x+y-a/x+y-b (dy/dx)=x+y+a/x+y+b
D Deleted member 4993 Guest Jul 14, 2016 #2 zeeshan said: x+y-a/x+y-b (dy/dx)=x+y+a/x+y+b Click to expand... Your post as written, turns out to be: y'= -2a/(bx) -1 ← This simple first order ODE. Please tell us where you are stuck! If your equation is: \(\displaystyle \displaystyle{\dfrac{x+y-a}{x+y-b} \left(\dfrac{dy}{dx}\right) \ = \ \dfrac{x+y+a}{x+y+b}}\) You should have written it as: (x+y-a)/(x+y-b) (dy/dx)=(x+y+a)/(x+y+b) ....those () are super important In that case, substitute: x + y = u and continue......
zeeshan said: x+y-a/x+y-b (dy/dx)=x+y+a/x+y+b Click to expand... Your post as written, turns out to be: y'= -2a/(bx) -1 ← This simple first order ODE. Please tell us where you are stuck! If your equation is: \(\displaystyle \displaystyle{\dfrac{x+y-a}{x+y-b} \left(\dfrac{dy}{dx}\right) \ = \ \dfrac{x+y+a}{x+y+b}}\) You should have written it as: (x+y-a)/(x+y-b) (dy/dx)=(x+y+a)/(x+y+b) ....those () are super important In that case, substitute: x + y = u and continue......
