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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 14 y^{^{'}} + 65 y = e^{7 x} sin (b x)

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

x^{^{''}} - 4 x^{^{'}} + 4 x = 2, x^{^{'}} (0) = 1

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

y^{^{''}} - y^{^{'}} - 2 y = \frac{( e ^{2 x} + e ^{- 2 x} )}{2}

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