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

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

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

linear

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

y = 3 e^{2 t} t

Solution steps

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

Solve linear ODE:

y = 3 e^{2 t} t

y = 3 e^{2 t} t

y^{^{''}} + 0.6 y^{^{'}} + 8 y = 0

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

4 \frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 26 y = 0

\frac{d ^{2} y}{d x ^{2}} + y = 0, y = cos (x) + sin (x)

64 y^{^{''}} + 96 y^{^{'}} + 36 y = 0, y (0) = 1, y^{^{'}} (0) = 8

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