What is the solution for y^{''}-2y+10y=0 ?

The solution for y^{''}-2y+10y=0 is y=c_{1}cos(2^{1+1/2}t)+c_{2}sin(2^{1+1/2}t)

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y^{''}-2y+10y=0

y^{^{''}} - 2 y + 10 y = 0

y = c_{1} cos (2^{1 + \frac{1}{2}} t) + c_{2} sin (2^{1 + \frac{1}{2}} t)

Solution steps

y^{''} (t) - 2 y + 10 y = 0

Solve linear ODE:

y = c_{1} cos (2^{1 + \frac{1}{2}} t) + c_{2} sin (2^{1 + \frac{1}{2}} t)

y = c_{1} cos (2^{1 + \frac{1}{2}} t) + c_{2} sin (2^{1 + \frac{1}{2}} t)

y^{^{''}} - 324 y = 0

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

y^{^{''}} + y^{^{'}} - 15 y = 0

6 y^{^{''''}} + 11 y^{^{''}} + 4 y = 0

y^{^{''''}} - 4 y^{^{'''}} + 5 y^{^{''}} - 4 y^{^{'}} + 4 y = 0

What is the solution for y^{''}-2y+10y=0 ?
The solution for y^{''}-2y+10y=0 is y=c_{1}cos(2^{1+1/2}t)+c_{2}sin(2^{1+1/2}t)