Answer to Question #122610 in Differential Equations for jse

Question

Solve the differential equation by variation of parameters.

y'' + 3y' + 2y = 1/ 1 + e^x

y(x) =____

Answer 1

y'' + 3y' + 2y = 1/( 1 + e^x )

Solution:

y'' + 3y' + 2y =0

k²+3k+2=0

k₁=-1,k₂=-2

y₀(x)=c₁e^-x+c₂e^-2x

y'' + 3y' + 2y = 1/( 1 + e^x )

y(x)=c₁(x)e^-x+c₂(x)e^-2x

Let

c₁'(x)e^-x+c₂'(x)e^-2x=0,

-c₁'(x)e^-x-2c₂'(x)e^-2x=1/( 1 + e^x).

Then

c₂'(x)=-c₁'(x)e^x => c₁'(x)=e^x/(1+e^x) => c₁(x)=$\int$e^x/(1+e^x) dx=$\int$ d(1+e^x)/(1+e^xl)=ln(1+e^x)+c₁

c₂'(x)=-e^2x/(1+e^x)=> c₂(x)= -∫e^2x/(1+e^x) dx=-$\int$(e^x+1-1)d(1+e^x)/(1+e^x)=-$\int$ (1-1/(1+e^x))d(1+e^x)=

=-(1+e^x)+ln(1+e^x)+c₂

y(x)=c₁(x)e^-x+c₂(x)e^-2x=(ln(1+e^x)+c₁)e^-x+(-(1+e^x)+ln(1+e^x)+c₂)e^-2x=

=(c₁-1)e^-x+(c₂-1)e^-2x+e^-xln(1+e^x)+e^-2xln(1+e^x).

Let c₁-1->c₁, c₂-1->c₂.

Answer:

y(x)=c₁e^-x+c₂e^-2x+e^-xln(1+e^x)+e^-2xln(1+e^x).