Extraordinary Differential Equations #2

Relevant wiki: Integrating factors in linear differential equations

We first solve the corresponding homogeneous equation $y''+6y'+9y=0$ . Using the roots of the auxiliary equation $x^2+6x+9=0$ which is $-3$ , we get the first component of the homogeneous solution $y_{h,1}=c_1 e^{-3x}$ . Trying $u(x)=x$ as a possible integrating factor gives the second linearly-independent component of the homogeneous solution $y_{h,2}=c_2 xe^{-3x}$ . Thus, the homogeneous solution is $y_h=c_1 e^{-3x}+c_2 xe^{-3x}$

For the particular solution we can use the method of undetermined coefficients by trialling $y_p=k_1 x^2+k_2 x+k_3+k_4 x^3 e^{-3x}+k_5 \sin{x}+k_6 \cos{x}$

Then we get $y_p''+6y_p'+9y_p=9k_1 x^2+(12k_1+9k_2)x+(2k_1+6k_2+9k_3)+6k_4 xe^{-3x}+(8k_5-6k_6)\sin{x}+(6k_5+8k_6)\cos{x}$

Comparing this with the RHS of the original ODE gives us the following system of linear equations:

$9k_1=1$

$12k_1+9k_2=0$

$2k_1+6k_2+9k_3=0$

$6k_4=1$

$8k_5-6k_6=0$

$6k_5+8k_6=1$

Therefore $k_1=\frac{1}{9}, k_2=-\frac{4}{27}, k_3=\frac{2}{27}, k_4=\frac{1}{6}, k_5=\frac{3}{50}, k_6=\frac{2}{25}$ .

(I'll leave it to the reader to determine why $k_4 x^3 e^{-3x}$ was used in the the trial solution instead of $k_4 x^2 e^{-3x}$ .)