Solve equation with variation of parameter.

Question

Solve d²y/dx²+y= tanx with variation of parameter

Answer 1

\(\frac{d^2y}{d\mathrm x^2}+y=\tan \mathrm x\)

Its auxiliary equation is

m² + 1 = 0 

⇒ m = \(\pm\)i.

Thus, complementary function of given differential equation is 

 y_h = C₁cos x + C₂ sin x

let y₁ = cos x, y₂ = sin x

W(y₁, y₂) \(=\begin{vmatrix}y_1 & y_2\\y_1' & y_2'\end{vmatrix}\)

\(=\begin{vmatrix}\cos \mathrm x & \sin \mathrm x\\-\sin \mathrm x & \cos \mathrm x\end{vmatrix}\)

\(=\cos^2\mathrm x\mathrm +\sin^2\mathrm x\)

= 1

Now, u₁ \(=-\int \frac{y_2\tan \mathrm x}{W(y_1, y_2)}\)

\(=-\int \frac{\sin \mathrm x\tan \mathrm x}{1}d\mathrm x\)

\(=-\int \frac{\sin^2\mathrm x}{\cos\mathrm x}d\mathrm x\)

\(=-\int \frac{1-\cos^2\mathrm x}{\cos\mathrm x}d\mathrm x\)

\(=-\int (\sec\mathrm x-\cos\mathrm x) d\mathrm x\)

\(=-\log (\sec\mathrm x+\tan \mathrm x) +\sin \mathrm x\)

u₂ \(=\int \frac{y_1\tan \mathrm x}{W(y_1, y_2)}\)

\(=\int \frac{\cos \mathrm x\frac{\sin \mathrm x}{\cos\mathrm x}}{1}\)

\(=\int \sin \mathrm x\)

\(=-\cos\mathrm x\)

Thus, particular integral is y_p = u₁y₁ + u₂y₂

₌ (–log(sec x + tan x) + sin x)cos x + (–cos x)sin x

= –log(sec x + tan x)cos x + sin x + cos x – sin xcos x

= –log(sec x + tan x)cos x

Thus, complete solution is y = y_p + y_h.

⇒ y = C₁cos x + C₂sin x – log(sec x + tan x)cos x