Re: Kartika's question at Yahoo! Answers regarding solvinga 2nd order inhomogeneous ODE
Hello Kartika,
We are given to solve:
$$y''-6y'+9y=4e^{3x}$$
and we are instructed not to use the method of undetermined coefficients. So, I will use variation of parameters instead.
First, we want to find a fundamental solution set for the corresponding homogeneous equation. We see that we have the repeated characteristic root $r=3$, and so our set is:
$$\{e^{3x},xe^{3x}\}$$
Hence, we take as our particular solution:
$$y_p(x)=v_1(x)e^{3x}+v_2xe^{3x}$$
Next, we determine $v_1(x)$ and $v_2(x)$ by solving the system:
$$e^{3x}v_1'+xe^{3x}v_2'=0$$
$$3e^{3x}v_1'+\left(3xe^{3x}+e^{3x} \right)v_2'=4e^{3x}$$
Dividing through both equations by $e^{3x}\ne0$, we obtain:
$$v_1'+xv_2'=0$$
$$3v_1'+\left(3x+1 \right)v_2'=4$$
The first equation gives us:
$$v_1'=-xv_2'$$
And substituting into the second equation, we find:
$$-3xv_2'+\left(3x+1 \right)v_2'=4$$
$$v_2'=4\,\therefore\,v_1'=-4x$$
Integrating with respect to $x$, we now find:
$$v_1(x)=-2x^2$$
$$v_2(x)=4x$$
And thus, our particular solution is:
$$y_p(x)=-2x^2e^{3x}+4x^2e^{3x}=2x^2e^{3x}$$
Thus, by superposition, we find the general solution is:
$$y(x)=y_h(x)+y_p(x)=c_1e^{3x}+c_2xe^{3x}+2x^2e^{3x}$$