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

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

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

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

y = 3 e^{2 t} + c_{2} t e^{2 t}

Solution steps

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

Solve linear ODE:

y = 3 e^{2 t} + c_{2} t e^{2 t}

y = 3 e^{2 t} + c_{2} t e^{2 t}

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

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

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

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

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

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