How Do I Convert a Bernoulli Differential Equation to Linear Form?

[stripedcat]

I know how to do a Linear First Order and I know how to do a Bernoulli (kind of).

The kind of part may be why I'm having a problem.

\(\displaystyle dy/dx+xy=xy^2\)

So I know in order for that to be a normal linear differential, that square on the last y has to go away somehow... I'm not sure how to do this. I don't need the whole solution, just start me off and (more importantly) explain this a bit if you can.

EDIT: I think this is the right track, can anyone confirm?

\(\displaystyle y^{-2} dy/dx + xy^{-1} = x\)

\(\displaystyle v=y^{-1}\)

This makes it...

\(\displaystyle dv/dx = (-1) y^{-2} dy/dx\)
\(\displaystyle -dv/dx = y^(-2) dy/dx\)

Do the subs

\(\displaystyle -dv/dx + xu = x\)
\(\displaystyle dv/dx - xu = -x\)

Does this work out..?

 

[Ackbach]

Hmm. You definitely have the right substitution, but I get
$$-\frac{dv}{dx}+xv=x.$$

 

[stripedcat]

Ah, I typo'd the answer with the - at the front.

Does this problem not flip back to x and y somehow? Rather than the x and v I mean.

 

[Prove It]

I know how to do a Linear First Order and I know how to do a Bernoulli (kind of).

The kind of part may be why I'm having a problem.

\(\displaystyle dy/dx+xy=xy^2\)

So I know in order for that to be a normal linear differential, that square on the last y has to go away somehow... I'm not sure how to do this. I don't need the whole solution, just start me off and (more importantly) explain this a bit if you can.

EDIT: I think this is the right track, can anyone confirm?

\(\displaystyle y^{-2} dy/dx + xy^{-1} = x\)

\(\displaystyle v=y^{-1}\)

This makes it...

\(\displaystyle dv/dx = (-1) y^{-2} dy/dx\)
\(\displaystyle -dv/dx = y^(-2) dy/dx\)

Do the subs

\(\displaystyle -dv/dx + xu = x\)
\(\displaystyle dv/dx - xu = -x\)

Does this work out..?

Even though this equation is Bernoulli, it is also separable...

$\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} + x\,y &= x\,y^2 \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= x\,y^2 - x\,y \\ \frac{\mathrm{d}x}{\mathrm{d}y} &= \frac{1}{x\,y^2 - x\,y} \\ x\,\frac{\mathrm{d}x}{\mathrm{d}y} &= \frac{1}{y^2 - y} \end{align*}$

Can you proceed?

 

[stripedcat]

Even though this equation is Bernoulli, it is also separable...

$\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} + x\,y &= x\,y^2 \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= x\,y^2 - x\,y \\ \frac{\mathrm{d}x}{\mathrm{d}y} &= \frac{1}{x\,y^2 - x\,y} \\ x\,\frac{\mathrm{d}x}{\mathrm{d}y} &= \frac{1}{y^2 - y} \end{align*}$

Can you proceed?

Yeah, going that way you would multiply out the dy on both sides

\(\displaystyle x dx = 1/(y-y^2) dy\)

\(\displaystyle x^2/2 = -ln(y-1/(y)\)

Yes?

 

[Prove It]

Yeah, going that way you would multiply out the dy on both sides

\(\displaystyle x dx = 1/(y-y^2) dy\)

First of all, the RHS was $\displaystyle \begin{align*} \frac{1}{y^2 - y} \end{align*}$, not $\displaystyle \begin{align*} \frac{1}{y - y^2} \end{align*}$.

As for the statement "multiply by dy", this is a VERY sloppy practice that only works computationally, but makes no mathematical sense.

Really what you should be doing is integrating both sides, and then recognising that the LHS is a chain rule, and so simplifies the same way an integration by substitution would...

$\displaystyle \begin{align*} \int{x\,\frac{\mathrm{d}x}{\mathrm{d}y} \, \mathrm{d}y} &= \int{ \frac{1}{y^2 - y}\,\mathrm{d}y} \\ \int{ x\,\mathrm{d}x } &= \int{ \frac{1}{y^2 - y}\,\mathrm{d}y} \end{align*}$

\(\displaystyle x^2/2 = -ln(y-1/(y)\)

Yes?

This will obviously have a mistake from the incorrect RHS.

 

[stripedcat]

I'm obviously still missing something there, I typo'd the 1/y^2-y as you saw but even still, that just changes it to positive instead of negative.

Then do I start the isolation of the y? Because these problems always seem to come out to y = something.

And what about the substitutions? I think you have to use them to do this right? Does the method you're using provide the same answer as the substitution method?

 

[Prove It]

I'm obviously still missing something there, I typo'd the 1/y^2-y as you saw but even still, that just changes it to positive instead of negative.

Then do I start the isolation of the y? Because these problems always seem to come out to y = something.

And what about the substitutions? I think you have to use them to do this right? Does the method you're using provide the same answer as the substitution method?

Well considering the RHS wouldn't be right even when made positive, you need to redo the integral as you have made a mistake somewhere.

When you have done that, then yes, you would try to solve for y in terms of x. Don't forget you need an integration constant as well.

When you say "what about the substitutions?" I assume you mean the substitution method to go from Bernoulli to First-Order Linear. Yes you will get an equivalent answer, though it may look a little different.

 

[stripedcat]

$\displaystyle \begin{align*} \int{x\,\frac{\mathrm{d}x}{\mathrm{d}y} \, \mathrm{d}y} &= \int{ \frac{1}{y^2 - y}\,\mathrm{d}y} \\ \int{ x\,\mathrm{d}x } &= \int{ \frac{1}{y^2 - y}\,\mathrm{d}y} \end{align*}$

This would become \(\displaystyle x^2/2 = ln(y-1/y)\)

Then...?

\(\displaystyle e^{x^2/2} = (y-1/y)\)

At that point I've got nowhere to go, I don't know what to do by that method.

 

[Prove It]

$\displaystyle \begin{align*} \int{x\,\frac{\mathrm{d}x}{\mathrm{d}y} \, \mathrm{d}y} &= \int{ \frac{1}{y^2 - y}\,\mathrm{d}y} \\ \int{ x\,\mathrm{d}x } &= \int{ \frac{1}{y^2 - y}\,\mathrm{d}y} \end{align*}$

This would become \(\displaystyle x^2/2 = ln(y-1/y)\)

Then...?

\(\displaystyle e^{x^2/2} = (y-1/y)\)

At that point I've got nowhere to go, I don't know what to do by that method.

The RHS should actually be $\displaystyle \begin{align*} \ln{ \left| \frac{1 - y}{y} \right| } \end{align*}$. Also where is your integration constant?

 

[stripedcat]

Yeah I know its + C, but I don't think I can use your method. It's not that I don't think it works, I'm sure it does, but there's nothing like it in my book or example problems. Unless I get a problem that is exactly the same, I don't know that it will work for any other problem.

Please bear in mind that although they moved this to the Differential Equation forum I'm only in Calc II and we just started it. All I have is the substitution method I was initially trying to do.

Even if we correct the RHS as you say, I would still have NO IDEA what to do next because the method you use is nothing at all like any method in this chapter. I've looked at all the example problems, there's nothing like it. Please understand that it's not that I don't appreciate the help, I really do, it's that I honestly have nowhere to go with the method you are using. No amount of hints will help, I can't use that method because I've never seen it before.

 

[Prove It]

Yeah I know its + C, but I don't think I can use your method. It's not that I don't think it works, I'm sure it does, but there's nothing like it in my book or example problems. Unless I get a problem that is exactly the same, I don't know that it will work for any other problem.

Please bear in mind that although they moved this to the Differential Equation forum I'm only in Calc II and we just started it. All I have is the substitution method I was initially trying to do.

Even if we correct the RHS as you say, I would still have NO IDEA what to do next because the method you use is nothing at all like any method in this chapter. I've looked at all the example problems, there's nothing like it. Please understand that it's not that I don't appreciate the help, I really do, it's that I honestly have nowhere to go with the method you are using. No amount of hints will help, I can't use that method because I've never seen it before.

I find it hard to believe that you have never encountered separable DEs when you are doing Bernoulli DEs and substitutions, a topic that is more difficult and comes much later.

Also, apart from the errors in your integrating and lack of integration constant, you are actually done. You should have

$\displaystyle \begin{align*} \frac{x^2}{2} + C &= \ln{ \left| \frac{1-y}{y} \right| } \end{align*}$

and now it's just a case of solving for y (if you really want to, it's not really necessary)...

$\displaystyle \begin{align*} \mathrm{e}^{ \frac{x^2}{2} + C } &= \left| \frac{1-y}{y} \right| \\ \mathrm{e}^C\,\mathrm{e}^{\frac{x^2}{2}} &= \left| \frac{1 - y}{y} \right| \\ A\,\mathrm{e}^{\frac{x^2}{2}} &= \frac{1 - y}{y} \textrm{ where } A = \pm \mathrm{e}^C \\ A\,y\,\mathrm{e}^{\frac{x^2}{2}} &= 1 - y \\ y + A\,y\,\mathrm{e}^{\frac{x^2}{2}} &= 1 \\ y \, \left( 1 + A\,\mathrm{e}^{\frac{x^2}{2}} \right) &= 1 \\ y &= \frac{1}{1 + A\,\mathrm{e}^{\frac{x^2}{2}}} \end{align*}$

 

[stripedcat]

Yeah, believe me, we have been... Less than enthusiastic... about some of these questions, which do not match our book, and are not explained by it.

We spent maybe an hour on just flat out Diff Equations then moved on to Bernoulli and Homogenous stuff. It was almost like Diff Equations was suppose to be a review or something.

 

[Prove It]

I know how to do a Linear First Order and I know how to do a Bernoulli (kind of).

The kind of part may be why I'm having a problem.

\(\displaystyle dy/dx+xy=xy^2\)

So I know in order for that to be a normal linear differential, that square on the last y has to go away somehow... I'm not sure how to do this. I don't need the whole solution, just start me off and (more importantly) explain this a bit if you can.

EDIT: I think this is the right track, can anyone confirm?

\(\displaystyle y^{-2} dy/dx + xy^{-1} = x\)

\(\displaystyle v=y^{-1}\)

This makes it...

\(\displaystyle dv/dx = (-1) y^{-2} dy/dx\)
\(\displaystyle -dv/dx = y^(-2) dy/dx\)

Do the subs

\(\displaystyle -dv/dx + xu = x\)
\(\displaystyle dv/dx - xu = -x\)

Does this work out..?

Just so that you can see we get the same answer by solving the equation with a substitution:

$\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} + x\,y = x\,y^2 \end{align*}$

Make the substitution $\displaystyle u = y^{1-2} = y^{-1} \implies y = u^{-1} $ so that $\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = -u^{-2}\,\frac{\mathrm{d}u}{\mathrm{d}x} \end{align*}$, and when we substitute into the DE we get

$\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} + x\,y &= x\,y^2 \\ -u^{-2}\,\frac{\mathrm{d}u}{\mathrm{d}x} + x\,u^{-1} &= x\,\left( u^{-1} \right) ^2 \\ -u^{-2}\,\frac{\mathrm{d}u}{\mathrm{d}x} + x\,u^{-1} &= x\,u^{-2} \\ \frac{\mathrm{d}u}{\mathrm{d}x} - x\,u &= -x \end{align*}$

This is now a first-order linear equation, and so can be solved with an integrating factor $\displaystyle \begin{align*} \mathrm{e}^{\int{ -x\,dx} } = \mathrm{e}^{-\frac{x^2}{2}} \end{align*}$. Multiplying both sides of the DE by the integrating factor gives

$\displaystyle \begin{align*} \mathrm{e}^{-\frac{x^2}{2}} \, \frac{\mathrm{d}u}{\mathrm{d}x} - x\,\mathrm{e}^{-\frac{x^2}{2}}\,u &= -x\,\mathrm{e}^{-\frac{x^2}{2}} \\ \frac{\mathrm{d}}{\mathrm{d}x} \left( \mathrm{e}^{-\frac{x^2}{2}} \, u \right) &= -x\,\mathrm{e}^{-\frac{x^2}{2}} \\ \mathrm{e}^{-\frac{x^2}{2}} \, u &= \int{ -x\,\mathrm{e}^{-\frac{x^2}{2}} \, \mathrm{d}x} \\ \mathrm{e}^{-\frac{x^2}{2}}\,u &= \mathrm{e}^{-\frac{x^2}{2}} + C \\ u &= 1 + C\,\mathrm{e}^{\frac{x^2}{2}} \\ y^{-1} &= 1 +C\,\mathrm{e}^{\frac{x^2}{2}} \\ y &= \left( 1 +C\,\mathrm{e}^{\frac{x^2}{2}} \right) ^{-1} \\ y &= \frac{1}{1 + C\,\mathrm{e}^{\frac{x^2}{2}}} \end{align*}$

And this is the same result we got through solving the equation by separation of variables :)

 

[stripedcat]

Oh I totally believed you had a correct answer. You're going way above the call of duty now and I appreciate that. My frustration stems from the fact that we had these problems dumped on us... And what do you do when your textbook is useless? You saw as far as I could get and that was after I spent several hours watching videos, reading websites, and then finally asking a question here. I try to follow the basic TOS guidelines of not just going "Hey what's the answer", I really do want to learn this stuff.

I really, really do appreciate the help, I might have come off as a bit snippy or unappreciative, please believe me that was not my intention.