${ message }
Your cart is empty
Discount (${discount_percentage}%) : - ${discount}KD
Need Help?
How was the speed of the video ?
How well did you understood the video ?
Was the video helpful?
Was the notes helpful?
Sign up to try our free practice
KD
19.500
1 month
Add to cart
39.500
4 months
Subscribe to Differential Equation
Practice (Free)
Practice
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 $$\quad y=v \cos x$$
$$y=\left(x \tan x-\ln |\sec x|+c_{1} \tan x+c_{2}\right) \cos x$$
$$=x \sin x-\cos x \ln |\sec x|+c_{1} \sin x+c_{2} \cos x$$
No comments yet
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 $$\quad y=v \cos x$$
$$y=\left(x \tan x-\ln |\sec x|+c_{1} \tan x+c_{2}\right) \cos x$$
$$=x \sin x-\cos x \ln |\sec x|+c_{1} \sin x+c_{2} \cos x$$
No comments yet
Join the conversation