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• Notes

By using the reduction of order method, Solve the following
differential equation
$\left(D^{2}+1\right) y=\sec x$

put $f(m)=0 \quad \longrightarrow m^{2}+1=0 \rightarrow m=\pm i$

$y_{c}=c_{1} \cos x+c_{2} \sin x \quad$ let $\quad y=v \cos x$

$y^\prime=v^{\prime} \cos x+-\sin x(v)$

$y^{\prime \prime}=v^{\prime \prime} \cos x-\sin x {v^\prime}-\cos x(v)-{v^\prime} \sin (x)$

$y^{\prime \prime}=v^{\prime \prime} \cos x-2 v^{\prime} \sin x-v \cos x$

Substitute :

$v^{\prime \prime} \cos x-2 v^\prime \sin x-v \cos x+v \cos x=\sec x$

$v^{\prime \prime} \cos x-2 v^{\prime} \sin x=\sec x \quad \div \cos x$

$v^{\prime \prime}-2 v^{\prime} \tan x=\sec ^{2} x$

$v^{\prime \prime}=w^{\prime}$
$v^{\prime}=w$

$w^{\prime}-2 \tan x w=\sec ^{2} x$ Linear in $w$

$\mu=e^{\int {-2} \tan x d x}=e^{-2 \ln |\sec x|}=e^{\ln sec^{-2} x}=\sec ^{-2} x=\frac{1}{\sec ^{2} x}$

$\mu=\cos ^{2} x$

$\mu \cdot w=\int \cos ^{2} x \sec ^{2} x d x$

$\cos ^{2} x \cdot w=x+c_{1} \quad \div \cos ^{2} x$

$\frac{1}{\cos x}=\sec x$

$w=x \cdot \sec ^{2} x+c_{1} \sec ^{2} x$

$v^{\prime}=x \cdot \sec ^{2} x+c_{1} \sec ^{2} x$ Integrate

$v=\int x \sec ^{2} x d x+c_{1} \tan x+c_{2} \cdots (1)$

$\int x \sec ^{2} x d x \longrightarrow$ by parts

$u=x \quad \longrightarrow d u=1 d x$
$v=\tan x \quad \leftarrow d v=\sec ^{2} x d x$

$\int x \sec ^{2} x d x=x \tan x-\int \tan x d x$

$=x \tan x-\ln |\sec x| \quad$ substitute in $(1)$

$v=x \tan x-\ln |\sec x|+c_{1} \tan x+c_{2} \quad$ But $\quad y=v \cos x$

$y=\left(x \tan x-\ln |\sec x|+c_{1} \tan x+c_{2}\right) \cos x$

$=x \sin x-\cos x \ln |\sec x|+c_{1} \sin x+c_{2} \cos x$