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

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

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

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

y = e^{2 t} + c_{2} e^{2 t} t

Solution steps

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

Solve linear ODE:

y = e^{2 t} + c_{2} e^{2 t} t

y = e^{2 t} + c_{2} e^{2 t} t

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

y^{^{''}} + 25 y = 0, y^{1} (0) = e^{5} x

\frac{1}{32} x^{^{''}} + 36 x = 0

y^{^{''}} - y^{^{'}} - 20 y = 0, y (0) = 3, y^{^{'}} (0) = 33

y^{^{''}} + 7 y^{^{'}} - 30 y = 0

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