Question

use the method of variation of parameters to find a particular solution of the following differential equation. y - 6y + 9y = 10e^{3x}
the particular solution is y_p(x) =

Explanation:

Step1: Find complementary function

The homogeneous equation is $y'' - 6y' + 9y=0$. The characteristic equation is $r^{2}-6r + 9 = 0$, which factors to $(r - 3)^{2}=0$. So, $r=3$ (repeated root). The complementary function $y_c(x)=(C_1 + C_2x)e^{3x}$, and we take $y_1=e^{3x}$ and $y_2=xe^{3x}$.

Step2: Calculate Wronskian

The Wronskian $W=

$$\begin{vmatrix}y_1&y_2\\y_1'&y_2'\end{vmatrix}$$

=

$$\begin{vmatrix}e^{3x}&xe^{3x}\\3e^{3x}&(3x + 1)e^{3x}\end{vmatrix}$$

=e^{6x}$.

Step3: Find $u_1'$ and $u_2'$

We know that for $y''+p(x)y'+q(x)y = f(x)$, here $p(x)=-6$, $q(x)=9$ and $f(x)=10e^{3x}$. Using the variation - of - parameters formulas $u_1'=-\frac{y_2f(x)}{W}$ and $u_2'=\frac{y_1f(x)}{W}$.
$u_1'=-\frac{xe^{3x}\cdot10e^{3x}}{e^{6x}}=- 10x$ and $u_2'=\frac{e^{3x}\cdot10e^{3x}}{e^{6x}} = 10$.

Step4: Integrate to find $u_1$ and $u_2$

Integrating $u_1'=-10x$, we get $u_1=-5x^{2}+C_3$. Integrating $u_2' = 10$, we get $u_2 = 10x+C_4$. We can choose $C_3 = C_4=0$ for a particular solution.

Step5: Find particular solution

The particular solution $y_p(x)=u_1y_1+u_2y_2$. Substituting $u_1=-5x^{2}$, $y_1 = e^{3x}$, $u_2 = 10x$ and $y_2=xe^{3x}$, we have $y_p(x)=-5x^{2}e^{3x}+10x\cdot xe^{3x}=5x^{2}e^{3x}$.

Answer:

$5x^{2}e^{3x}$