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

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

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

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (1) = 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^{^{''}} + 15 y^{^{'}} + 56 y = 0, y (0) = 11, y^{^{'}} (0) = - 85

4 y^{^{''}} = - 25 y

2500 y^{^{''}} + 729 y = 0, y (0) = 2, y^{^{'}} (0) = 3

x^{^{''}} + 8 x^{^{'}} + 10 x = 0, x (0) = 10, x^{^{'}} (0) = 0

2500 y^{^{''}} + 729 y = 0, y (0) = 2, y^{^{'}} (0) = 5

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