# Number of integration constants

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• LagrangeEuler
LagrangeEuler
If we have system of 3 ordinary differential equation in mechanics and we have two initial condition ##\vec{r}(t=0)=0## and ##\vec{v}(t=0)=\vec{v}_0 \vec{i}##. If we somehow get
$$\frac{d^2v_x}{dt^2}=-\omega^2v_x$$
then $$v_x(t)=A\sin(\omega t)+B\cos(\omega t)$$
Two integration constants and one initial condition for velocity. What to do? Should we put that one constant is equal to zero? So ##A=0##, ##B=v_0##?

Gold Member
cooridinate:
$$x=A\frac{1-\cos \omega t}{\omega}+B \frac{\sin \omega t}{\omega}$$
accerelation:
$$a=A\omega \cos \omega t - B\omega \sin \omega t$$
So
$$A= \frac{a_0}{\omega}$$
$$B= v_0$$

LagrangeEuler
Yes, I agree. But suppose that I have only two initial conditions in the beginning. I do not know acceleration. When I integrate ##v_x(t)## I will get
$$x(t)=C\cos(\omega t)+D\sin (\omega t)+E$$
so 3 integration constants. And because ##x(t)## satisfy some differential equation of second order, general solution should have only 2 integration constants. And I can put ##D=0##. Is it good reasoning?

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Gold Member
My formula of x in #2 which includes initial coordinate condition is different from yours. Please criticize it.

LagrangeEuler
I made a typo. From ##v_x(t)=A\sin \omega t+B \cos \omega t## you can write
##\frac{dx}{dt}=A\sin \omega t+B \cos \omega t##
by integrating
##x(t)=-\frac{A}{\omega}\cos \omega t+\frac{B}{\omega}\sin \omega t+E##.
Of course I can say that ##\frac{A}{\omega}=C## and ##\frac{B}{\omega}=D##
Initial condition is ##x(0)=0##. From that
##0=-\frac{A}{\omega}+E##.
I will get the same result as you.

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LagrangeEuler
cooridinate:
$$x=A\frac{1-\cos \omega t}{\omega}+B \frac{\sin \omega t}{\omega}$$
accerelation:
$$a=A\omega \cos \omega t - B\omega \sin \omega t$$
So
$$A= \frac{a_0}{\omega}$$
$$B= v_0$$
I do not think also that this is a general solution. Because by putting ##t=0## first term is always zero, regardless of value of ##A##.

Gold Member
by integrating
sin and cos are to be exchanged.

I agree with your observation that initial coordinate condition does not decide A nor B. As said in #2 initial velocity and initial acceleration decide them.

Popular oscillation equation for x has two constnts to decide the motion, initial cooridinate and initial velocity. Your oscillation equation for v, which has three times derivative of coordinates thus is not Newton's equation of motion, has two constants to decide the motion, initial velocity and initial acceleration. All one rank up in time derivative.

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