What is the solution for y^{''}+6.2832y=0 ?

The solution for y^{''}+6.2832y=0 is y=c_{1}cos((sqrt(3927)t)/(25))+c_{2}sin((sqrt(3927)t)/(25))

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y^{''}+6.2832y=0

y^{^{''}} + 6.2832 y = 0

y = c_{1} cos (\frac{3927 t}{25}) + c_{2} sin (\frac{3927 t}{25})

Solution steps

y^{''} (t) + 6.2832 y = 0

Solve linear ODE:

y = c_{1} cos (\frac{3927 t}{25}) + c_{2} sin (\frac{3927 t}{25})

y = c_{1} cos (\frac{3927 t}{25}) + c_{2} sin (\frac{3927 t}{25})

n = 0 \sum \infty (\frac{4}{5})^{n}

\int 18 t e^{t} d t

5 x^{2} y^{^{'}} = y^{^{'}} + 7 x e - y

\int x^{2} (1 - x)^{\frac{3}{2}} d x

t y^{^{'}} = 2 y - 3 t^{2} cos (4 t)

What is the solution for y^{''}+6.2832y=0 ?
The solution for y^{''}+6.2832y=0 is y=c_{1}cos((sqrt(3927)t)/(25))+c_{2}sin((sqrt(3927)t)/(25))