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

The solution for y^{''}-3y^'+2y=sin(t) is y=c_{1}e^{2t}+c_{2}e^t+1/10 sin(t)+3/10 cos(t)

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y^{''}-3y^'+2y=sin(t)

y^{^{''}} - 3 y^{^{'}} + 2 y = sin (t)

y = c_{1} e^{2 t} + c_{2} e^{t} + \frac{1}{10} sin (t) + \frac{3}{10} cos (t)

Solution steps

y^{''} (t) - 3 y^{'} (t) + 2 y = sin (t)

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{t} + \frac{1}{10} sin (t) + \frac{3}{10} cos (t)

y = c_{1} e^{2 t} + c_{2} e^{t} + \frac{1}{10} sin (t) + \frac{3}{10} cos (t)

y^{^{''}} + 2 y^{^{'}} - 8 y = 2 e^{- 2 x} - e^{- x}

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{- 4 t}

y^{^{''}} - 4 y^{^{'}} - 5 y = 2 e^{3 x}

y^{^{''}} - 5 y^{^{'}} + 6 y = x^{2}

y^{^{''}} - 4 y^{^{'}} - 12 y = 2 x + 6

What is the solution for y^{''}-3y^'+2y=sin(t) ?
The solution for y^{''}-3y^'+2y=sin(t) is y=c_{1}e^{2t}+c_{2}e^t+1/10 sin(t)+3/10 cos(t)