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

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

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

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

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

Solution steps

y^{''} (t) - 4 y^{'} (t) + 4 y = 0

Solve linear ODE:

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

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

y^{^{''''}} + 3 y^{^{'''}} - 4 y^{^{''}} = 0

y^{^{''}} - 7 y^{^{'}} + 3 y = 0

y^{^{'''}} + y^{^{'}} - 2 y = 0

2 y^{^{''}} + 2 y^{^{'}} + 2 y = 0

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

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