Find the general solution of the given Differential Equation:?

Question

y" + 2y' + y = e^-t / (1 + t^2)

Answer 1

First find the homogeneous solution:

y"+2y'+y=0

the characteristic polynomial is

r^2+2r+1

factor and you get

(r+1)(r+1) so the roots are r1=-1 r2=-1

so the homogeneous solution is

yh=Ae^(-x)+Bxe^(-x) [A & B are constants]

y1=e^(-x), y2=xe^-x

Now find the particular solution

it will be in the form

yp=u(x)y1+v(x)y2

find the Wronskian

W= (y1)(y2)'-(y2)(y1)'

= [e^(-x)] [e^(-x) - xe^(-x)] - [xe^(-x)] [-e^(-x)]

cancel like terms and you get

W = e^(-2x)

u(x) is the integral of -y2 * e^(-x) / { (1+x^2) W}dx

u(x) = integral (-xe^(-2x)/[e^(-2x) (1+x^2)])dx

cancel and you get

integral -x/(1+x^2) dx which is

solve in terms of z=1+x^2; dz=2dx

multiply inside by -2 outside by -1/2

u(x) = -1/2 integral [(2xdx)/(1+x^2)]

substitute

u(x)= -1/2 integral (dz/z)

u(x) = -1/2 ln (z) ---absolute value of z, but z is always positive so doesn't matter

u(x)= -1/2 ln (1+x^2)

now find v(x) = integral of y1 * e^(-x) / { (1+x^2) W}dx

(sorry just noticed i used x instead of t, oh well same thing)

v(x) = integral (e^(-2x)/[e^(-2x) (1+x^2)])dx

cancel and you get

integral 1/(1+x^2) dx which is

v(x)=arctanx

now put these functions into yp

the general solution is yp+yh

y = Ae^(-x) + Bxe^(-x) - 1/2 ln (1+x^2) e^(-x)+ xe^(-x) arctan(x)

with A and B as constants