What is the solution for y^{''}-6y^'+9y=t^2e^{3t} ?

The solution for y^{''}-6y^'+9y=t^2e^{3t} is y=c_{1}e^{3t}+c_{2}te^{3t}+(e^{3t}t^4)/(12)

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y^{''}-6y^'+9y=t^2e^{3t}

y^{^{''}} - 6 y^{^{'}} + 9 y = t^{2} e^{3 t}

y = c_{1} e^{3 t} + c_{2} t e^{3 t} + \frac{e ^{3 t} t ^{4}}{12}

Solution steps

y^{''} (t) - 6 y^{'} (t) + 9 y = t^{2} e^{3 t}

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} t e^{3 t} + \frac{e ^{3 t} t ^{4}}{12}

y = c_{1} e^{3 t} + c_{2} t e^{3 t} + \frac{e ^{3 t} t ^{4}}{12}

y^{^{''}} - 25 y = cosh (5 x)

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

3 y^{^{''}} + 27 y = 2 cos (3 t)

y^{^{''}} + 16 y = 32 t^{2} - 48 t + 100, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + y^{^{'}} = cos (x), y (0) = 0, y^{^{'}} (0) = 2

What is the solution for y^{''}-6y^'+9y=t^2e^{3t} ?
The solution for y^{''}-6y^'+9y=t^2e^{3t} is y=c_{1}e^{3t}+c_{2}te^{3t}+(e^{3t}t^4)/(12)