What is the solution for y^{''}-5y^'+4y=1 ?

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

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y^{''}-5y^'+4y=1

y^{^{''}} - 5 y^{^{'}} + 4 y = 1

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

Solution steps

y^{''} (t) - 5 y^{'} (t) + 4 y = 1

Solve linear ODE:

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

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

y^{^{''}} - 8 y^{^{'}} - 9 y = - 10 e^{- x} - 425 cos (2 x)

y^{^{''}} - 100 y = 9 t^{2} e^{10 t}

y^{^{''}} - 16 y = cosh (4 x)

y^{^{''}} + 9 y = \frac{csc ( 3 x )}{4}

y^{^{''}} + 9 y = \frac{3}{2} sin (2 t), y (0) = 0, y^{^{'}} (0) = 0

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