Use the method of undetermined coefficients to solve :?

Question

y'' -16y = 2e^(2x), y(0) = -1/6, y'(0) = 2/3

Particularly concentrate on what am supposed to do with initial values / where to substitute them etc...

Answer 1

The homogeneous equation has solution
y=C1 e^4x+ C2e^-4x
Try to find a particular solution of type y= a*e^2x
y´´-16y =( 4a -16a)e^2x =2 e^2x so a= -1/6
The general solution is

y= C1 e^4x+C2 e^-4x-1/6 *e^2x
y(0)=-1/6 =C1+C2-1/6 so
C1+C2=0
y´= 4C1e^4x-4C2e^-4x-1/3e^2x
y´(0) = 2/3 =4C1-4C2 -1/3


C1-C2=1/4 so C1 = 1/8 and C2 =-1/8

y= 1/8(e^4x-e^4x)-1/16 e^2x