What is the solution for 2x^{''}+7x=0 ?

The solution for 2x^{''}+7x=0 is x=c_{1}cos(sqrt(7/2)t)+c_{2}sin(sqrt(7/2)t)

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2x^{''}+7x=0

2 x^{^{''}} + 7 x = 0

x = c_{1} cos (\frac{7}{2} t) + c_{2} sin (\frac{7}{2} t)

Solution steps

2 x^{''} (t) + 7 x = 0

Solve linear ODE:

x = c_{1} cos (\frac{7}{2} t) + c_{2} sin (\frac{7}{2} t)

x = c_{1} cos (\frac{7}{2} t) + c_{2} sin (\frac{7}{2} t)

(D^{2} + 4 D + 29)^{2} y = 0

y^{^{''}} - 2 y^{^{'}} + 4 y = 0, y (0) = 2, y^{^{'}} (0) = 0

x^{^{''}} + 2 x^{^{'}} + 50 x = 0, x (0) = - \frac{3}{2}, x^{^{'}} (0) = - 2

49 y^{^{''}} - y = 0, y (0) = 0, y^{^{'}} (0) = 1

x^{^{''}} - c x = 0

What is the solution for 2x^{''}+7x=0 ?
The solution for 2x^{''}+7x=0 is x=c_{1}cos(sqrt(7/2)t)+c_{2}sin(sqrt(7/2)t)