Answer to Question #218033 in Differential Equations for burji

Question

solve the following equation

1. d²y/dx²-2dy/dx+5y=e^xtan2x

Answer 1

Solution

To get general solution of homogeneous equation y₀’’-2y₀’+5y₀=0 let’s solve characteristic equation r²-2r+5=0.

r_1,2=1±2i

So y₀(x) = C₁e^xsin(2x)+C₂e^xcos(2x), where C₁ and C₂ are arbitrary constants.

Partial solution x₁(t) may be find in the form y₁(x) = A(x)ϕ₁(x)+ B(x)ϕ₂(x), where ϕ₁(x)= e^xsin(2x),  ϕ₂(x)= e^xcos(2x) Substituting into equation according to the method of undetermined coefficients  we’ll get

A’ϕ₁(x)+ B’ϕ₂(x)=0, A’ϕ’₁(x)+ B’ϕ’₂(x)= e^xtan2x

From this system

A(x)=-∫ϕ₂(x) e^xtan2x dx/W(x), B(x)=∫ϕ₁(x) e^xtan2x dx/W(x), where W(x) = ϕ₁(x) ϕ’₂(x) - ϕ₂(x) ϕ’₁(x)

ϕ’₁(x) = e^xsin(2x)+2 e^xcos(2x), ϕ’₂(x) = e^xcos(2x) - 2 e^xsin(2x)  =>  W(x) = -2e^2x  

A(x) = -0.5∫cos(2x)tan(2x)dx = -0.25∫tan(2x)dsin(2x) = 0.25cos(2x)

B(x) = -0.5∫sin(2x)tan(2x)dx = 0.25∫tan(2x)dcos(2x) = 0.25[sin(2x)-ln|(1+sin(2x))/cos(2x)|]

So y₁(x) = A(x)ϕ₁(x)+ B(x)ϕ₂(x) = -0.25 e^xcos(2x) ln|(1+sin(2x))/cos(2x)|]

and solution of DE is

y(x) = y₀(x) + y₁(x) = C₁e^xsin(2x)+C₂e^xcos(2x) - 0.25 e^xcos(2x) ln|(1+sin(2x))/cos(2x)|]