Help required plz...?

Question

Consider the differential equation
y dx = (ye^(y) - 2x) dy
(i) Find the general solution.
(ii) Give the largest interval I over which the general solution is defined.
(iii) Determine whether there are any transient terms in the general solution.

Answer 1

Best Answer - Chosen by Voters

Assuming y is nonzero, let u = x y^2 be the unknown function of the variable y. du = y(y dx + 2x dy). Multiplying both sides of the equation by y, we obtain:

du = y^2 exp(y) dy

Therefore, u = C + (y^2 -2y + 2) exp(y) [using integration by parts]

x y^2 = C + (y^2 -2y + 2) exp(y)

Now, this solution is valid only over any domain in which y does not vanish...There's a singularity on the line y=0 unless C=-2. It's only for this particular value of the constant C that x is a smooth function of y across y=0.

x = [ (y^2 -2y + 2) exp(y) - 2 ] / y^2

For small values of y, x(y) is infinitrly smooth:
x = y/3 + y^2/4 + y^3/10 + y^4/36 + y^5/168 + ...

Now, what about y as a function of x? Well, the above function x(y) achives a single NEGATIVE minimum x=--A=-0.169998... when y=-1.45123...

For x greater than -A, the solution has two branches, one corresponding to y less than -1.45123... which has an asymptote at x=0 and the other (for which y is above -1.45123) which is valid from x=-A to infinity, which goes through x=0, y=0 without a glitch.

So, the largest interval over which a particular solution (C=-2) is defined for y(x) goes from x=-A... to infinity. (This is a much more interesting question that the "largest interval over which the GENERAL solution (any value of C) is defined, which is simply 0 (excluded) to unfinity.)