# Math Help - Need help with homogeneous dif. eq.

1. ## Need help with homogeneous dif. eq.

Hello i got some problems with one dif. eq. mainly this is it:

y'[x] = (x+y[x])/(x-y[x])

the problem im having is that i dont know how to split the ratio on two ratios with simple dominator e.g. y/x or x/y .. other than that i need to use u=y/x when im done with spliting this expression and then it's easy just solve dif. eq with separate variables.

thanks

2. You have (x-y) dy = (x+y) dx, with substitution y = ux, and hence dy = u dx + x du. Plug into your DE and see what you get.

3. dude i found solution and i think im doing it right, this is what i did:

so the eq. is y' = (x+y)/(x-y) , i took (x+y)/(x-y) and found that this is equal to (1+(y/x))/(1-(y/x)) so now => y' = (1+(y/x))/(1-(y/x)), now i make sub. u=y/x
from here y=ux and y' = u'x + u so now => u'x+u=(1+u)/(1-u) with little fixation i get => (du/dx)x + u = (1+u)/(1-u) now => (du/dx)x = (1+u)/(1-u) - u
=> (du/dx)=((1+u)/(1-u) - u)/x swap and i get this => du/((1+u)/(1-u) - u) = dx/x now i got the from where i get my variables separated now little fixation on the left side so i can get easyer integral, so du/((1+u)/(1-u) - u) = du/((1+u-u+u^2)/(1-u)) = du/((1+u^2)/(1-u)) now this is du((u^2+1)/(1-u)) = dx/x now i separate left side on two integrals and i get this Integral((1/(u^2+1))du) - Integral((u/(u^2+1)du) = Integral(dx/x) now the leftmost is arctg(u)+c
the next one i solve with sub. for t=u^2+1 and now i get dt=2udu => du=(1/2)dt => this one is (1/2)*Integral(dt/t) => (1/2)ln|t| +c => (1/2)ln|u^2+1| + c which is ln(u^2+1) + c cause of the square i get all positive, and the rightmost side is ln|x| + c now i combine all and i get : arctg(u) - (1/2)ln(u^2+1) = ln|x| now i get back to the first substitution u=y/x and i get final result : arctg(y/x) - (1/2)ln((y/x)^2 +1) = ln|x| + C, now i put only one C cause with combination of all the constants i surly get a constant so i put it with capital C and i guess this solution is ok, please revise it and tell me if im wrong somewhere, thanks!!

4. That was a little hard to read but I think it is correct! Notice, by the way, that $1+ \frac{y^2}{x^2}= \frac{x^2+ y^2}{x^2}$ so that $\frac{1}{2}ln(1+ \frac{y^2}{x^2})= \frac{1}{2}ln\left(\frac{x^2+ y^2}{x^2}$ $= \frac{1}{2}ln(x^2+ y^2}- \frac{1}{2}ln(x^2}= \frac{1}{2}ln(x^2+ y^2}+ ln|x|$ so that you can cancel the ln(|x|) on both sides of the equation.

5. ye that's good note but it's only so that i get the expression bit more clearer to read, but one can solve it to the point where one just have some more work to do if one wants to get the expression clearer but the result is already there in more complex form , except if one is asked to give sober solution or to express y in explicit form and not in implicit.. etc..