1
$\begingroup$

So I'm struggling to come to grips with a method that my prof introduced in a lecture. It's basically reducing ODEs so that they're separable and easier to work with.

Apparently the trick in this case is if you have an ODE that doesn't appear separable you can use some form of subtitution to make it separable.

For example

$$y'=f(ax+by+c)$$ you can define $$z(x)=ax+by+c)$$ and then $$z'=a+by' =a+bf(z)$$

Okay, fairly straightforward to follow but at this point I don't understand why this is necessary. The confusing part comes when, at the end of solving the ODE, you have to subsitute back in for various things and then calculate an inverse of something for some reason.

The example was as follows: $$y'=(4y-x-6)^2$$

So we do the substitution and we have $$z(x)=4y-x-6 \implies y'=z^2$$ Then $$z'=4y-1 \implies z'=4z-1$$ Applying separation of variables and integrating we have $$\int \frac{dz}{4z^2-1} = \int dx$$ Up to this point, super cool, nothing seems too strange, but this is where I start to get a little confused.

Solving the integral gives us (and I have no idea where $H(z)$ or the $u$ comes from or why it's needed) $$u=H(z)=\frac{1}{4} ln \lvert \frac{2z-1}{2z+1} \rvert = x+c$$

Then apparently we need to calculate the inverse $z=H^{-1}(u)$ so we have the two solutions as $$(i) z> \frac{1}{2} , z< -\frac{1}{2} : z(u) = \frac{1}{2} \frac{1+e^{4u}}{1-e^{4u}}$$ $$(ii)-\frac{1}{2}<z<\frac{1}{2} : z(u)= \frac{1}{2} \frac{1-e^{4u}}{1+e^{4u}}$$

Now replacing $u$ with $x+c$ $$z(u(x))=z(x)=H^{-1}(x+c)=\frac{1}{2} \frac{1+Ae^{4x}}{1-Ae^{4x}}, A = \pm e^{4c}$$ And finally $$z(x) =4y-x-6 \implies y(x) = \frac{1}{4}(z+x+6)$$

Yielding the final result of $$y(x)=\frac{1}{4}(x+6+\frac{1}{2} \frac{1+Ae^{4x}}{1-Ae^{4x}})$$

I think my issue lies with all the substitutions and then going back in and removing the substitutions to get the final result.

Is there any way someone can perhaps refine this method and explain some of the substitution steps, or perhaps there is a good resource that I could be directed towards.

Thanks.

$\endgroup$
  • $\begingroup$ There is a mistake in $\quad z'=4y-1.\quad$ See my answer. $\endgroup$ – JJacquelin Feb 12 '17 at 10:30
0
$\begingroup$

$$y'=(4y-x-6)^2$$

$$z(x)=4y-x-6 \implies y'=z^2$$ $$z'=4y'-1 \implies z'=4z^2-1$$ $$\int \frac{dz}{4z^2-1}=\int dx$$ $$x=\frac{1}{4}\ln\left|\frac{1-2z}{1+2z} \right|+c$$ $$z=\frac{1}{2}\:\:\frac{1-e^{4(x-c)}}{1+e^{4(x-c)}}$$ $$y=\frac{1}{8}\:\:\frac{1-e^{4(x-c)}}{1+e^{4(x-c)}}+\frac{x+6}{4}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.