# Solve the following differential equation; $(D^{2}+5)$$y=\cos^{2}$$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$
• Where $X$ is a function of $x$
• The solution is obtained in two parts.
• 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 $m_1$
• PI :When $x$ is of the form $\cos\alpha x$ or $\sin\alpha x$
• Case 1 : When $f$ is a function of $D^2$
• PI =$\large\frac{1}{f(D)} $$\cos\alpha x(or \sin\alpha x)=\large\frac{1}{\phi(D^2)}$$\cos\alpha x=\large\frac{1}{\phi(-\alpha ^2)}$$\cos\alpha x • Case 2:When f=\phi(D,D^2) • PI is obtained by replacing D^2 by -\alpha^2 • PI=\large\frac{1}{-\alpha ^2+bD+1}$$\cos\alpha x(or \sin\alpha x)$
• $\;\;\;=\large\frac{1}{bD+P}$$\cos\alpha x(or \sin\alpha x) • \;\;\;=\large\frac{bD-P}{(bD+P)(bD-P)}$$\cos\alpha x(or \sin\alpha x)$
• $\;\;\;=\large\frac{bD-P}{(bD^2-P^2)}$$\cos\alpha x(or \sin\alpha x) • \;\;\;=\large\frac{bD-P}{(b(-\alpha^2)-P^2)}$$\cos\alpha x(or \sin\alpha x)$
• The denominator is a constant and the numerator represents a differentiation
• Case 3: If $\phi(-\alpha^2)=0$ then
• PI=$\large\frac{1}{\phi(D^2)}$$\cos \alpha x=\large\frac{1}{D^2+\alpha^2}$$\cos\alpha x$
• $\Rightarrow$ R.P of $\large\frac{1}{(D+i\alpha)(D-i\alpha)}$$e^{i\alpha x} • \Rightarrow R.P of \large\frac{1}{\theta(i\alpha)}$$e^{i\alpha x}$
• $\Rightarrow$ $\large\frac{-x}{2\alpha}$$(-\sin \alpha x)=\large\frac{x\sin\alpha x}{2\alpha} • or PI=\large\frac{1}{\phi(D^2)}$$\sin\alpha x$=I.P of $\large\frac{1}{(D+i\alpha)(D-i\alpha)}$$e^{i\alpha x} • PI=\large\frac{-x}{2\alpha}$$\cos\alpha x$
Step 1:
CE :$m^2+5=0$
$m=\pm\sqrt{5}i$
$\alpha =0,\beta=\sqrt 5$
CF: $y=A\cos\sqrt 5x+B\sin\sqrt 5x$
Step 2:
PI :$\large\frac{1}{D^2+5}$$\cos^2x \;\;\;=\large\frac{1}{D^2+5}\frac{1+\cos 2x}{2} \;\;\;=\large\frac{1}{D^2+5}\frac{1}{2}+\frac{1}{2}\frac{1}{D^2+5}$$\cos 2x$
$\;\;\;=\large\frac{1}{0+5}\frac{1}{2}+\frac{1}{2}\frac{1}{-4+5}$$\cos 2x \;\;\;=\large\frac{1}{10}+\large\frac{1}{2}$$\cos 2x$
Step 3: