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Determine whether the given equation is linear in the dependent variable $v$ and the dependent variable $u$:

$$u dv+(v+uv-ue^u)du=0$$

I'm confused by this because I've never seen $dv$ and $du$ just floating around like this.
I also have no idea how to proceed. Any help is greatly appreciated.

So here is what I have got now

Think: is linear if have the form $$a_1(x)\frac{dy}{dx}+a_o(x)y=g(x)$$

1)Divide through by du $$\frac{udv+(v+uv-ue^u)du}{du}=\frac{0}{du}$$ 2)$$u \cdot \frac{dv}{du}+v+uv-ue^u=0$$ so $$u \cdot \frac{dv}{du}$$ looks like $$a_1(x)\frac{dy}{dx}$$ from the general equation for linear ODEs

3)Move ue^u to right side $$u \cdot \frac{dv}{du}+v+uv=ue^u$$ 4)Divide both sides by u to get $$\frac{dv}{du}+\frac{v}{u}+v=e^u$$ 5)Group like terms i.e. factor $$\frac{dv}{du}+v(\frac{1}{u}+1)=e^u$$

Now I don't know what to do. Does e^u=y from the general equation?

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  • 1
    $\begingroup$ It could be a little confusing to see this at first, but you can start by dividing through with $du$ or $dv$ to make sense of the equation. $\endgroup$ – Tunococ Sep 10 '13 at 0:35
  • $\begingroup$ @Celeritas: Have you learned integrating factor or exact equations? $\endgroup$ – Amzoti Sep 10 '13 at 0:39
  • $\begingroup$ @Amzoti I'm not sure but I think I'm supposed to rewrite the equation to see if it fits the form of a linear ODE. $\endgroup$ – Celeritas Sep 10 '13 at 1:47
  • $\begingroup$ @Celeritas: Got it! Added a hint as an answer, see if that works for you. Regards $\endgroup$ – Amzoti Sep 10 '13 at 1:57
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Hints: What if we could re-write this as:

  • $uv + v + u \dfrac{dv}{du} - e^u u = 0$
  • What if we bring the exponential term to the right and divide by $u$?
  • What if we now group like terms and re-write this new system?

Now that it is in this form (and all we are learning is to look for forms), can you see what type of equation it is and how to approach it?

Update

Good job getting the correct form! We have:

$$\tag 1 \frac{dv}{du}+v(\frac{1}{u}+1)=e^u$$

Write:

$\mu = e^{\int (1/u +1)du} = e^u u$

Multiply both side of $(1)$ by $\mu$, yielding:

  • $e^u u \dfrac{dv}{du} + e^u u\left(\dfrac{1}{u} + 1\right)v = e^{2u}u$, but since:
  • $e^u\left(\dfrac{1}{u}+1\right)u = \dfrac{d}{du}(e^u u)$, we have:
  • $e^u u \dfrac{dv}{du} + \dfrac{d}{du}(e^u u)v = e^{2u}u$ (this form is the reverse of the product rule), so we can write it as:
  • $\dfrac{d}{du}(e^u u)v = e^{2u}u$, we can integrate both sides in respect to $u$ to write it as:
  • $\displaystyle \int \dfrac{d}{du} (e^u u)v~du = \int e^{2u}u~ du$

Upon integrating, we get:

$e^u u v = e^{2u}\left(\dfrac{u}{2}-\dfrac{1}{4}\right) + c$, and solving for $v$ yields:

$$v(u) = \dfrac{e^u}{u}\left(\dfrac{u}{2}-\dfrac{1}{4}\right) + c\dfrac{e^{-u}}{u}$$

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  • $\begingroup$ The equation you have would be linear if v is the dependent variable since dv/du is a first order derivative. $\endgroup$ – Celeritas Sep 10 '13 at 2:16
  • $\begingroup$ @Celeritas: That is correct. Now, can you also solve it! You are almost there! Regards $\endgroup$ – Amzoti Sep 10 '13 at 2:17
  • $\begingroup$ I updated the question to show how far I got $\endgroup$ – Celeritas Sep 10 '13 at 8:04
  • $\begingroup$ @Celeritas: See update to answer. Go through each step and make sure you understand. Regards $\endgroup$ – Amzoti Sep 10 '13 at 12:31
  • $\begingroup$ Needs a TU!! +1 $\endgroup$ – Namaste Sep 10 '13 at 14:04

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