Question

Solve y'=(-6x+y-3)/(2x-y-1) implicitly.

Answer 1

@SithsAndGiggles

Answer 2

Equations like these tend to simplify nicely when you try a substitution...

Answer 3

What substitution?

Answer 4

Currently trying to figure that out. I've tried a few already but they don't seem to help.

Answer 5

I'll be back in a while... maybe some idea will crop up.

Answer 6

Any ideas?

Answer 7

Actually, yes. I found a neat resource online for dealing with equations like this.

First we let \(X=x-x_0\) and \(Y=y-y_0\). Then we have \(y'=Y'\).
\[\begin{align*}Y'&=\frac{-6(X+x_0)+(Y+y_0)-3}{2(X+x_0)-(Y+y_0)-1}\\\\
Y'&=\frac{-6X+Y+(-6x_0+y_0-3)}{2X-Y+(2x_0-y_0-1)}\end{align*}\]
What we want to do is find constants \(x_0\) and \(y_0\) such that the constant terms in the numerator and denominator are both zero; the resulting equation will be easier to deal with.
\[\begin{cases}-6x_0+y_0-3=0\\
2x_0-y_0-1=0\end{cases}~~\implies~~x_0=-1,~y_0=-3\]
You then have
\[\begin{align*}Y'&=\frac{-6X+Y}{2X-Y}\\\\
Y'&=\frac{-6+\dfrac{Y}{X}}{2-\dfrac{Y}{X}}\end{align*}\]
Substitute \(Y=uX\), so \(Y'=Xu'+u\).
\[\begin{align*}Xu'+u&=\frac{-6+u}{2-u}\\\\
Xu'&=\frac{-6+u}{2-u}-u\\\\
Xu'&=\frac{-6+u-u(2-u)}{2-u}\\\\
Xu'&=\frac{-6-u+u^2}{2-u}\\\\
\frac{2-u}{u^2-u-6}~du&=\frac{dX}{X}\end{align*}\]

Answer 8

u=x+h
v=y+k
dy/dx=dv/du
dv/du=(-6(u-h)+v-k-3)/(2(u-h)-(v-k)-1)
dv/du=(-6u+v+(6h-k-3))/(2u-v+(-2h+k-1))
6h-k-3=0
-2h+k-1=0
so h=1, k=3
dv/du=(-6u+v)/(2u-v)
dv/du=(-6+v/u)/(2-v/u)
v=uz
dv/du=u*dz/du+z
u*dz/du+z=(-6+z)/(2-z)
u*dz/du=(-6+z)/(2-z)-z
u*dz/du=(-6+z)/(2-z)-(z(2-z))/(2-z)
=(-6+z-2z+z^2)/(2-z)
=(-6-z+z^2)/(2-z)
=(z^2-z-6)/(2-z)
=((z-3)(z+2))/(2-z)
(2-z)/((z-3)(z+2)) dz=du/u
am I right so far?

Answer 9

Yes

Answer 10

Then can you tell me the right answer? Because the answer I got is (y-3x)(y+2x+5)^4=C.

Answer 11

Here's what I'm getting:\[\begin{align*}\frac{2-u}{u^2-u-6}~du&=\frac{dX}{X}\\\\
\int\frac{2-u}{(u+2)(u-3)}~du&=\int\frac{dX}{X}\\\\
\int\frac{2-u}{(u+2)(u-3)}~du&=\ln|X|+C\\\\
-\frac{1}{5}\int\frac{du}{u-3}-\frac{4}{5}\int\frac{du}{u+2}&=\ln|X|+C\\\\
-\frac{1}{5}\left(\ln|u-3|+4\ln|u+2|\right)&=\ln|X|+C\\\\
\left((u-3)(u+2)^2\right)^{-1/5}&=CX\\\\
(u-3)(u+2)^2&=CX^{-5}\\\\
(u-3)(u+2)^2&=CX^{-5}\\\\
\left(\frac{Y}{X}-3\right)\left(\frac{Y}{X}+2\right)^2&=CX^{-5}\\\\
\left(\frac{y+3}{x+1}-3\right)\left(\frac{y+3}{x+1}+2\right)^2&=C(x+1)^{-5}\\\\
\left(\frac{y-3x}{x+1}\right)\left(\frac{y+2x+5}{x+1}\right)^2&=C(x+1)^{-5}\\\\
\left(y-3x\right)\left(y+2x+5\right)^2&=C(x+1)^{-2}
\end{align*}\]

Answer 12

I got (z-3)(z+2)^4=C/u^5

Answer 13

Let me see if I get a different result on paper... I tend to make slight mistakes when I only use LaTeX.

Answer 14

I think you got it wrong in the middle. You have (u+2)^2 but I have ^4 since it's 4*ln abs(z+2).

Answer 15

Ah there it is. Right, so it should be
\[(u-3)(u+2)^4=CX^{-5}\]
where you use \(z\) in place of my \(u\) and \(u\) in place of my \(X\), right?

Answer 16

Yes.

Answer 17

\[\left(y-3x\right)\left(y+2x+5\right)^4=C\]

Answer 18

So I got it right?