Find the particular solution of the differential equation :
`(1+x^(2))(dy)/(dx)=e^(m tan^(-1)x)-y`, given that `y=1` when `x=0`.

Answer

To find the particular solution of the differential equation

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

given that \( y = 1 \) when \( x = 0 \), we will follow these steps:

### Step 1: Rearranging the Equation
First, we can rearrange the differential equation into standard linear form:

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

### Step 2: Identifying \( p(x) \) and \( q(x) \)
From the standard form \( \frac{dy}{dx} + p(x)y = q(x) \), we identify:

- \( p(x) = \frac{1}{1+x^2} \)
- \( q(x) = \frac{e^{m \tan^{-1} x}}{1+x^2} \)

### Step 3: Finding the Integrating Factor
The integrating factor \( \mu(x) \) is given by:

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

### Step 4: Multiplying the Equation by the Integrating Factor
We multiply the entire differential equation by the integrating factor:

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

This simplifies to:

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

### Step 5: Integrating Both Sides
Now we integrate both sides:

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

The left side becomes:

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

### Step 6: Solving the Integral on the Right Side
Let \( t = \tan^{-1} x \), then \( dt = \frac{1}{1+x^2} \, dx \). The integral becomes:

\[ \int e^{(m+1)t} \, dt = \frac{1}{m+1} e^{(m+1)t} + C \]

Substituting back, we have:

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

### Step 7: Solving for \( y \)
Thus, we can express \( y \) as:

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

### Step 8: Applying Initial Condition
Using the initial condition \( y(0) = 1 \):

\[ 1 = \frac{1}{m+1} + C e^{0} \]

This simplifies to:

\[ 1 = \frac{1}{m+1} + C \]

From this, we find:

\[ C = 1 - \frac{1}{m+1} = \frac{m}{m+1} \]

### Final Solution
Substituting \( C \) back into the equation for \( y \):

\[ y = \frac{1}{m+1} + \frac{m}{m+1} e^{-\tan^{-1} x} \]