Differential Equation Problem?

Help with Differential Equation Problem?

• Im having some trouble solving a differential equation. Any Suggestions would be appreciated. The problem is shown as: Find all solutions to the differential equation dy/dx = sin(x-y). Hint: there are infinitely many singular solutions but they are easy to list in a single formula. Here is what I have so far: dy/dx = sin(x-y) dy/dx = sinxcosy - cosxsiny I am thinking that this problem is a seperable ODE but I am unable to seperate the equation to look like the form f(x)dx = g(x)dy.

• Answer:

  Hi, there, as promised I am back. The first steps I leave as they were, for reference, and add from the point we stopped. *)dy/dx = sin(x-y) x-y = v => dx-dy= dv (dx-dy)/dx = dv/dx 1-dy/dx=dv/dx => dy/dx = 1-dv/dx dy/dx = sin(v)= 1-dv/dx sin(v)dx= dx - dv [sin(v)-1]dx= -dv dv/[1-sin(v)] = dx 1)Int(dx)= x+C [integration of r.h.s in *)] 2)Integration of l.h.s in *) : Int [dv/(1-sin v)]=? Let us multiply both the numerator and the denominator of the fraction dv/1-sin v by the expression 1+sin v. We could do it of course only provided (1+sin v)/=0. But what happens if 1+sin v =0? In that case sin v= -1. Solve, using trig, for v, and obtain v = 3Pi/2+ 2PI*K, K belongs to Z. Let us check, if it satisfies the original condition. v= x-y = 3Pi/2+2PIK => y= x-3PI/2-2PIK dy/dx=1; sin (x-y)= sin v = -1; dy/dx /= sin (x-y) That is why we can afford to multiply by 1+sin v, it doesn't interfere with the solutions. Remember, both dv and 1-sin v, get multiplied by 1+sin v, so the whole fraction was really multiplied by 1. Int(dv/1-sinv)= Int[(1+sin v)dv/(1+sin v)(1-sin v)] Now, according to (a+b)(a-b)=a^2 - b^2 we have (1+sin v)(1-sin v)= 1 - sin^2 v. But according to the rules of trig, sin^2+cos^2 = 1 so 1-sin^2 = cos^2 Use this identity to get, in the denominator of the fraction (1-sin v)(1+sin v)= cos^v And the numerator remains as it was after the multiplication, (1+sin v)dv. The whole fraction then becomes (1+sin v)dv/ cos^2 v So, Int[(1+sin v) dv]/[1-sin^2(v)]= =Int [1+sin(v)]dv/cos^2(v) Seperate the numerator and get two integrals out of one Int[(1+sinv)dv/cos^2v] = Int[dv/cos^2v)]+ Int[sinv dv/cos^2v] A) Int (dv/cos^2v)= ? From the intergal tablets (tan v)' = 1/cos^2 v So A)= tan v ( performing the reverse action, of "anti derivative") B)Int(sinv dv/cos^2v)=? Put t = cos v. Then dt = -sinv dv; cos^2v = t^2 Our B) becomes Int(-dt/t^2) = 1/t + K Putting back t = cos v we get: 1/cos v + K Summary: A)= tan v B)= (1/cos v)+K 2) = A)+B) = tan v + 1/cos v + K That was our left hand side, which was equal to r.h.s, which was x + c Let's equate them again: tan v + (1/cos v) + K = x + c Putting back v = x - y we get tan(x-y) + [1/cos (x-y)]+ K = x + c put c - K= C, a different constant of integration, and obtain: tan(x-y) + [1/cos(x-y)] - y = C. That is the fullest solution of the ODE, given in the implicit form, a perfectly valid form, as you know from Calculus 3 and all the implicit function theorems. If you need more help, I'll get back to this question later. Good Luck.