What is the solution for y^{''}-10y^'+25y=6t^2e^{5t} ?

The solution for y^{''}-10y^'+25y=6t^2e^{5t} is y=c_{1}e^{5t}+c_{2}te^{5t}+(e^{5t}t^4)/2

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y^{''}-10y^'+25y=6t^2e^{5t}

y^{^{''}} - 10 y^{^{'}} + 25 y = 6 t^{2} e^{5 t}

y = c_{1} e^{5 t} + c_{2} t e^{5 t} + \frac{e ^{5 t} t ^{4}}{2}

Solution steps

y^{''} (t) - 10 y^{'} (t) + 25 y = 6 t^{2} e^{5 t}

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} t e^{5 t} + \frac{e ^{5 t} t ^{4}}{2}

y = c_{1} e^{5 t} + c_{2} t e^{5 t} + \frac{e ^{5 t} t ^{4}}{2}

y^{^{''}} - 12 y^{^{'}} + 36 y = 18 x + 2

y^{^{''}} - 5 y = 2 x^{3} - 4 x^{2} - x + 6

y^{^{''}} + 4 y^{^{'}} + 5 y = 10 e^{- 2 x} cos (x)

y^{^{''}} + 6 y^{^{'}} + 9 y = cos (x)

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{3 x} cos (e^{x})

What is the solution for y^{''}-10y^'+25y=6t^2e^{5t} ?
The solution for y^{''}-10y^'+25y=6t^2e^{5t} is y=c_{1}e^{5t}+c_{2}te^{5t}+(e^{5t}t^4)/2