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

The solution for y^{''}-5y^'+4y=2t^2 is y=c_{1}e^{4t}+c_{2}e^t+(t^2)/2+(5t)/4+21/16

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

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

y = c_{1} e^{4 t} + c_{2} e^{t} + \frac{t ^{2}}{2} + \frac{5 t}{4} + \frac{21}{16}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} e^{t} + \frac{t ^{2}}{2} + \frac{5 t}{4} + \frac{21}{16}

y = c_{1} e^{4 t} + c_{2} e^{t} + \frac{t ^{2}}{2} + \frac{5 t}{4} + \frac{21}{16}

x^{^{''}} - x^{^{'}} - 12 x = e^{4 t}

y^{^{''}} - 60 y^{^{'}} - 900 y = 5 e^{10 x}

y^{^{''}} - 16 y = x e^{x}

y^{^{''}} - 2 y^{^{'}} - 8 y = e^{- 2 x} + 2 x^{2}

\frac{d ^{2} y}{d x ^{2}} + c^{2} y = sec^{2} (c x)

What is the solution for y^{''}-5y^'+4y=2t^2 ?
The solution for y^{''}-5y^'+4y=2t^2 is y=c_{1}e^{4t}+c_{2}e^t+(t^2)/2+(5t)/4+21/16