What is the solution for x^{''}-3x^'+2x=2sin(t)-6cos(t) ?

The solution for x^{''}-3x^'+2x=2sin(t)-6cos(t) is x=c_{1}e^{2t}+c_{2}e^t+2sin(t)

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x^{''}-3x^'+2x=2sin(t)-6cos(t)

x^{^{''}} - 3 x^{^{'}} + 2 x = 2 sin (t) - 6 cos (t)

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

Solution steps

x^{''} (t) - 3 x^{'} (t) + 2 x = 2 sin (t) - 6 cos (t)

Solve linear ODE:

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

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

x^{^{''}} + 8 x^{^{'}} + 16 x = 10 sin (4 t)

y^{^{''}} - 7 y^{^{'}} + 12 y = 6 e^{x}

y^{^{''}} + 7 y^{^{'}} + 10 y = 3 x^{2} + x - 2

y^{^{''}} - 4 y = 8 x^{3}, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + a y^{^{'}} + b y = sin (x) + 1

What is the solution for x^{''}-3x^'+2x=2sin(t)-6cos(t) ?
The solution for x^{''}-3x^'+2x=2sin(t)-6cos(t) is x=c_{1}e^{2t}+c_{2}e^t+2sin(t)