Solve the following initial value equations :
`(1+e^(2x))dy+(1+y^(2))e^(x)dx=0`, given that `x=0, y=1`.

Options

A. `e^(y)x=1`
B. `e^(y)=1`
C. `e^(x)=1`
D. `e^(x)y=1`

Answer

To solve the initial value equation

\[ (1 + e^{2x}) dy + (1 + y^2)e^x dx = 0 \]

with the initial condition \( x = 0, y = 1 \), we can follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the equation to separate the variables \(y\) and \(x\):

\[ (1 + e^{2x}) dy = - (1 + y^2)e^x dx \]

### Step 2: Separating Variables
Now, we can separate the variables:

\[ \frac{dy}{1 + y^2} = -\frac{e^x}{1 + e^{2x}} dx \]

### Step 3: Integrating Both Sides
Next, we integrate both sides. The left side integrates to:

\[ \int \frac{dy}{1 + y^2} = \tan^{-1}(y) + C_1 \]

For the right side, we simplify the integrand:

\[ -\int \frac{e^x}{1 + e^{2x}} dx \]

Using the substitution \( t = e^x \), we have \( dt = e^x dx \) or \( dx = \frac{dt}{t} \). The integral becomes:

\[ -\int \frac{t}{1 + t^2} \frac{dt}{t} = -\int \frac{1}{1 + t^2} dt = -\tan^{-1}(t) + C_2 \]

Thus, we have:

\[ \tan^{-1}(y) = -\tan^{-1}(e^x) + C \]

### Step 4: Applying the Initial Condition
Now we apply the initial condition \( x = 0, y = 1 \):

\[ \tan^{-1}(1) = -\tan^{-1}(e^0) + C \]

This gives:

\[ \frac{\pi}{4} = -\frac{\pi}{4} + C \]

Solving for \( C \):

\[ C = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2} \]

### Step 5: Final Equation
Substituting \( C \) back into our equation gives:

\[ \tan^{-1}(y) + \tan^{-1}(e^x) = \frac{\pi}{2} \]

### Step 6: Rearranging the Final Result
Using the identity \( \tan^{-1}(a) + \tan^{-1}(b) = \frac{\pi}{2} \) when \( ab = 1 \), we can express this as:

\[ y \cdot e^x = 1 \]

Thus, the final solution to the initial value problem is:

\[ y = \frac{1}{e^x} \]