Using second order differentiation how do you solve y2y3yex3

Solution

The characteristic equation is:

r^2 - 2r - 3 = 0

(r-3)(r+1) = 0

roots are -1 , 3.

So the homogeneous solution is:

yh = Ae^(-x) + Be^(3x)

To find particular solution we have:

p(x) = e^x + 3e^3x -> yp = Ce^x + Dx.e^(3x)

We have:

yp\' = Ce^x + (3Dx+D)e^(3x)

yp\" = Ce^x + (9Dx+6D)e^(3x)

Therefore:

yp\" - 2yp\' - 3yp = -4Ce^x + D.e^(3x) = e^x + 3e^3x  

-4C = 1 -> C = -1/4

D = 3

So the final solution is:

y = yh + yp = Ae^(-x) + Be^(3x) -1/4 e^x + 3x.e^(3x)