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

The solution for y^{''}+3y^'-y=t is y=c_{1}e^{((-3+sqrt(13))t)/2}+c_{2}e^{((-3-sqrt(13))t)/2}-t-3

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y^{''}+3y^'-y=t

y^{^{''}} + 3 y^{^{'}} - y = t

y = c_{1} e^{\frac{( - 3 + 13 ) t}{2}} + c_{2} e^{\frac{( - 3 - 13 ) t}{2}} - t - 3

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{( - 3 + 13 ) t}{2}} + c_{2} e^{\frac{( - 3 - 13 ) t}{2}} - t - 3

y = c_{1} e^{\frac{( - 3 + 13 ) t}{2}} + c_{2} e^{\frac{( - 3 - 13 ) t}{2}} - t - 3

y^{^{''}} - 5 y^{^{'}} + 4 y = 3 e^{x} sin (x)

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

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} = x - 8 e^{x}

y^{^{'''}} - y^{^{''}} - 16 y^{^{'}} + 16 y = 9 - e^{x} + e^{4 x}

y^{^{''}} + y = t^{2} + 2

What is the solution for y^{''}+3y^'-y=t ?
The solution for y^{''}+3y^'-y=t is y=c_{1}e^{((-3+sqrt(13))t)/2}+c_{2}e^{((-3-sqrt(13))t)/2}-t-3