For differential equation `(1 + x^2) (dy)/(dx) + y = e^(tan^(-1)x`. Let y(x) is solution if y(0) =2 then value of y(1) is

Options

A. `frac{e^(pi/4)}{2} + frac{3}{2e^(pi/4)}`
B. `frac{e^(pi/4)}{2} + frac{3}{2e^(-pi/4)}`
C. `frac{e^(pi/4)}{2} - frac{3}{2e^(pi/4)}`
D. `frac{e^(pi/4)}{2} - frac{3}{2e^(-pi/4)}`

Answer

To solve the differential equation

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

with the initial condition \( y(0) = 2 \), we will follow these steps:

### Step 1: Rewrite the Differential Equation
First, we rewrite the equation to isolate \(\frac{dy}{dx}\):

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

### Step 2: Identify the Integrating Factor
The standard form of a linear first-order differential equation is

\[ \frac{dy}{dx} + P(x)y = Q(x) \]

where \( P(x) = \frac{1}{1 + x^2} \) and \( Q(x) = \frac{e^{\tan^{-1}(x)}}{1 + x^2} \).

To find the integrating factor \( \mu(x) \), we compute:

\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int \frac{1}{1 + x^2} \, dx} = e^{\tan^{-1}(x)} \]

### Step 3: Multiply the Equation by the Integrating Factor
Now, we multiply the entire differential equation by the integrating factor:

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

This simplifies to:

\[ \frac{d}{dx}\left( y e^{\tan^{-1}(x)} \right) = \frac{e^{2 \tan^{-1}(x)}}{1 + x^2} \]

### Step 4: Integrate Both Sides
Next, we integrate both sides with respect to \( x \):

\[ \int \frac{d}{dx}\left( y e^{\tan^{-1}(x)} \right) \, dx = \int \frac{e^{2 \tan^{-1}(x)}}{1 + x^2} \, dx \]

The left side simplifies to:

\[ y e^{\tan^{-1}(x)} = \int \frac{e^{2 \tan^{-1}(x)}}{1 + x^2} \, dx + C \]

### Step 5: Solve the Integral
To solve the integral on the right side, we can use the substitution \( t = \tan^{-1}(x) \), which gives \( x = \tan(t) \) and \( dx = \sec^2(t) dt \).

This results in:

\[ \int e^{2t} dt = \frac{e^{2t}}{2} + C \]

Substituting back, we have:

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

### Step 6: Solve for \( y \)
Now, we can solve for \( y \):

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

### Step 7: Apply Initial Condition
Using the initial condition \( y(0) = 2 \):

\[ 2 = \frac{e^{\tan^{-1}(0)}}{2} + C e^{-\tan^{-1}(0)} \]

Since \( \tan^{-1}(0) = 0 \), we have:

\[ 2 = \frac{1}{2} + C \implies C = \frac{3}{2} \]

### Step 8: Final Solution
Thus, the particular solution is:

\[ y = \frac{e^{\tan^{-1}(x)}}{2} + \frac{3}{2} e^{-\tan^{-1}(x)} \]

### Step 9: Find \( y(1) \)
Now, we calculate \( y(1) \):

\[ y(1) = \frac{e^{\tan^{-1}(1)}}{2} + \frac{3}{2} e^{-\tan^{-1}(1)} \]

Since \( \tan^{-1}(1) = \frac{\pi}{4} \):

\[ y(1) = \frac{e^{\frac{\pi}{4}}}{2} + \frac{3}{2} e^{-\frac{\pi}{4}} \]

### Final Answer
Thus, the value of \( y(1) \) is:

\[ y(1) = \frac{e^{\frac{\pi}{4}}}{2} + \frac{3}{2} e^{-\frac{\pi}{4}} \]

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