Answer to Question #145519 in Differential Equations for jericho

Question

y′′′−2y′′−y′+2y=e^4x

variation of paramaters

Answer 1

Let us solve the differential equation $y'''−2y''−y'+2y=e^{4x}$ using the method of variation of paramaters. For this, firstly solve the characteristic equation $k^3-2k^2-k+2=0$ of $y'''−2y''−y'+2y=0$. This characteristic eqution is equivalent to $k^2(k-2)-(k-2)=0$ which is equivalent to $(k^2-1)(k-2)=0$. Therefore, $k_1=1,k_2=-1,k_3=2.$ According to the method of variation of paramaters, the solution of $y'''−2y''−y'+2y=e^{4x}$ is

$y(x)=C_1(x)e^x+C_2(x)e^{-x}+C_3(x)e^{2x}.$

Then we have the following system:

$\begin{cases} C_1'(x)e^x+C_2'(x)e^{-x}+C_3'(x)e^{2x}=0\\ C_1'(x)e^x-C_2'(x)e^{-x}+2C_3'(x)e^{2x}=0\\ C_1'(x)e^x+C_2'(x)e^{-x}+4C_3'(x)e^{2x}=e^{4x} \end{cases}$

which is equivalent to

$\begin{cases} C_1'(x)e^x+C_2'(x)e^{-x}+C_3'(x)e^{2x}=0\\ 2C_1'(x)e^x+3C_3'(x)e^{2x}=0\\ 3C_3'(x)e^{2x}=e^{4x} \end{cases}$

$\begin{cases} C_2'(x)=-C_1'(x)e^{2x}-C_3'(x)e^{3x}\\ C_1'(x)=-\frac{3}{2}C_3'(x)e^{x}\\ C_3'(x)=\frac{1}{3}e^{2x} \end{cases}$

$\begin{cases} C_2'(x)=\frac{1}{2}e^{5x}-\frac{1}{3}e^{5x}\\ C_1'(x)=-\frac{1}{2}e^{3x}\\ C_3'(x)=\frac{1}{3}e^{2x} \end{cases}$

$\begin{cases} C_2'(x)=\frac{1}{6}e^{5x}\\ C_1'(x)=-\frac{1}{2}e^{3x}\\ C_3'(x)=\frac{1}{3}e^{2x} \end{cases}$

$\begin{cases} C_2(x)=\frac{1}{30}e^{5x}+C_2\\ C_1(x)=-\frac{1}{6}e^{3x}+C_1\\ C_3(x)=\frac{1}{6}e^{2x}+C_3 \end{cases}$

Therefore, the solution is the following:

$y(x)=(-\frac{1}{6}e^{3x}+C_1)e^x+(\frac{1}{30}e^{5x}+C_2)e^{-x}+(\frac{1}{6}e^{2x}+C_3)e^{2x}=$

$=C_1e^x+C_2e^{-x}+C_3e^{2x}+\frac{1}{30}e^{4x}.$