What is the solution for y^{''}-4y^'+4y=0,y(0)=5 ?

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

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

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

y = 5 e^{2 t} + c_{2} t e^{2 t}

Solution steps

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

Solve linear ODE:

y = 5 e^{2 t} + c_{2} t e^{2 t}

y = 5 e^{2 t} + c_{2} t e^{2 t}

y^{^{''}} + 8 y^{^{'}} + 41 y = 0, y (0) = 4, y^{^{'}} (0) = - 13

y^{^{''}} - 20 y^{^{'}} + 136 y = 0

y^{^{'''}} + 13 y^{^{''}} + 45 y^{^{'}} + 33 y = 0

\frac{d ^{2} x}{d t ^{2}} - 49 x = 0

9 y^{^{''}} - 12 y^{^{'}} + 4 y = 0, y (0) = e^{\frac{2 x}{3}}

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