What is the solution for x^{''}-4x^'+4x=2 ?

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

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x^{''}-4x^'+4x=2

x^{^{''}} - 4 x^{^{'}} + 4 x = 2

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 8 y^{^{'}} + 15 y = 3 e^{- 3 t}

y^{^{''}} - y^{^{'}} - 12 y = - 5 e^{3 x}

y^{^{''}} + 5 y^{^{'}} + 4 = 0

\frac{d ^{2} y}{d x ^{2}} + y = 3 cos (2 x)

y^{^{''}} + 2 y^{^{'}} = 3 + sin (t)

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