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

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

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

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

y = - 2 e^{2 t} t

Solution steps

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

Solve linear ODE:

y = - 2 e^{2 t} t

y = - 2 e^{2 t} t

y^{^{''}} + 20 y^{^{'}} + α y = 0, y (0) = 1, y^{^{'}} (0) = - 24

y^{^{'''}} - 3 y^{^{''}} - 54 y^{^{'}} = 0

77 y^{^{''''}} + 123 y^{^{''}} + 16 y = 0

1.6 x^{^{''}} + 40 x = 0, x (0) = - \frac{1}{3}, x^{^{'}} (0) = \frac{5}{4}

y^{^{''}} - 2 a y^{^{'}} + (a^{2} - e^{2}) y = 0

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