What is the solution for y^{''}-6y^'+13y=0,y(0)=0,y^'(0)=-5 ?

The solution for y^{''}-6y^'+13y=0,y(0)=0,y^'(0)=-5 is y=-5/2 e^{3t}sin(2t)

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y^{''}-6y^'+13y=0,y(0)=0,y^'(0)=-5

y^{^{''}} - 6 y^{^{'}} + 13 y = 0, y (0) = 0, y^{^{'}} (0) = - 5

y = - \frac{5}{2} e^{3 t} sin (2 t)

Solution steps

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

Solve linear ODE:

y = - \frac{5}{2} e^{3 t} sin (2 t)

y = - \frac{5}{2} e^{3 t} sin (2 t)

\frac{d ^{2} x}{d t ^{2}} + 4 \frac{d x}{d t} + 2 x = 0

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

y^{^{'''}} - 4 y^{^{''}} - 3 y^{^{'}} + 18 y = 0

100 y^{^{''}} - y = 0, y (- 10) = 1, y^{^{'}} (- 10) = - 1

y^{^{''}} + 12 y = 0

What is the solution for y^{''}-6y^'+13y=0,y(0)=0,y^'(0)=-5 ?
The solution for y^{''}-6y^'+13y=0,y(0)=0,y^'(0)=-5 is y=-5/2 e^{3t}sin(2t)