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 (t) = 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)})

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

\int 1 - cos (6 x) d x

t \to 1 lim (\frac{\frac{1}{t} - 1}{t - 1})

n = 0 \sum \infty n^{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)