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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 4 y = 2 e^{2 x}

y^{^{''}} + y^{^{'}} - 2 y = 1 + t

y^{^{'''}} + 25 y^{^{'}} = tan (5 x)

y^{^{''}} - 5 y^{^{'}} + 6 y = (x + 1)^{2} e^{2 x}

y^{^{''}} - 2 y^{^{'}} + 5 y = - 8

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