Solve the first order linear initial value problem

dydx+3y=3x3−4x2−5xy(0)=0

y(x)

dy/dx + 3y = 3x^3 -4x^2 -5x, y(0)=0

We are asked to solve the first-order linear initial value problem:

\[
\frac{dy}{dx} + 3y = 3x^3 - 4x^2 - 5x, \quad y(0) = 0
\]

### Step 1: Write the equation in standard linear form

The given equation is already in standard form:

\[
\frac{dy}{dx} + P(x)y = Q(x)
\]
where \( P(x) = 3 \) and \( Q(x) = 3x^3 - 4x^2 - 5x \).

### Step 2: Find the integrating factor

The integrating factor is found using the formula:

\[
\mu(x) = e^{\int P(x)\,dx} = e^{\int 3 \,dx} = e^{3x}
\]

### Step 3: Multiply through by the integrating factor

Multiply both sides of the differential equation by \( e^{3x} \):

\[
e^{3x}\frac{dy}{dx} + 3e^{3x}y = \left(3x^3 - 4x^2 - 5x\right)e^{3x}
\]

The left-hand side becomes:

\[
\frac{d}{dx} \left( y e^{3x} \right)
\]

Thus, the equation becomes:

\[
\frac{d}{dx} \left( y e^{3x} \right) = \left(3x^3 - 4x^2 - 5x\right)e^{3x}
\]

### Step 4: Integrate both sides

Now integrate both sides with respect to \(x\):

\[
y e^{3x} = \int \left(3x^3 - 4x^2 - 5x\right)e^{3x} \, dx
\]

We need to integrate the right-hand side, which involves integrating by parts. I'll compute this step.The integral of \( \left(3x^3 - 4x^2 - 5x\right)e^{3x} \) is:

\[
\int \left(3x^3 - 4x^2 - 5x\right)e^{3x} \, dx = \frac{(27x^3 - 63x^2 - 3x + 1)}{27} e^{3x}
\]

### Step 5: Write the general solution

Now we have:

\[
y e^{3x} = \frac{(27x^3 - 63x^2 - 3x + 1)}{27} e^{3x} + C
\]

Dividing both sides by \( e^{3x} \), we get:

\[
y(x) = \frac{(27x^3 - 63x^2 - 3x + 1)}{27} + Ce^{-3x}
\]

### Step 6: Apply the initial condition \( y(0) = 0 \)

Substitute \( x = 0 \) and \( y(0) = 0 \) into the solution:

\[
0 = \frac{(27(0)^3 - 63(0)^2 - 3(0) + 1)}{27} + C e^{0}
\]
\[
0 = \frac{1}{27} + C
\]
\[
C = -\frac{1}{27}
\]

### Step 7: Final solution

Substitute \( C = -\frac{1}{27} \) into the general solution:

\[
y(x) = \frac{(27x^3 - 63x^2 - 3x + 1)}{27} - \frac{1}{27}e^{-3x}
\]

Simplifying:

\[
y(x) = \frac{27x^3 - 63x^2 - 3x}{27} + \frac{1}{27}(1 - e^{-3x})
\]

Thus, the solution to the differential equation is:

\[
y(x) = x^3 - \frac{7x^2}{3} - \frac{x}{9} + \frac{1}{27}(1 - e^{-3x})
\]

Let me know if you would like more details or have questions!

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Here are 5 related questions:
1. How is the integrating factor method applied to general linear differential equations?
2. What is the significance of the initial condition \( y(0) = 0 \) in determining the constant?
3. Can integrating factors be applied to non-linear differential equations?
4. How does one perform integration by parts for more complex functions?
5. What is the general solution form for homogeneous first-order linear differential equations?

**Tip**: Always check that your final solution satisfies the original equation and any given initial conditions.

Solve the first order linear initial value problem dy/dx + 3y = 3x^3 - 4x^2 - 5x, y(0) = 0.

Explore a step-by-step solution to the first-order linear differential equation dy/dx + 3y = 3x^3 - 4x^2 - 5x, including the integrating factor method and the initial condition y(0) = 0. Suitable for students in Grades 11-12 or college-level differential equations courses.

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