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

The solution for x^{''}-4x^'+4x=e^{2t},x(0)=2,x^'(0)=1 is x=2e^{2t}-3e^{2t}t+(e^{2t}t^2)/2

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

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

x = 2 e^{2 t} - 3 e^{2 t} t + \frac{e ^{2 t} t ^{2}}{2}

Solution steps

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

Solve linear ODE:

x = 2 e^{2 t} - 3 e^{2 t} t + \frac{e ^{2 t} t ^{2}}{2}

x = 2 e^{2 t} - 3 e^{2 t} t + \frac{e ^{2 t} t ^{2}}{2}

y^{^{'''}} + 256 y^{^{'}} = tan (16 x)

y^{^{''}} - 4 y = 3 x^{2} e^{3 x}

y^{^{''}} + 6 y^{^{'}} + 5 y = 10

y^{^{''}} + 6 y^{^{'}} + 5 y = 15

y^{^{''}} + 3 y^{^{'}} + 2 y = 1 + 3 x + x^{2}

What is the solution for x^{''}-4x^'+4x=e^{2t},x(0)=2,x^'(0)=1 ?
The solution for x^{''}-4x^'+4x=e^{2t},x(0)=2,x^'(0)=1 is x=2e^{2t}-3e^{2t}t+(e^{2t}t^2)/2