Question

solve the given equation by variation of parameters.
y - 6y + 9y = -8e^{3x}

the solution is y = \boxed{}

Explanation:

Step1: Find the complementary solution

The homogeneous equation is \(y'' - 6y' + 9y=0\). The characteristic equation is \(r^{2}-6r + 9=(r - 3)^{2}=0\). So, \(r = 3\) (repeated root). The complementary solution \(y_{c}(x)=c_{1}e^{3x}+c_{2}xe^{3x}\), where \(y_{1}=e^{3x}\) and \(y_{2}=xe^{3x}\).

Step2: Calculate the 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^{3x}\cdot(3x + 1)e^{3x}-xe^{3x}\cdot3e^{3x}=e^{6x}\).

Step3: Find \(u_{1}\) and \(u_{2}\)

We have \(f(x)=- 8e^{3x}\).
\(u_{1}=-\int\frac{y_{2}f(x)}{W}dx=-\int\frac{xe^{3x}\cdot(-8e^{3x})}{e^{6x}}dx=\int8xdx = 4x^{2}\)
\(u_{2}=\int\frac{y_{1}f(x)}{W}dx=\int\frac{e^{3x}\cdot(-8e^{3x})}{e^{6x}}dx=\int(-8)dx=-8x\)

Step4: Construct the particular solution

The particular solution \(y_{p}=u_{1}y_{1}+u_{2}y_{2}=4x^{2}e^{3x}-8x\cdot xe^{3x}=-4x^{2}e^{3x}\)

Step5: Write the general solution

The general solution \(y = y_{c}+y_{p}=c_{1}e^{3x}+c_{2}xe^{3x}-4x^{2}e^{3x}\)

Answer:

\(y = c_{1}e^{3x}+c_{2}xe^{3x}-4x^{2}e^{3x}\)