Solve using the method of undetermined coefficients yy2ye3xs

Solve using the method of undetermined coefficients: y\'\'-y\'-2y=e^3xsin(2x)

Solution

Solution :

y\"- y\' - 2y = 0

Characteristic equation is :

m² - m - 2 = 0

(m - 2)(m + 1) = 0

m = - 1 or m = 2

y(x) = Ae^{(- x)} + Be^(2x)

use parametric variation method to solve the particular solution,

wronskeian determinant,

|W| = |{{e^{(- x)}, e^(2x)},{- e^{(- x)}, 2e^(2x)}}|

|W| = 3e^x

Particular solution is :

y(x) = - y1 y2 r(x)/|W|dx + y2 y1 r(x)/|W|dx

y(x) = - e^{(- x)} e^(2x) (e^(3x))sin(2x) / (3e^x)dx + e^(2x) e^(-x) (e^(3x))sin(2x) /(3e^x)dx

y(x) = - e^{(- x)} 1/3 e^(4x)sin(2x) dx + e^(2x) 1/3 e^x sin(2x)dx

y(x) = 1/30 e^(3x)(cos(2x) - 2sin(2x)) + 1/15 e^(3x)(sin(2x) - 2cos(2x))

y(x) = - 1/10 e^(3x)cos(2x)

General solution is :

y(x) = y(x) homogen + y(x) particular

y(x) = Ae^{(- x)} + Be^(2x) - 1/10 e^(3x)cos(2x)

Here A and B is an arbitrary constant...