1

Correct Answer: 3

Starting with the differential equation:
\[(e^y + 1) \cos x \, dx + e^y \sin x \, dy = 0\]
Rewrite as:
\[\implies d \left( (e^y + 1) \sin x \right) = 0\]
Integrating, we get:
\[(e^y + 1) \sin x = C\]
Since the solution passes through \( \left( \frac{\pi}{2}, 0 \right) \), substitute \( x = \frac{\pi}{2} \) and \( y = 0 \):
\[e^0 + 1 = C \implies C = 2\]
Now, let \( x = \frac{\pi}{6} \):
\[(e^y + 1) \sin \frac{\pi}{6} = 2\]
\[\implies \frac{e^y + 1}{2} = 2\]
\[\implies e^y = 3\]
Thus, \( e^{y \left( \frac{\pi}{6} \right)} = 3 \).

Answer 2

The problem asks for the value of \(e^{y(\frac{\pi}{6})}\) where \(y(x)\) is the solution to the differential equation \((e^y + 1) \cos x \, dx + e^y \sin x \, dy = 0\), given the initial condition that the solution passes through the point \((\frac{\pi}{2}, 0)\).

Concept Used:

The given differential equation is a first-order, first-degree differential equation. We can solve it using the method of separation of variables. This method involves rearranging the equation so that all terms involving \(y\) are on one side with \(dy\), and all terms involving \(x\) are on the other side with \(dx\). After separation, we integrate both sides to find the general solution.

\[ \int g(y) \, dy = \int f(x) \, dx + C \]

where \(C\) is the constant of integration, which can be found using the given initial condition.

Step-by-Step Solution:

Step 1: Separate the variables in the differential equation.

The given equation is:

\[ (e^y + 1) \cos x \, dx + e^y \sin x \, dy = 0 \]

Rearranging the terms to separate \(x\) and \(y\):

\[ e^y \sin x \, dy = -(e^y + 1) \cos x \, dx \]

Divide both sides by \((e^y + 1)\) and \(\sin x\) to group the variables:

\[ \frac{e^y}{e^y + 1} \, dy = -\frac{\cos x}{\sin x} \, dx \]

Step 2: Integrate both sides of the separated equation.

Integrating the left side with respect to \(y\) and the right side with respect to \(x\):

\[ \int \frac{e^y}{e^y + 1} \, dy = \int -\frac{\cos x}{\sin x} \, dx \]

For the left integral, let \(u = e^y + 1\), so \(du = e^y dy\). The integral becomes \(\int \frac{1}{u} du = \ln|u| = \ln(e^y + 1)\) (since \(e^y+1\) is always positive).

For the right integral, we have \(-\int \cot x \, dx = -\ln|\sin x|\).

Combining the results, we get the general solution:

\[ \ln(e^y + 1) = -\ln|\sin x| + C \]

Step 3: Use the initial condition to find the constant \(C\).

The solution passes through the point \((\frac{\pi}{2}, 0)\), so we substitute \(x = \frac{\pi}{2}\) and \(y = 0\) into the general solution.

\[ \ln(e^0 + 1) = -\ln\left|\sin\left(\frac{\pi}{2}\right)\right| + C \] \[ \ln(1 + 1) = -\ln|1| + C \] \[ \ln(2) = -0 + C \implies C = \ln(2) \]

Step 4: Write the particular solution and simplify.

Substituting \(C = \ln(2)\) back into the general solution:

\[ \ln(e^y + 1) = -\ln|\sin x| + \ln(2) \]

Using the properties of logarithms, we can combine the terms on the right side:

\[ \ln(e^y + 1) = \ln\left(\frac{2}{|\sin x|}\right) \]

By taking the exponential of both sides, we get:

\[ e^y + 1 = \frac{2}{|\sin x|} \]

Step 5: Calculate the value of \(e^{y(\frac{\pi}{6})}\).

We need to find the value of \(e^y\) when \(x = \frac{\pi}{6}\). Substitute \(x = \frac{\pi}{6}\) into the particular solution. Note that \(\sin(\frac{\pi}{6}) = \frac{1}{2}\), which is positive.

\[ e^{y(\frac{\pi}{6})} + 1 = \frac{2}{\sin(\frac{\pi}{6})} \] \[ e^{y(\frac{\pi}{6})} + 1 = \frac{2}{1/2} = 4 \]

Now, solve for \(e^{y(\frac{\pi}{6})}\):

\[ e^{y(\frac{\pi}{6})} = 4 - 1 = 3 \]

The value of \(e^{y\left(\frac{\pi}{6}\right)}\) is 3.