What is the solution for (d^2y)/(dt^2)-6(dy)/(dt)+13y=0 ?

The solution for (d^2y)/(dt^2)-6(dy)/(dt)+13y=0 is y=e^{3t}(c_{1}cos(2t)+c_{2}sin(2t))

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(d^2y)/(dt^2)-6(dy)/(dt)+13y=0

\frac{d ^{2} y}{d t ^{2}} - 6 \frac{d y}{d t} + 13 y = 0

y = e^{3 t} (c_{1} cos (2 t) + c_{2} sin (2 t))

Solution steps

\frac{d ^{2} y}{d t ^{2}} - 6 \frac{d y}{d t} + 13 y = 0

Solve linear ODE:

y = e^{3 t} (c_{1} cos (2 t) + c_{2} sin (2 t))

y = e^{3 t} (c_{1} cos (2 t) + c_{2} sin (2 t))

y^{^{''}} - \frac{3}{10} y^{^{'}} + \frac{1}{50} y = 0

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

y^{^{'''}} - 18 y^{^{''}} + 95 y^{^{'}} - 150 y = 0

\frac{d ^{2} y}{d x ^{2}} + 11 \frac{d y}{d x} + 28 y = 0

y^{^{''''}} + y^{^{'''}} - 5 y^{^{''}} - 3 y^{^{'}} + \frac{1}{2} y = 0

What is the solution for (d^2y)/(dt^2)-6(dy)/(dt)+13y=0 ?
The solution for (d^2y)/(dt^2)-6(dy)/(dt)+13y=0 is y=e^{3t}(c_{1}cos(2t)+c_{2}sin(2t))