# Solve the following differential equation;$(D^{2}-13D+12)$y$=e^{-2x}+5e^x$

Toolbox:
• A general second-order homogeneous equation is of the form $a\large\frac{d^2y}{dx^2}$$+b\large\frac{dy}{dx}$$+cy=X$
• The GS is $y=CF+PI$
• Case 1: $m_1,m_2$ are real numbers and distinct
• $CF=Ae^{m_1x}+Be^{m_2x}$
• The PI when $X=e^{\alpha x},\alpha$ is a constant
•  $f(\alpha)\neq 0$($\alpha=m_1,$ one of the roots of the CE)
• $\large\frac{1}{f(D)}e^{\alpha x}=\large\frac{1}{(D-\alpha)\theta(D)}e^{\alpha x}$ where $\theta(\alpha)\neq 0$
• $\Rightarrow \large\frac{1}{(D-\alpha)\theta(\alpha)}e^{\alpha x}=\large\frac{xe^{\alpha x}}{\theta(\alpha)}$
Step 1:
CE :$m^2-13m+12=0$
On Soving this we get,
$(m-12)(m-1)=0$
$m_1=1,m_2=12$
Hence the Complimentary function is
$CF=Ae^x+Be^{12x}$
Step2:
Now let us find the Particular Integral,
PI=$\large\frac{1}{D^2-13D+12}e^{-2x}+\frac{1}{D^2-13D+12}5e^x$

$PI_1=\large\frac{1}{D^2-13D+12}e^{-2x}$ [$\alpha\neq m_1\;or\; m_2$]
$\Rightarrow \large\frac{e^{-2x}}{4+26+12}=\frac{e^{-2x}}{42}$
$PI_2=5.\large\frac{1}{D^2-13D+12}e^x$ [$\alpha=m_1=1$]
$\Rightarrow \large\frac{5}{(D-1)(D-12)}e^x=\frac{5xe^x}{-11}$
Step 3:
Hence the General Solution is CF + PI
GS :$y=Ae^x+Be^{12x}+\large\frac{e^{-2x}}{42}-\frac{5xe^x}{11}$

edited Sep 10, 2013