How can I solve the Trigonometric First Order DE \frac{dx}{dt} = \cos(x+t)?

[Matuku]

\frac{dx}{dt} = \cos(x+t)

I'm having real troubles with this; I tried a substitution of u=x+t but it just ends up as,

t=\tan{u} + \csc{u} + C

And I can't see where to go from there. (the middle function is cosec; it didn't come out very clearly on my screen).

 

[Dick]

I keep getting t=csc(u)-cot(u)+C. You could then put u=x+t. But I don't think you can go much further than that as far as explicitly solving for x(t). It looks like kind of a mess.

 

[Matuku]

You're right, it is -cot(u) rather than tan(u); don't know where I got that from. But you don't believe there's anyway to get it into a form x(t) [or even t(x)]?

 

[Dick]

I don't think so. You can try expressing everything in terms of sin and cos and then apply the addition formulas and hope things magically sort out. But it looks to me like x and t are pretty thoroughly mixed up.

 

[Mark44]

I don't think so. You can try expressing everything in terms of sin and cos and then apply the addition formulas and hope things magically sort out. But it looks to me like x and t are pretty thoroughly mixed up.

That was my thought, too, looking at dx/dt = cos(x + t). I tried the cosine sum formula but didn't find any magic there. cos(x + t) apparently has x and t inextricably twined.