What is the solution for 1x^{''}-3x^'+2x=20sin(2t) ?

The solution for 1x^{''}-3x^'+2x=20sin(2t) is x=c_{1}e^{2t}+c_{2}e^t-sin(2t)+3cos(2t)

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1x^{''}-3x^'+2x=20sin(2t)

1 x^{^{''}} - 3 x^{^{'}} + 2 x = 20 sin (2 t)

x = c_{1} e^{2 t} + c_{2} e^{t} - sin (2 t) + 3 cos (2 t)

Solution steps

1 x^{''} (t) - 3 x^{'} (t) + 2 x = 20 sin (2 t)

Solve linear ODE:

x = c_{1} e^{2 t} + c_{2} e^{t} - sin (2 t) + 3 cos (2 t)

x = c_{1} e^{2 t} + c_{2} e^{t} - sin (2 t) + 3 cos (2 t)

What is the solution for 1x^{''}-3x^'+2x=20sin(2t) ?
The solution for 1x^{''}-3x^'+2x=20sin(2t) is x=c_{1}e^{2t}+c_{2}e^t-sin(2t)+3cos(2t)