The discussion centers on solving the differential equation x' = sec(x) with the initial condition x(0) = 2. The solution leads to the equation sin(x) = t + c, where c is a constant. Participants clarify that while the solution can be expressed in terms of arcsin, the problem constraints suggest avoiding this function. Additionally, a function ϕ(θ) = sin(θ) - sin(2) is proposed as a solution to the initial value problem (IVP), with ϕ'(θ) = cos(θ) and ϕ(2) = 0.
PREREQUISITESMathematics students, educators, and professionals dealing with differential equations, particularly those interested in initial value problems and trigonometric solutions.
(i) What is the initial condition? x(0) = ??simo said:I was given an DE equa x'=sec(x) : x(0)
I then solved it to
sin(x)= t + c(c0nstant)
but then I can't solve for x cause they say we shouldn't use arcsin. what can I do
Opalg said:(i) What is the initial condition? x(0) = ??
(ii) There may be some trickery in this question. The sec function is undefined (or "infinite") at some points, so the equation needs to be interpreted cautiously at any such point.
In that case I don't see any way of avoiding an inverse sin function.simo said:it it x(0)=2
Opalg said:In that case I don't see any way of avoiding an inverse sin function.
You solution $\sin x = t +c$ is correct. You can find the constant from the initial condition. Then you can either leave the solution in that form, or you can write it as $x = \arcsin(t+c)$. There is no way of writing the arcsin function in terms of other functions.simo said:are you saying the only way possible to solve this is by the arcsin
Opalg said:You solution $\sin x = t +c$ is correct. You can find the constant from the initial condition. Then you can either leave the solution in that form, or you can write it as $x = \arcsin(t+c)$. There is no way of writing the arcsin function in terms of other functions.