_{1}

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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.

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 [

Question 6. Integrate the differential equation

Solutions:

leads to the solution:

due to the reason for right hand side to be zero what are in curly brackets together must be zero.

are the expansions of Cosine and Sine trigonometric functions.

Therefore

where

two equations of straight lines passing through the origin (0, 0) one inclined at

then 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

For

by integration

is 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 [

For hold of equality

since

Then by integration, the solution for y, a constant, is obtained. There are two tangents to the circle at the origin.

If

s is the arc length along the curve.

if

then by integration,

If

by integration where G is a constant with

the series expansion

to the first order approximation.

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).

by differentiation

So that

by squaring

So it produces the differential equation

by squaring

for

so it produces the differential equation

The analysis done would be sufficient.

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 are there any Singular Solutions?

The results that we have derived are in concurrence with the solutions of Leonard Euler in his archives.

We acknowledge W. Nadun, S. Susitha, A. C. Ranjith De Alwis for the assistance given. Present author [alternative E-Mail: dealwis_a@yahoo.com] is staying in Lanka [No. 299, Galle Road, Gorakana, Moratuwa] the Historical Golden Lanka (TammaPanni) in the midst of the country, you can find Sri Parwatha according to Eragudi Rock Edict of King Dharmashoka where Buddha Gotama kept his foot step in his visit to Lanka.

De Alwis, A.C.W.L. (2017) Solution to Stokes-Maxwell-Euler Dif- ferential Equation. Applied Mathematics, 8, 410-416. https://doi.org/10.4236/am.2017.83033