Solve the following initial value problems :
`(dy)/(dx)=2x+y`, given that `x=0, y=0`.

Answer

To solve the initial value problem given by the differential equation

\[ \frac{dy}{dx} = 2x + y \]

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

### Step 1: Rearranging the Equation
We start by rearranging the equation into a standard linear form:

\[ \frac{dy}{dx} - y = 2x \]

Here, we identify \( p = -1 \) and \( q = 2x \).

### Step 2: Finding the Integrating Factor
The integrating factor \( \mu(x) \) is given by:

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

### Step 3: Multiplying Through by the Integrating Factor
Next, we multiply the entire differential equation by the integrating factor:

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

This simplifies to:

\[ \frac{d}{dx}(y e^{-x}) = 2x e^{-x} \]

### Step 4: Integrating Both Sides
Now, we integrate both sides:

\[ \int \frac{d}{dx}(y e^{-x}) \, dx = \int 2x e^{-x} \, dx \]

The left side simplifies to:

\[ y e^{-x} \]

For the right side, we use integration by parts. Let \( u = 2x \) and \( dv = e^{-x} dx \). Then, \( du = 2 dx \) and \( v = -e^{-x} \).

Using integration by parts:

\[ \int 2x e^{-x} \, dx = -2x e^{-x} - \int -2 e^{-x} \, dx = -2x e^{-x} + 2 e^{-x} + C \]

### Step 5: Putting It All Together
Now we have:

\[ y e^{-x} = -2x e^{-x} + 2 e^{-x} + C \]

Multiplying through by \( e^{x} \) gives:

\[ y = -2x + 2 + Ce^{x} \]

### Step 6: Applying the Initial Condition
Now we apply the initial condition \( y(0) = 0 \):

\[ 0 = -2(0) + 2 + Ce^{0} \]
\[ 0 = 2 + C \]

Thus, \( C = -2 \).

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

\[ y = -2x + 2 - 2e^{x} \]

### Summary
The solution to the initial value problem is:

\[ y = -2x + 2 - 2e^{x} \]