What is the solution for y^{''}-4y^'+4y=0,y(0)=3,y^'(0)=1 ?

The solution for y^{''}-4y^'+4y=0,y(0)=3,y^'(0)=1 is y=3e^{2t}-5e^{2t}t

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y^{''}-4y^'+4y=0,y(0)=3,y^'(0)=1

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (0) = 3, y^{^{'}} (0) = 1

y = 3 e^{2 t} - 5 e^{2 t} t

Solution steps

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

Solve linear ODE:

y = 3 e^{2 t} - 5 e^{2 t} t

y = 3 e^{2 t} - 5 e^{2 t} t

(D^{2} - 1) y = 0

y^{^{''}} + 16 y = 0, y^{^{'}} (\frac{π}{4}) = - 3, y^{^{'}} (0) = 2

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (0) = - 1, y^{^{'}} (0) = 0

y^{^{''}} - 3 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = 5

9 y^{^{''}} - 5 y^{^{'}} - 4 y = 0

What is the solution for y^{''}-4y^'+4y=0,y(0)=3,y^'(0)=1 ?
The solution for y^{''}-4y^'+4y=0,y(0)=3,y^'(0)=1 is y=3e^{2t}-5e^{2t}t