What is the solution for y^{''}-8y+20y=0 ?

The solution for y^{''}-8y+20y=0 is y=c_{1}cos(2sqrt(3)t)+c_{2}sin(2sqrt(3)t)

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y^{''}-8y+20y=0

y^{^{''}} - 8 y + 20 y = 0

y = c_{1} cos (23 t) + c_{2} sin (23 t)

Solution steps

y^{''} (t) - 8 y + 20 y = 0

Solve linear ODE:

y = c_{1} cos (23 t) + c_{2} sin (23 t)

y = c_{1} cos (23 t) + c_{2} sin (23 t)

(D^{4} + 6 D^{3} + 9 D^{2}) y = 0

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

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

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

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

What is the solution for y^{''}-8y+20y=0 ?
The solution for y^{''}-8y+20y=0 is y=c_{1}cos(2sqrt(3)t)+c_{2}sin(2sqrt(3)t)