Question

Find all solutions of xy'=2-x+(2x-2)y-xy^2.

Answer 1

Use the homogeneous sub, \(y=vx\).

Answer 2

But do I divide by x first?

Answer 3

Yes, it should give you a linear equation in \(v\).

Answer 4

Oh wait it might not be linear...

Answer 5

This is what I did.
xy'=2-x+2xy-2y-xy^2
xy'=(xy^2-2xy+x)+2(1-y)
xy'=x(y-1)^2-2(y-1)
y'=(y-1)^2-2(y-1)/x
So from there, do I plug in y=vx?

Answer 6

Just a sec. The homogeneous sub might not work here...

Answer 7

I use the substitution u=y-1 and u'=y' and u'=u^2-2u/x, v=1/u, v'=-1/u^2*u' and v'=-1/u^2(u^2-2u/x)=-1+2/ux=2v/x-1 and w=2v/x-1 and w'=-2/x^2*v' but it doesn't work.

Answer 8

The answer in the book is y=1-1/(x(1-cx)).

Answer 9

I think I might have it, but I need to change computers. This one is being very slow.

Answer 10

@SithsAndGiggles

Answer 11

Sorry, I can't see anything. Have you posted your reply?

Answer 12

Hey, sorry for the wait. I was checking to see if this is a particular type of ODE, and it's called a Ricatti equation, which has the form
\[y'=f_0(x)+f_1(x)y+f_2(x)y^2\]
i.e. a nonlinear equation that's quadratic in the dependent variable \(y(x)\). We want to be able to reduce this into a linear 2nd order equation.

Divide by \(x\):
\[y'=\frac{2-x}{x}+\frac{2x-2}{x}y-y^2\]
Substitute \(v=yf_2(x)=-y\), so \(v'=-y'\).
\[\begin{align*}-v'&=\frac{2-x}{x}-\frac{2x-2}{x}v-v^2\\\\
v'&=\frac{x-2}{x}+\frac{2x-2}{x}v+v^2\end{align*}\]
Next you substitute \(v=-\dfrac{t'}{t}\), which gives \(v'=\dfrac{(t')^2-t~t''}{t^2}\).
\[\begin{align*}
v'&=\frac{x-2}{x}+\frac{2x-2}{x}v+v^2\\\\
\frac{(t')^2-t~t''}{t^2}&=\frac{x-2}{x}+\frac{2x-2}{x}\frac{t'}{t}+\left(\frac{t'}{t}\right)^2\\\\
-\frac{t''}{t}&=\frac{x-2}{x}+\frac{2x-2}{x}\frac{t'}{t}\\\\
-t''&=\frac{x-2}{x}t+\frac{2x-2}{x}t'\\\\
t''+\frac{2x-2}{x}t'+\frac{x-2}{x}t&=0\end{align*}\]
To give the equation constant coefficients, we'll substitute \(t=\dfrac{p}{x}\), so \(t'=\dfrac{xp'-p}{x^2}\) and \(t''=\dfrac{x^2p''-2xp'+2p}{x^3}\).
\[\begin{align*}\frac{1}{x}p''-\frac{2}{x^2}p'+\frac{2}{x^3}p-\frac{2x-2}{x^2}p'+\frac{2x-2}{x^3}p+\frac{x-2}{x^2}p&=0\\\\
\frac{1}{x}p''-\frac{2}{x}p'+\frac{1}{x}p&=0\\\\
p''-2p'+p&=0\end{align*}\]

Answer 13

That's fine. We all know that technology has always been an issue at times. Can you please take a snapshot for what you just typed because my computer currently has 0% for processing math.

Answer 14

So p^2-2p+1=0
(p-1)(p-1)=0
p=1

Answer 15

No, that's not an equation that's quadratic in \(p\); there are derivatives involved. It's a homogeneous linear equation. Find the characteristic equation and its roots.

Answer 16

But if v=-t'/t, shouldn't \[v'=\frac{ t(-t'')+t' }{ t^2 }\]

Answer 17

\[\frac{d}{dx}\left[-\frac{t'}{t}\right]=-\left(\frac{t\dfrac{d}{dx}[t']-t'\dfrac{d}{dx}[t]}{t^2}\right)=-\frac{t~t''-(t')^2}{t^2}=\frac{(t')^2-t~t''}{t^2}\]

Answer 18

Screenshot?

Answer 19

I see. So after finding the characteristic equation and the roots, you have to go back to all the substitutions to find the equation of y, right?

Answer 20

Yes, might take a while...

Answer 21

But I found that\[-\frac{ t" }{ t }=\frac{ x-2 }{ x }-\frac{ 2x-2 }{ x }*\frac{ t' }{ t }\]

Answer 22

What you have is \[-\frac{ t" }{ t }=\frac{ x-2 }{ x }-\frac{ 2x-2 }{ x }*\frac{ t' }{ t }\]

Answer 23

Yes you're right. It looks like I made that error but corrected it in the end... Didn't even notice it. The equation in \(p\) and its derivatives is correct though.

Answer 24

But I wonder how y=ce^x+cxe^x is the solution. The root is 1...

Answer 25

I'll be back in a few minutes. The hard part's over.

Answer 26

Okay.

Answer 27

Okay so\[p''-2p'+p=0\]
gives the characteristic equation,
\[r^2-2r+1=0\]
with root \(r=1\), multiplicity 2, hence the general solution
\[p=C_1e^x+C_2xe^x\]
Back-subbing to \(t\) gives
\[\begin{align*}xt&=C_1e^x+C_2xe^x\\\\
t&=C_1\frac{e^x}{x}+C_2e^x
\end{align*}\]
Back-subbing to \(v\):
\[\begin{align*}
t&=C_1\frac{e^x}{x}+C_2e^x\\\\
\color{gray}{t'}&\color{gray}{=C_1\frac{x-1}{x^2}e^x+C_2e^x}\\\\
v&=-\frac{C_1\dfrac{x-1}{x^2}e^x+C_2e^x}{C_1\dfrac{e^x}{x}+C_2e^x}
\end{align*}\]
and finally back-subbing to \(y\):
\[\begin{align*}
y&=\frac{C_1\dfrac{x-1}{x^2}e^x+C_2e^x}{C_1\dfrac{e^x}{x}+C_2e^x}
\end{align*}\]
And you should be able to simplify that a bit.

Answer 28

THANK YOU SO MUCH! YOU'RE AWESOME!

Answer 29

You're welcome!