Solve the first order linear initial value problem

dy/dx +2y = 3x^2 -2x,  y(0)=0

To solve the first-order linear initial value problem:

\[
\frac{dy}{dx} + 2y = 3x^2 - 2x, \quad y(0) = 0
\]

we can proceed as follows:

### Step 1: Identify the integrating factor
The given equation is a first-order linear differential equation of the form:

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

where \( P(x) = 2 \) and \( Q(x) = 3x^2 - 2x \).

The integrating factor \( \mu(x) \) is found using the formula:

\[
\mu(x) = e^{\int P(x) dx} = e^{\int 2 dx} = e^{2x}.
\]

### Step 2: Multiply through by the integrating factor
Multiplying the entire equation by \( e^{2x} \):

\[
e^{2x} \frac{dy}{dx} + 2 e^{2x} y = (3x^2 - 2x) e^{2x}.
\]

The left-hand side is now the derivative of \( e^{2x} y \), so we can rewrite the equation as:

\[
\frac{d}{dx}(e^{2x} y) = (3x^2 - 2x) e^{2x}.
\]

### Step 3: Integrate both sides
Integrating both sides with respect to \( x \):

\[
e^{2x} y = \int (3x^2 - 2x) e^{2x} dx.
\]

### Step 4: Solve the integral
We need to solve the integral:

\[
\int (3x^2 - 2x) e^{2x} dx.
\]

We will do this by integration by parts.

1. **First term:** \( \int 3x^2 e^{2x} dx \)
2. **Second term:** \( - \int 2x e^{2x} dx \)

Using integration by parts for each, this can be solved as follows:

Let me calculate the solution.The integral of \((3x^2 - 2x) e^{2x}\) is:

\[
\int (3x^2 - 2x) e^{2x} dx = \frac{(6x^2 - 10x + 5) e^{2x}}{4}.
\]

### Step 5: General solution
Thus, the solution for \( e^{2x} y \) is:

\[
e^{2x} y = \frac{(6x^2 - 10x + 5) e^{2x}}{4} + C,
\]

where \( C \) is the constant of integration.

### Step 6: Solve for \( y \)
Now, divide both sides by \( e^{2x} \) to solve for \( y \):

\[
y = \frac{(6x^2 - 10x + 5)}{4} + C e^{-2x}.
\]

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

\[
0 = \frac{(6(0)^2 - 10(0) + 5)}{4} + C e^{0},
\]

\[
0 = \frac{5}{4} + C.
\]

Solving for \( C \):

\[
C = -\frac{5}{4}.
\]

### Step 8: Final solution
Substitute \( C = -\frac{5}{4} \) back into the solution for \( y \):

\[
y = \frac{6x^2 - 10x + 5}{4} - \frac{5}{4} e^{-2x}.
\]

Thus, the solution to the initial value problem is:

\[
y(x) = \frac{6x^2 - 10x + 5}{4} - \frac{5}{4} e^{-2x}.
\]

Would you like more details or have any questions?

Here are some related questions:
1. How do we determine an integrating factor for different forms of linear equations?
2. What are the steps for solving linear first-order differential equations?
3. How can you apply integration by parts to different integrals?
4. What are other methods to solve first-order differential equations?
5. How does the initial condition affect the constant of integration?

**Tip:** Always verify the solution by substituting it back into the original equation!

Solve the first order linear initial value problem dy/dx + 2y = 3x^2 - 2x, y(0)=0

Solve the first-order linear differential equation dy/dx + 2y = 3x^2 - 2x with the initial condition y(0) = 0. Learn how to apply the integrating factor method and use integration by parts to find the solution. Suitable for advanced high school or early college students.

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