What is the solution for y^{''}-6y^'+12y=0,y(0)=0,y^'(0)=-7 ?

The solution for y^{''}-6y^'+12y=0,y(0)=0,y^'(0)=-7 is y=-7/(sqrt(3))e^{3t}sin(sqrt(3)t)

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y^{''}-6y^'+12y=0,y(0)=0,y^'(0)=-7

y^{^{''}} - 6 y^{^{'}} + 12 y = 0, y (0) = 0, y^{^{'}} (0) = - 7

y = - \frac{7}{3} e^{3 t} sin (3 t)

Solution steps

y^{''} (t) - 6 y^{'} (t) + 12 y = 0

Solve linear ODE:

y = - \frac{7}{3} e^{3 t} sin (3 t)

y = - \frac{7}{3} e^{3 t} sin (3 t)

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

y^{^{''}} + 3 y^{^{'}} = 0, y (0) = - 2, y^{^{'}} (0) = - 3

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

y^{^{''}} + 64 y = 0, y (π) = 0, y (0) = 0

\frac{\frac{d ^{2} f}{d x ^{2}}}{f ( x )} = a^{2}

What is the solution for y^{''}-6y^'+12y=0,y(0)=0,y^'(0)=-7 ?
The solution for y^{''}-6y^'+12y=0,y(0)=0,y^'(0)=-7 is y=-7/(sqrt(3))e^{3t}sin(sqrt(3)t)