What is the solution for 2x^{''}+40x=0 ?

The solution for 2x^{''}+40x=0 is x=c_{1}cos(2sqrt(5)t)+c_{2}sin(2sqrt(5)t)

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2x^{''}+40x=0

2 x^{^{''}} + 40 x = 0

x = c_{1} cos (25 t) + c_{2} sin (25 t)

Solution steps

2 x^{''} (t) + 40 x = 0

Solve linear ODE:

x = c_{1} cos (25 t) + c_{2} sin (25 t)

x = c_{1} cos (25 t) + c_{2} sin (25 t)

\frac{x - 5}{x + 5}

f (t) = (1 + e^{2 t})^{2}

\frac{d}{d x} (4 - 6 x)

\frac{d}{d x} (x - 9)

\int 14 arctan (x) d x

What is the solution for 2x^{''}+40x=0 ?
The solution for 2x^{''}+40x=0 is x=c_{1}cos(2sqrt(5)t)+c_{2}sin(2sqrt(5)t)