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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 :

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

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

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

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

$$w^{\prime}-2 \tan x w=\sec ^{2} x$$ 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 $$\quad y=v \cos x$$

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

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

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