What is the solution for y^{''}-4y^'+4y=0,y(0)=8,y(1)=0 ?

The solution for y^{''}-4y^'+4y=0,y(0)=8,y(1)=0 is y=8e^{2t}-8e^{2t}t

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y^{''}-4y^'+4y=0,y(0)=8,y(1)=0

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (0) = 8, y (1) = 0

y = 8 e^{2 t} - 8 e^{2 t} t

Solution steps

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

Solve linear ODE:

y = 8 e^{2 t} - 8 e^{2 t} t

y = 8 e^{2 t} - 8 e^{2 t} t

\frac{d ^{2} y}{d x ^{2}} - \frac{11}{2} \frac{d y}{d x} + 6 y = 0

y^{^{'''}} - 27 y = 0

(D^{2} - 6 D + 9) y = 0

4 y^{^{''}} + 4 y^{^{'}} + 512 y = 0, y (0) = 0, y^{^{'}} (0) = 7

\frac{d ^{2} y}{d x ^{2}} - 5 (\frac{d y}{d x}) + 6 y = 0

What is the solution for y^{''}-4y^'+4y=0,y(0)=8,y(1)=0 ?
The solution for y^{''}-4y^'+4y=0,y(0)=8,y(1)=0 is y=8e^{2t}-8e^{2t}t