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

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

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5y^{''}+y=0

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

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

Solution steps

5 y^{''} (t) + 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})

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

4 y^{^{''}} - 8 y^{^{'}} + 3 y = 0

12 \frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} - 3 y = 0

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

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

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