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# Solve the following differential equation;$(D^{2}-4D+13)$y$=e^{-3x}$

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A)
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$
• Where $X$ is a function of $x$
• The solution is obtained in two parts.
• The first part is the complementary function CF
• This is obtained by solving the equation $am^2+bm+c=0$.The second part is called the particular integral or PI
• The GS is $y=CF+PI$
• Let $m_1,m_2$ be the roots of the CE
• Case 1: $m_1,m_2$ are real numbers and distinct
• $CF=Ae^{m_1x}+Be^{m_2x}$
• Case 2: $m_1,m_2$ are complex (i.e)$m_1=\alpha+i\beta$ and $m_2=\alpha-i\beta$
• Then $CF=e^{\alpha x}[A\cos\beta x+B\sin \beta x]$
• Case 3: $m_1,m_2$ are real and equal say $m_1$
• $CF=(A+Bx)e^{m_1x}$
• The PI when $X=e^{\alpha x},\alpha$ is a constant
• Case 1: $f(\alpha)\neq 0$
• $PI=\large\frac{1}{f(D)}e^{\alpha x}=\frac{e^{\alpha x}}{f(\alpha)}$
• Case 2: $f(\alpha)\neq 0$($\alpha=m_1,$ 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 $\theta(\alpha)\neq 0$
• $\Rightarrow \large\frac{1}{(D-\alpha)\theta(\alpha)}e^{\alpha x}=\large\frac{xe^{\alpha x}}{\theta(\alpha)}$
• Case 3: $f(\alpha)=0$ and $m_1=m_2=\alpha$ then
• $\large\frac{1}{f(D)}e^{\alpha x}=\large\frac{x^2}{2}e^{\alpha x}$
Step 1:
Characterstic equation :$m^2-4m+13=0$
$m=\large\frac{4\pm\sqrt{16-52}}{2}=\frac{4\pm 6i}{2}$$=2\pm 3i$
$\alpha=2,\beta=3$
$CF =e^{2x}[A\cos 3x+B\sin 3x]$
Step 2:
P.I=$\large\frac{1}{D^2-4D+13}e^{-3x}=\frac{e^{-3x}}{9+12+13}=\frac{e^{-3x}}{34}$
Step 3:
GS : $y=Ae^{2x}[A\cos 3x+B\sin 3x]+\large\frac{e^{-3x}}{34}$