Question

Does anyone know: y''+2y'-3y
=e^3x thank you!

Answer 1

r^2 +2r-3=0

(r-2)(r-1)=0

r=2,1

check my algebra,

Answer 2

The solution of the associated homogeneous differential equation can be found from the auxiliary equation \(m^2+2m-3=0 \implies (m+3)(m-1)=0 \implies m=-3,1\) . So \(y_c=c_1e^{-3x}+c_2e^x\).

Answer 3

e(X)+e(-3x)+1/4e(3x)

Answer 4

the first two terms are the complementary solution while the last one is the particular integral

Answer 5

As for the particular solution, let's assume a solution of the form \(y_p=Ae^{3x} \implies y'=3Ae^{3x} \) and \(y''=9Ae^{3x}\). Plug in the original equation to find A: \(9Ae^{3x}+6Ae^{3x}-3Ae^{3x}=e^{3x} \implies A=\frac{1}{12}\). So, \(y_p=\frac{1}{12}e^{3x}\).

Answer 6

The general solution \(y=y_c+y_p\).

Answer 7

thanks!!

Answer 8

You're welcome!