What is the solution for x^{''}-4x^'+4x=2,x^'(0)=1 ?

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

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x^{''}-4x^'+4x=2,x^'(0)=1

x^{^{''}} - 4 x^{^{'}} + 4 x = 2, x^{^{'}} (0) = 1

x = \frac{e ^{2 t} ( 1 - c _{2} )}{2} + c_{2} e^{2 t} t + \frac{1}{2}

Solution steps

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

Solve linear ODE:

x = \frac{e ^{2 t} ( 1 - c _{2} )}{2} + c_{2} e^{2 t} t + \frac{1}{2}

x = \frac{e ^{2 t} ( 1 - c _{2} )}{2} + c_{2} e^{2 t} t + \frac{1}{2}

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

y^{^{''}} - y^{^{'}} - 2 y = \frac{( e ^{2 x} + e ^{- 2 x} )}{2}

y^{^{''}} + 4 y^{^{'}} + 5 y = - 10 + e^{- x}

y^{^{''}} - 9 y = e^{3 x}, y (0) = 1, y^{^{'}} (0) = - 5

y^{^{''}} + y^{^{'}} - 6 y = e^{- 5 x} (- 102 + 146 x - 70 x^{2})

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