**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 :$y=Ae^x+Be^{12x}+\large\frac{e^{-2x}}{42}-\frac{5xe^x}{11}$