What is the solution for y^{''}-4y^'+4y=0,y(0)=12,y^'(0)=-3 ?

The solution for y^{''}-4y^'+4y=0,y(0)=12,y^'(0)=-3 is y=12e^{2t}-27e^{2t}t

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y^{''}-4y^'+4y=0,y(0)=12,y^'(0)=-3

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (0) = 12, y^{^{'}} (0) = - 3

y = 12 e^{2 t} - 27 e^{2 t} t

Solution steps

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

Solve linear ODE:

y = 12 e^{2 t} - 27 e^{2 t} t

y = 12 e^{2 t} - 27 e^{2 t} t

\int (\frac{1}{x ^{2}} - \frac{1}{x x} + 5) d x

4 x sin (x)

- 4 y^{^{'}} + 12 x^{2} = 0

\frac{\partial}{\partial x} (3 y^{2} cos (5 x))

\frac{\partial}{\partial x} (2 x y e^{x y^{2}} - 3 y^{2})

What is the solution for y^{''}-4y^'+4y=0,y(0)=12,y^'(0)=-3 ?
The solution for y^{''}-4y^'+4y=0,y(0)=12,y^'(0)=-3 is y=12e^{2t}-27e^{2t}t