What is the solution for y^{''}-4y^'+4y=0,y(0)=8.1,y^'(0)=3.9 ?

The solution for y^{''}-4y^'+4y=0,y(0)=8.1,y^'(0)=3.9 is y=8.1e^{2t}-12.3e^{2t}t

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y^{''}-4y^'+4y=0,y(0)=8.1,y^'(0)=3.9

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (0) = 8.1, y^{^{'}} (0) = 3.9

y = 8.1 e^{2 t} - 12.3 e^{2 t} t

Solution steps

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

Solve linear ODE:

y = 8.1 e^{2 t} - 12.3 e^{2 t} t

y = 8.1 e^{2 t} - 12.3 e^{2 t} t

y^{^{'''}} + y^{^{''}} - 24 y^{^{'}} + 36 y = 0

36 y^{^{''}} - 132 y^{^{'}} + 121 y = 0, y (0) = 5, y^{^{'}} (0) = 5

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

625 y^{^{''}} - 9 y = 0, y (0) = 9, y^{^{'}} (0) = 8

\frac{5}{8} \frac{d ^{2} x}{d t ^{2}} + 40 x = 0

What is the solution for y^{''}-4y^'+4y=0,y(0)=8.1,y^'(0)=3.9 ?
The solution for y^{''}-4y^'+4y=0,y(0)=8.1,y^'(0)=3.9 is y=8.1e^{2t}-12.3e^{2t}t