What is the solution for ax^{''}+bx=0 ?

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

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ax^{''}+bx=0

a x^{^{''}} + b x = 0

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

Solution steps

a x^{''} (t) + b x = 0

Solve linear ODE:

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

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

\frac{d}{d x} (e^{t a n h (2 x)})

\int 1 - cos (6 x) d x

(0.5) (1.6)

\frac{d y}{d x} x^{3} + y^{3} = 6 x y

y^{^{'}} = \frac{x ( x ^{2} + 1 )}{4 y ^{3}}, y (0) = - \frac{1}{2}

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