What is the solution for y^{''}-4y^'+4y=0,y^1(0)=e^{2x} ?

The solution for y^{''}-4y^'+4y=0,y^1(0)=e^{2x} is y=c_{1}e^{2x}+c_{2}xe^{2x}

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y^{''}-4y^'+4y=0,y^1(0)=e^{2x}

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y^{1} (0) = e^{2 x}

y = c_{1} e^{2 x} + c_{2} x e^{2 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x}

y = c_{1} e^{2 x} + c_{2} x e^{2 x}

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

y^{^{'''}} - 5 y^{^{''}} + 2 y^{^{'}} + 8 y = 0

y^{^{''}} - 4 y^{^{'}} - 3 y = 0

y^{^{''}} + 9 y^{^{'}} + 8 y = 0

y^{^{''}} + k y^{^{'}} = 0

What is the solution for y^{''}-4y^'+4y=0,y^1(0)=e^{2x} ?
The solution for y^{''}-4y^'+4y=0,y^1(0)=e^{2x} is y=c_{1}e^{2x}+c_{2}xe^{2x}