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}(1-c_{2}))/2+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 = \frac{e ^{2 t} ( 1 - c _{2} )}{2} + c_{2} e^{2 t} t

Solution steps

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

Solve linear ODE:

y = \frac{e ^{2 t} ( 1 - c _{2} )}{2} + c_{2} e^{2 t} t

y = \frac{e ^{2 t} ( 1 - c _{2} )}{2} + c_{2} e^{2 t} t

y^{^{''}} - 2 y^{^{'}} - 17 y = 0

5 y^{^{''}} - 11 y^{^{'}} + 2 y = 0, y (0) = - 5, y^{^{'}} (0) = - 6.4

y^{^{''}} + 6 y^{^{'}} + 8 y = 0, y (0) = - 6, y^{^{'}} (0) = 18

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

y^{^{''}} + \frac{8}{100} y^{^{'}} + \frac{12}{10000} 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}(1-c_{2}))/2+c_{2}e^{2t}t