What is the solution for y^{''}+4y^'-5=2t^2-3t+6 ?

The solution for y^{''}+4y^'-5=2t^2-3t+6 is y=c_{1}+c_{2}e^{-4t}+(t^3)/6-(t^2)/2+3t

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y^{''}+4y^'-5=2t^2-3t+6

y^{^{''}} + 4 y^{^{'}} - 5 = 2 t^{2} - 3 t + 6

y = c_{1} + c_{2} e^{- 4 t} + \frac{t ^{3}}{6} - \frac{t ^{2}}{2} + 3 t

Solution steps

y^{''} (t) + 4 y^{'} (t) - 5 = 2 t^{2} - 3 t + 6

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 t} + \frac{t ^{3}}{6} - \frac{t ^{2}}{2} + 3 t

y = c_{1} + c_{2} e^{- 4 t} + \frac{t ^{3}}{6} - \frac{t ^{2}}{2} + 3 t

y^{^{''}} + 25 y = 6 x, y (0) = 1, y^{^{'}} (0) = 5

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

y^{^{''}} + 16 y = 3, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + y^{^{'}} = x, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - 12 y^{^{'}} + 36 y = 42 x + 2

What is the solution for y^{''}+4y^'-5=2t^2-3t+6 ?
The solution for y^{''}+4y^'-5=2t^2-3t+6 is y=c_{1}+c_{2}e^{-4t}+(t^3)/6-(t^2)/2+3t