y"+3y'+2y=sin(e^x)

Answer 1

Characterstic equation of homogenius ODE: r² + 3r + 2 = 0, (r + 2)(r + 1) = 0, r = - 2 or r = - 1.

So, general solution of Homogenius ODE is y_h = C₁e^-2x + C₂e^-x.

Fundamental set of solutions: y₁ = e^-2x, y₂ = e^-x.

We will use variation of parameters formula.

First Wronskian W = det[(e^-2x, e^-x) (-2e^-2x, - e^-x)] = - e^-3x + 2e^-3x = e^-3x.

Then particular solution of given ODE: y_p = - y₁∫y₂sin(e^x)/Wdx + y₂∫y₁sin(e^x)/Wdx =

= - e^-2x∫e^-xsin(e^x)/e^-3xdx + e^-x∫e^-2xsin(e^x)/e^-3xdx = - e^-2x∫e^2xsin(e^x)dx + e^-x∫e^xsin(e^x)dx =

1. - e^-2x∫e^xsin(e^x)d(e^x) = e^-2x cos(e^x) - cos(e^x) + e^-xsin(e^x).

So, general solution of given ODE: y = C₁e^-2x + C₂e^-x + e^-2xcos(e^x) - cos(e^x) + e^-xsin(e^x).