What is the solution for y^{''}-4y^'+12y=0,y(0)=0,y^'(0)=1 ?

The solution for y^{''}-4y^'+12y=0,y(0)=0,y^'(0)=1 is y=(e^{2t}sin(2^{1/2+1}t))/(2sqrt(2))

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y^{''}-4y^'+12y=0,y(0)=0,y^'(0)=1

y^{^{''}} - 4 y^{^{'}} + 12 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y = \frac{e ^{2 t} sin ( 2 ^{\frac{1}{2} + 1} t )}{2 2}

Solution steps

y^{''} (t) - 4 y^{'} (t) + 12 y = 0

Solve linear ODE:

y = \frac{e ^{2 t} sin ( 2 ^{\frac{1}{2} + 1} t )}{2 2}

y = \frac{e ^{2 t} sin ( 2 ^{\frac{1}{2} + 1} t )}{2 2}

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

3 x^{^{''}} - 7 x^{^{'}} + 4 x = 0

y^{^{'''}} + 6 y^{^{''}} - 27 y^{^{'}} = 0

\frac{d ^{2} y}{d t ^{2}} - 7 \frac{d y}{d t} = 0

y^{^{''}} + 4 y^{^{'}} - 5 y = 0, y (ln (2)) = 3, y (0) = 33

What is the solution for y^{''}-4y^'+12y=0,y(0)=0,y^'(0)=1 ?
The solution for y^{''}-4y^'+12y=0,y(0)=0,y^'(0)=1 is y=(e^{2t}sin(2^{1/2+1}t))/(2sqrt(2))