What is the solution for y^{''}-6y^'+13y=0,y(0)=0,y^'(0)=-9 ?

The solution for y^{''}-6y^'+13y=0,y(0)=0,y^'(0)=-9 is y=-9/2 e^{3t}sin(2t)

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y^{''}-6y^'+13y=0,y(0)=0,y^'(0)=-9

y^{^{''}} - 6 y^{^{'}} + 13 y = 0, y (0) = 0, y^{^{'}} (0) = - 9

y = - \frac{9}{2} e^{3 t} sin (2 t)

Solution steps

y^{''} (t) - 6 y^{'} (t) + 13 y = 0

Solve linear ODE:

y = - \frac{9}{2} e^{3 t} sin (2 t)

y = - \frac{9}{2} e^{3 t} sin (2 t)

What is the solution for y^{''}-6y^'+13y=0,y(0)=0,y^'(0)=-9 ?
The solution for y^{''}-6y^'+13y=0,y(0)=0,y^'(0)=-9 is y=-9/2 e^{3t}sin(2t)