Differential Equations?

Question

Ok I was trying to explain this to my friend who is in differential equations, but i couldn't remember how to exactly do it myself.

The problem looks like it should be super easy

y" + 2y' + y = e^x

I know you have to find the solutions of the homogeneous part, which would be y(x)=ce^x (its a double root when you work it out)

But the problem i can't remember is how to figure out the "particular" solution. Could somebody refresh my memory on this?

Answer 1

Actually the homogeneous part should be Ce^x+Dxe^x

There are a number of ways to find a particular soln but for this problem you could guess a sol of the form
(Ax^2+Bx+Z)e^x. Plug it into the original equation and solve for A,B,Z.

Answer 2

The solution of the homogeneous is
y=C1 e^-x +C2 x*e^-x as -1 is a double root
Now to find a particular solution take C1 and C2 as functions
(I know this method as "variation of constants")
If you call y1 and y2 the two solutions
y´= C´1y1+C´2 y2 +C1y´1+C2y´2
We impose the condition
C´1y1+C´2y2=0 (arbitrarily)
If you derivate again and put the results into your equation you get finally
C´1= P*y2/(y2y´1-y1y´2)
C´2= Py1/(y1y´2-y2y´1) where P is the second side
If you integrate
C1=q1(x)+A
and C2 = q2(x)+B
so the general solution is
q1(x)*y1+q2(x)*y2 +Ay1+By2 where A and B are constants