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

The solution for x^{''}+4x^'-5x=4t^2 is x=c_{1}e^t+c_{2}e^{-5t}-(4t^2)/5-(32t)/(25)-168/125

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x^{''}+4x^'-5x=4t^2

x^{^{''}} + 4 x^{^{'}} - 5 x = 4 t^{2}

x = c_{1} e^{t} + c_{2} e^{- 5 t} - \frac{4 t ^{2}}{5} - \frac{32 t}{25} - \frac{168}{125}

Solution steps

x^{''} (t) + 4 x^{'} (t) - 5 x = 4 t^{2}

Solve linear ODE:

x = c_{1} e^{t} + c_{2} e^{- 5 t} - \frac{4 t ^{2}}{5} - \frac{32 t}{25} - \frac{168}{125}

x = c_{1} e^{t} + c_{2} e^{- 5 t} - \frac{4 t ^{2}}{5} - \frac{32 t}{25} - \frac{168}{125}

What is the solution for x^{''}+4x^'-5x=4t^2 ?
The solution for x^{''}+4x^'-5x=4t^2 is x=c_{1}e^t+c_{2}e^{-5t}-(4t^2)/5-(32t)/(25)-168/125