Solution to Stokes-Maxwell-Euler Differential Equation

Solutions to the differential equation in Smith’s Prize Examination taken by Maxwell are discussed. It was a competitive examination using which skill full students were identified and James Clerk Maxwell was one of them. He later formulated the theory of Electromagnetism and predicted the light speed & its value was subsequently confirmed by experiments. Light travel in a direction perpendicular to oscillating electric and magnetic field through a vacuum from sun. In the same exam paper, Maxwell answered the question related to Stokes Theorem of vector calculus which was used in the formalism of Electromagnetic theory.


Introduction
Question 6 was a differential equation in the Smith's prize exam.Stokes asked it to integrate.The exam was taken by James Clerk Maxwell at Cambridge in February 1854.Stokes was a personal friend of Maxwell (George Gabriel Stokes [1], EsQ.M.A., was the Lucasian Professor at Cambridge).Maxwell completed the exam and tied for first.There is a solution to this problem in the Mathematical Archives of Leonard Euler [2] [3] [4].The solution to this differential equation appears with a tangent to a circle whose center is at the origin of the coordinate system ( ) 0, 0 .But our discussion is not the Euler's way of solutions to the problem that is geometrical.But fresh independent solutions that can be given in

Smith's Prize Exam
leads to the solution: due to the reason for right hand side to be zero what are in curly brackets together must be zero.
( ) = ± to the first order approximation then for A = 0, C = 0, two equations of straight lines passing through the origin (0, 0) one inclined at Sin a A to the positive side of the x-axis and other inclined 45˚ to the negative side of the x-axis.That is only when A = 0.For non zero A, then 1 Cos Sin A A + = is not a correct result in general but under certain specific conditions it can be true.
1 0 1 0 1 1 1 + = + = = = satisfy the given rela- tion very successfully.The most generalized solution is π 2 A = ± so that For     the applied restriction on x. ( ) then by integration, ( ) is the arc length along the curve under the restriction applied on x y product.by integration.So that equation of a circle exactly fit to the given differential equation, it is the perfect solution and it is circle of radius a whose origin at (0, 0).

Tangent to a Circle
x y a + = a circle of radius a whose center is at the origin (0, 0) meet the requirements of Euler's formulation of the same problem geometrically.By squaring and reversing d d 0 x x y y + = it leads to the given differential equation as started.
= constants, so it produces the given differential equation   The analysis done would be sufficient.

Conclusion
The results that we have derived are in concurrence with the solutions of Leonard Euler in his archives.
a mathematics prize exam without knowing what Euler has said before.Solutions were expected from Maxwell by Stokes in the exam.But in the Field of Mathematics, Leonard Euler was another historical figure.On the other hand, Euler's Solutions are in concurrence with us.
two real roots exists cutting the circle by the straight line.The most important observation is if a 2 (1 + B 2 ) = C 2 the two real roots coincide and the straight line touch the circle and it becomes a tangent to the circle.But unfortunate fact is this cannot be correct due to the reason ( ) x = 0 there are tangents to the circle.
the solution a straight line passes through the origin depending on the value and sign of D a constant.Solutions given by Maxwell are not easily accessible by internet web other than his published volumes of Collected Scientific Papers[5].

For
+ == the equation of a circle.
Then by integration, the solution for y, a constant, is obtained.There are two tangents to the circle at the origin.
the arc length along the curve.d d y xis the limiting tangential gradient to the curve at any point.

α
are constants assumed.Whether Maxwell answered problem 6 is not stated but he answered the problem 8 of the question paper as well noted which is Stokes Theorem popular in Vector Calculus.What is presented could have been the answers if Maxwell attempted the question 6, these are scattered thoughts.There are seventeen problems.Maxwell answered Stokes Theorem very successfully.So it has become a celebrated proof subsequently.
the gradient of a tangent to the circle & if the point of contact is ( ) of a rectangular hyperbola.PV k = the Boyle's curve has the same shape the relationship in between pressure and volume of a gas discovered by Robert Boyle in the sixteenth century.d d 0 x y y x + = by differentiation.
Euler in 1758 paper (E 236) Explanation of Certain Paradoxes In Integral Calculus, states the same problem as: Given the point A, find the curve EM such that the perpendicular AV, derived from point A onto some tangent of the curve MV, is the same size everywhere & solution to it is the given differential equation by Stokes in the Smiths Prize Exam Paper as Question 6 Sat by Maxwell that must be from Euler's Mathematical Literature.Further Stokes asked there are such cases that was questioned by Stokes when the diffe- rential equation is attempted for solutions.But no reference to Euler appears in the Exam.Mean while Euler has very long detail solution that Maxwell might have not seen before but Stokes must have known them prior to the Exam problem that was set by him.