What is the solution for y^{''}-4y+13y=0 ?

The solution for y^{''}-4y+13y=0 is y=c_{1}cos(3t)+c_{2}sin(3t)

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y^{''}-4y+13y=0

y^{^{''}} - 4 y + 13 y = 0

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

Solution steps

y^{''} (t) - 4 y + 13 y = 0

Solve linear ODE:

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

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

49 y^{^{'''}} + 56 y^{^{''}} + 16 y^{^{'}} = 0

\frac{d ^{3} y}{d x ^{3}} + 8 \frac{d y}{d x} = 0

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = 2, y^{^{'}} (0) = 5

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = 2, y^{^{'}} (0) = 4

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

What is the solution for y^{''}-4y+13y=0 ?
The solution for y^{''}-4y+13y=0 is y=c_{1}cos(3t)+c_{2}sin(3t)