## UnkleRhaukus 4 years ago ${dy \over dx} + {2x-y-4 \over x-y+5}=0$

1. JamesJ

You'll want to change variables to help you get to a simpler equation. Something like u = x - y + 5 Then u' = 1 - y' and your equation becomes 1 - u' + (x + u-9)/u = 0 This isn't quite simple enough, but you get the idea.

2. UnkleRhaukus

yeah ill show you what i got up to

3. UnkleRhaukus

${dy \over dx} +{2x-y-4 \over 2y-x+5}=0$${dy \over dx} = {y-2x+4 \over 2y-x+5}$${dY \over dX} = {(Y+p)-2(X+q)+4 \over 2(Y+p)-(X+q)+5}$${dY \over dX} = {Y-2X+(p-2q+4) \over 2Y-X+(2p-q+5)}$

4. UnkleRhaukus

to find p and q we use $p-2q+4=0$$2p-q+5=0$ $p=-2$$q=1$

5. JamesJ

Right. And now you can write the new equation as a homogeneous equation.

6. UnkleRhaukus

${dY \over dX}={Y-2X \over 2Y-X}$${dY \over dX}={Y/X-2 \over 2Y/X-1}$$X{dV \over dX}+V={V-2 \over 2V-1}$$X{dV \over dX}={-V^2+2V-2 \over 2V-1}$${ 2V-1 \over V^2-V+1} {dV}=-2{dX \over X}$$\int{ 2V-1 \over V^2-V+1} {dV}=-2 \int {dX \over X}$$ln(V^2-V+1)=-2lnX+c$$V^2-V+1=kX^{-2}$$Y^2-YX+X^2=k$$(y+2)^2-(y+2) \times (x-1)+(x-1)-k=0$$x^2+y^2-xy-4x+5y-k=0$

7. UnkleRhaukus

is there quicker way?

8. JamesJ

I don't think so.

9. JamesJ

What is that btw ... a skewed ellipse I think

10. UnkleRhaukus

yes a skewed ellipse something like |dw:1327457371333:dw| how can you tell that , ?

11. JamesJ

You can see how might factor is as something like a^2(x-ky)^2 + b^2y^2 = 1

12. UnkleRhaukus

kinda,

13. UnkleRhaukus

I have some similar questions that i can not seem to solve for p,q their seems be no solution ${dy \over dx} ={x-y+2 \over x-y+1}$

14. JamesJ

For this one, I'd set u = x - y + 1 u' = 1 -y' ode => 1 - u' = (u+1)/u This equation is separable.

15. JamesJ

In fact, it's u' = -1/u

16. UnkleRhaukus

$u=-\int{1 \over u}du$$u=-ln|u|+c$$e^u=-ln(u)+c$$e^u=k/u$$ue^u=k$$(x-y+1)e^{x-y+1}-k=0$ i went wrong somewhere

17. JamesJ

must be tired. This one's easy $u du = -dx$

18. UnkleRhaukus

$u^2=-2x+c$$u^2+2x-c=0$$(x−y+1)^2+2x-c=0$ √ got there thank you and yes i am tired