Differential equation help please?

Question

what is the general solution of the differential equation

dy/dx = sin(y-x)




help me please
thanks

Answer 1

So simple, and so very, very nasty!

I have managed to find the answer, I think. There's probably a numerical error somewhere! Be warned: I'm going to switch to complex variables temporarily. I promise you we will be back in R by the time we finish. ;-)

Let u = e^(i(y-x)) = cos(y-x) + i sin(y-x).
Then u' = e^(i(y-x)).i(y'-1) = ui(y'-1), and
sin(y-x) = 1/2i (e^i(y-x) - e^(-i(y-x))) = 1/2i (u - 1/u) by definition.
So y' = 1 + u'/(ui) = 1/2i (u - 1/u).
Multiplying through by 2ui we get
2ui + 2u' = u^2 - 1
=> 2u' = u^2 - 2ui - 1 = (u - i)^2
So let v = u-i, then v' = v^2 / 2.
So ∫dv/v^2 = ∫dx/2
=> -1/v = (x + c)/2
=> v = -2/(x+c)
=> u = -2/(x+c) + i.

Now remember that u = e^i(y-x) has magnitude 1 since x and y are real; also note that c is a complex constant, not a real one. Let c = a + bi. Then
u = -2/(x + a + bi) + i
= -2(x+a-bi) / [(x+a)^2 + b^2] + i
The real part of u is -2(x+a)/[(x+a)^2+b^2] and the imaginary part is 2b/[(x+a)^2+b^2] + 1. Since ||u|| = 1 we have
[-2(x+a)]^2 + [2b + (x+a)^2+b^2]^2 = [(x+a)^2+b^2]^2
=> 4(x+a)^2 + 4b^2 + 2(2b)[(x+a)^2+b^2] = 0
=> 4(x+a)^2 (1+b) + 4b^2(1+b) = 0.
Remember this must hold everywhere, so we must have b = -1.

Now the real part of u (given above) is also cos(y-x), and the imaginary part is sin(y-x). So we get two formulae:

cos(y-x) = -2(x+a)/[(x+a)^2+1]
and
sin(y-x) = 1 - 2/[(x+a)^2+1]

(These are nearly equivalent, but not 100% equivalent because of quadrant differences. Use both together to determine the correct quadrant for y-x. Note that solutions will be periodic in y with a period of 2π, as we could have predicted from the original ODE.)

From this we can derive tan(y-x) = -[(x+a)^2+1]/[2(x+a)] + 1/(x+a)
= 1/(x+a) - (x+a)/2 - 1/(2(x+a))
= (1/2)[1/(x+a) - (x+a)].
You may very well prefer this formula, but half the solutions will be incorrect because y-x will be in the wrong quadrant, so make sure you pick the right ones. If you are using a computer or calculator that has a two-input arctan function, you can use that with the cos and sin formulae given above to get the correct values, though y will still jump by 2π when the arctan function wraps around from -π to π. Note that in this case you can scale the inputs by any desired positive amount, so I'd scale them by [(x+a)^2+1].

Note that in the limit a-> ∞ we get cos(y-x) = 0, sin(y-x) = 1 which leads to the additional solutions y = x + (2k + 1/2)π, k any integer, which is obviously valid since dy/dx = 1 and sin(y-x) = sin(2kπ+π/2) = 1. Also, you can show easily that y is asymptotic to these lines as x->∞ or x->-∞, for any value of a; the interesting region in these graphs is when (x+a) is not very large in magnitude.

Answer 2

dy/dx = (x^2)(8 + y) First, we separate the variables by using multiplying by using dx and dividing by using (8 + y): dy/(8 + y) = (x^2)dx combine the two factors, remembering the left will use a organic log and the wonderful might have a continuing: ln(8 + y) = (a million/3)x^3 + C develop e to the means of the two area: 8 + y = Ae^((a million/3)x^3) Subtract 8 and we've the final type: y = Ae^((a million/3)x^3) - 8 Now, we've the factor (0, 3), so plug those in to discover a and the particular answer: 3 = Ae^((a million/3)(0)^3) - 8 clean up for A: 11 = Ae^((a million/3)(0)) 11 = Ae^(0) A = 11 So, the particular answer is: y = 11e^((a million/3)x^3) - 8

Answer 3

I don't know the answer myself as I have only done linear diff equations, however, I would probably try using addition formulae to split sin(y-x) into sinycosx-sinxcosy then maybe laplace transforms?

Sorry I can't help more, I want to see the answer myself