Math Snap

STEP 1

What is this asking?
We're given a differential equation and two linearly independent solutions.
We need to use the variation of parameters method to find a particular solution $y_p$ for a non-homogeneous version of the equation.
Watch out!
Don't mix up the homogeneous and non-homogeneous equations!
Also, be careful with the signs when integrating.

STEP 2

1. Set up the Variation of Parameters
2. Calculate the Wronskian
3. Find $u_1'$ and $u_2'$
4. Integrate to find $u_1$ and $u_2$
5. Construct the particular solution $y_p$

STEP 3

We're given $y_1 = x$ and $y_2 = x^3$ as solutions to the homogeneous equation: $y'' - \frac{3}{x}y' + \frac{3}{x^2}y = 0$.
Our goal is to find a particular solution $y_p$ to the non-homogeneous equation: $y'' - \frac{3}{x}y' + \frac{3}{x^2}y = 2x^2e^x$.

STEP 4

The variation of parameters method tells us to look for a solution of the form $y_p = u_1y_1 + u_2y_2$, where $u_1$ and $u_2$ are functions we need to determine.

STEP 5

The Wronskian, $W(y_1, y_2)$, is a crucial part of this method!
It's given by the determinant:
$$W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix}$$

STEP 6

With $y_1 = x$ and $y_2 = x^3$, we have $y_1' = 1$ and $y_2' = 3x^2$.
So, the Wronskian is:
$$W(x, x^3) = \begin{vmatrix} x & x^3 \\ 1 & 3x^2 \end{vmatrix} = x \cdot 3x^2 - x^3 \cdot 1 = 3x^3 - x^3 = 2x^3$$

STEP 7

Now we need to find $u_1'$ and $u_2'$.
The formulas are:
$$u_1' = \frac{-y_2f(x)}{W(y_1, y_2)} \quad \text{and} \quad u_2' = \frac{y_1f(x)}{W(y_1, y_2)}$$ where $f(x)$ is the right-hand side of the non-homogeneous equation, which is $2x^2e^x$.

STEP 8

Let's plug in our values:
$$u_1' = \frac{-x^3(2x^2e^x)}{2x^3} = \mathbf{-x^2e^x}$$ $$u_2' = \frac{x(2x^2e^x)}{2x^3} = \mathbf{e^x}$$

STEP 9

Now, we integrate $u_1'$ and $u_2'$ to find $u_1$ and $u_2$:
$$u_1 = \int -x^2e^x \, dx = -(x^2 - 2x + 2)e^x$$ (using integration by parts twice)
$$u_2 = \int e^x \, dx = \mathbf{e^x}$$

STEP 10

Finally, we put it all together to find our particular solution $y_p$:
$$y_p = u_1y_1 + u_2y_2 = -(x^2 - 2x + 2)e^x \cdot x + e^x \cdot x^3$$ $$y_p = -x^3e^x + 2x^2e^x - 2xe^x + x^3e^x = \mathbf{2x^2e^x - 2xe^x}$$

The particular solution is $2x^2e^x - 2xe^x$.