Question

\[{dy \over dx} + {2x-y-4 \over x-y+5}=0\]

Answer 1

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.

Answer 2

yeah ill show you what i got up to

Answer 3

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

Answer 4

to find p and q we use
\[p-2q+4=0\]\[2p-q+5=0\]
\[p=-2\]\[q=1\]

Answer 5

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

Answer 6

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

Answer 7

is there quicker way?

Answer 8

I don't think so.

Answer 9

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

Answer 10

yes a skewed ellipse
something like

how can you tell that , ?

Answer 11

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

Answer 12

kinda,

Answer 13

I have some similar questions that i can not seem to solve for p,q
their seems be no solution

\[{dy \over dx} ={x-y+2 \over x-y+1}\]

Answer 14

For this one, I'd set
u = x - y + 1
u' = 1 -y'

ode
=> 1 - u' = (u+1)/u

This equation is separable.

Answer 15

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

Answer 16

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

Answer 17

must be tired. This one's easy
\[ u du = -dx \]

Answer 18

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