What is the solution for 25y^{''}+5y=0 ?

The solution for 25y^{''}+5y=0 is y=c_{1}cos(t/(sqrt(5)))+c_{2}sin(t/(sqrt(5)))

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25y^{''}+5y=0

25 y^{^{''}} + 5 y = 0

y = c_{1} cos (\frac{t}{5}) + c_{2} sin (\frac{t}{5})

Solution steps

25 y^{''} (t) + 5 y = 0

Solve linear ODE:

y = c_{1} cos (\frac{t}{5}) + c_{2} sin (\frac{t}{5})

y = c_{1} cos (\frac{t}{5}) + c_{2} sin (\frac{t}{5})

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What is the solution for 25y^{''}+5y=0 ?
The solution for 25y^{''}+5y=0 is y=c_{1}cos(t/(sqrt(5)))+c_{2}sin(t/(sqrt(5)))