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

The solution for y^{''}-4y^'+4y=0,y(0)=-2,y^'(0)=2 is y=-2e^{2t}+6e^{2t}t

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

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

y = - 2 e^{2 t} + 6 e^{2 t} t

Solution steps

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

Solve linear ODE:

y = - 2 e^{2 t} + 6 e^{2 t} t

y = - 2 e^{2 t} + 6 e^{2 t} t

\frac{d}{d x} (x^{3} - 2 x^{2} + 1)

e^{(- 2 s)}

\frac{d y}{d x} = 4 x e^{6 y}

\int x ln (5 x) d x

\frac{d}{d x} ((x^{2} - \frac{5}{4}) e^{x})

What is the solution for y^{''}-4y^'+4y=0,y(0)=-2,y^'(0)=2 ?
The solution for y^{''}-4y^'+4y=0,y(0)=-2,y^'(0)=2 is y=-2e^{2t}+6e^{2t}t