What is the solution for y^{''}-4y^'+4y=0,y(0)=5,y^'(0)=3 ?

The solution for y^{''}-4y^'+4y=0,y(0)=5,y^'(0)=3 is y=5e^{2t}-7e^{2t}t

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y^{''}-4y^'+4y=0,y(0)=5,y^'(0)=3

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (0) = 5, y^{^{'}} (0) = 3

y = 5 e^{2 t} - 7 e^{2 t} t

Solution steps

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

Solve linear ODE:

y = 5 e^{2 t} - 7 e^{2 t} t

y = 5 e^{2 t} - 7 e^{2 t} t

(D^{4} + 4 D^{3} + 2 D^{2} - 8 D - 8) y = 0

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 y = 0

y^{^{''}} + 4 y^{^{'}} - 2 y = 0

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

(D^{4} - D^{3} - 3 D^{2} + D + 2) y = 0

What is the solution for y^{''}-4y^'+4y=0,y(0)=5,y^'(0)=3 ?
The solution for y^{''}-4y^'+4y=0,y(0)=5,y^'(0)=3 is y=5e^{2t}-7e^{2t}t