What is the solution for 2500y^{''}+729y=0,y(0)=2,y^'(0)=5 ?

The solution for 2500y^{''}+729y=0,y(0)=2,y^'(0)=5 is y=2cos((27t)/(50))+250/27 sin((27t)/(50))

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2500y^{''}+729y=0,y(0)=2,y^'(0)=5

2500 y^{^{''}} + 729 y = 0, y (0) = 2, y^{^{'}} (0) = 5

y = 2 cos (\frac{27 t}{50}) + \frac{250}{27} sin (\frac{27 t}{50})

Solution steps

2500 y^{''} (t) + 729 y = 0

Solve linear ODE:

y = 2 cos (\frac{27 t}{50}) + \frac{250}{27} sin (\frac{27 t}{50})

y = 2 cos (\frac{27 t}{50}) + \frac{250}{27} sin (\frac{27 t}{50})

35 y^{^{''''}} + 52 y^{^{''}} + 9 y = 0

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

\frac{d ^{4} y}{d t ^{4}} - y^{^{'''}} - y^{^{''}} + y^{^{'}} = 0

y^{^{''}} + 5 y^{^{'}} + 4 y = 0, y (0) = 0, y^{^{'}} (0) = 0

2 u^{^{''}} + u^{^{'}} + 3 H (t) = 0

What is the solution for 2500y^{''}+729y=0,y(0)=2,y^'(0)=5 ?
The solution for 2500y^{''}+729y=0,y(0)=2,y^'(0)=5 is y=2cos((27t)/(50))+250/27 sin((27t)/(50))