Math / Calculus

QuestionGiven that $y_{1}=x, y_{2}=x^{3}$ are linearly independent solutions for the differential equation $y^{\prime \prime}-\frac{3}{x} y^{\prime}+\frac{3}{x^{2}} y=0$ . Using Variation of parameter for $y^{\prime \prime}-\frac{3}{x} y^{\prime}+\frac{3}{x^{2}} y=2 x^{2} e^{x}$ , we get $y_{p}=$ $2 x^{2} e^{x}-2 x e^{x}$ the above option $2 x^{2} e^{x}+2 x e^{x}$ $4 x^{2} e^{x}-4 x e^{x}$ the above option None of the mentioned

Studdy Solution

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 $y_p$ : $y_p = u_1y_1 + u_2y_2 = -(x^2 - 2x + 2)e^x \cdot x + e^x \cdot x^3$ $y_p = -x^3e^x + 2x^2e^x - 2xe^x + x^3e^x = \mathbf{2x^2e^x - 2xe^x}$

STEP 11

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

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Studdy Solution

STEP 1

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 $y_p$ : $y_p = u_1y_1 + u_2y_2 = -(x^2 - 2x + 2)e^x \cdot x + e^x \cdot x^3$ $y_p = -x^3e^x + 2x^2e^x - 2xe^x + x^3e^x = \mathbf{2x^2e^x - 2xe^x}$

STEP 11

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

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Studdy Solution

STEP 1

STEP 2

1. Set up the Variation of Parameters2. Calculate the Wronskian3. Find u1′u_1'u1′​ and u2′u_2'u2′​4. Integrate to find u1u_1u1​ and u2u_2u2​5. Construct the particular solution ypy_pyp​

STEP 3

STEP 4

The variation of parameters method tells us to look for a solution of the form yp=u1y1+u2y2y_p = u_1y_1 + u_2y_2yp​=u1​y1​+u2​y2​, where u1u_1u1​ and u2u_2u2​ are functions we need to determine.

STEP 5

The **Wronskian**, W(y1,y2)W(y_1, y_2)W(y1​,y2​), is a crucial part of this method!It's given by the determinant: W(y1,y2)=∣y1y2y1′y2′∣ W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} W(y1​,y2​)=∣∣​y1​y1′​​y2​y2′​​∣∣​

STEP 6

STEP 7

STEP 8

Let's plug in our values: u1′=−x3(2x2ex)2x3=−x2ex u_1' = \frac{-x^3(2x^2e^x)}{2x^3} = \mathbf{-x^2e^x} u1′​=2x3−x3(2x2ex)​=−x2ex u2′=x(2x2ex)2x3=ex u_2' = \frac{x(2x^2e^x)}{2x^3} = \mathbf{e^x} u2′​=2x3x(2x2ex)​=ex

STEP 9

STEP 10

STEP 11

The particular solution is 2x2ex−2xex2x^2e^x - 2xe^x2x2ex−2xex.

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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 $y_p$

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.

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

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

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