Question

What are the characteristics roots of a lde having 4+xe^2x as its particular solution

Answer 1

Step-by-step explanation:

Example: Find the particular solution for the d.e. y '' - 4 y = 2 e 2x ... So the characteristic equation is r2 - 4 = 0, and roots r = ± 2, ... of the d.e. with its right hand side as follows: 4 C 3 e .

Answer 2

γe2x+8γxe2x+4γx2e2x−4(2γxe2x+2γx2e2x)+4γx2e2x=2e2x

⇒γ=1 and yp=x2e2x

The general solution is: yg=yc+yp

yg=e2x(αx+β)+x2e2x

=e2x(x2+αx+β)

Now applying the IV's:

y(0)=1