What is the solution for 2y^{''}+26y=0,y(0)=9,y^'(0)=5 ?

The solution for 2y^{''}+26y=0,y(0)=9,y^'(0)=5 is y=9cos(sqrt(13)t)+5/(sqrt(13))sin(sqrt(13)t)

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2y^{''}+26y=0,y(0)=9,y^'(0)=5

2 y^{^{''}} + 26 y = 0, y (0) = 9, y^{^{'}} (0) = 5

y = 9 cos (13 t) + \frac{5}{13} sin (13 t)

Solution steps

2 y^{''} (t) + 26 y = 0

Solve linear ODE:

y = 9 cos (13 t) + \frac{5}{13} sin (13 t)

y = 9 cos (13 t) + \frac{5}{13} sin (13 t)

What is the solution for 2y^{''}+26y=0,y(0)=9,y^'(0)=5 ?
The solution for 2y^{''}+26y=0,y(0)=9,y^'(0)=5 is y=9cos(sqrt(13)t)+5/(sqrt(13))sin(sqrt(13)t)