What is the solution for x^{''}-4x^'+4x=4e^{2t},x^0=-1,x^'0=-4 ?

The solution for x^{''}-4x^'+4x=4e^{2t},x^0=-1,x^'0=-4 is x=c_{1}e^{2t}+c_{2}te^{2t}+2t^2e^{2t}

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x^{''}-4x^'+4x=4e^{2t},x^0=-1,x^'0=-4

x^{^{''}} - 4 x^{^{'}} + 4 x = 4 e^{2 t}, x^{0} (0) = - 1, x^{^{'}} 0 = - 4

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 12 y^{^{'}} + 36 y = 10 e^{6 x}

y^{^{''}} - 2 y^{^{'}} + y = 14 x^{\frac{2}{3}} e^{x}

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 6 y = 4 e^{2 x}

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = t^{2} e^{t}

11 x^{^{''}} + 112.5 x = 5 e^{- t}, x (0) = 0, x^{^{'}} (0) = 0

What is the solution for x^{''}-4x^'+4x=4e^{2t},x^0=-1,x^'0=-4 ?
The solution for x^{''}-4x^'+4x=4e^{2t},x^0=-1,x^'0=-4 is x=c_{1}e^{2t}+c_{2}te^{2t}+2t^2e^{2t}