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In my ODE class lecture notes there's a section on solving I-order ODE using substitution method and there's one step which I couldn't really figure out, I'd really appreciate it if somebody could help me out here:

So there's this transformation, from: $${dv\over du}=f\left({au+bv\over mu+nv}\right) \;\;\;\;\;\; \mathbf{(A)}$$ to $${dz\over {f\left({a+bz\over m+nz}\right)-z}}={du\over u} \;\;\;\;\;\;\;\; \mathbf{(B)}$$

using substitution $z = \frac vu$

Here $a, b, m, n$ are all constants. How do I go from $\mathbf{(A)}$ to $\mathbf{(B)}$ using this substitution exactly? I've been looking at this for hours now but still couldn't figure out the magic. What I have after replace $v$ using $z*u$ is: $${d(z*u)\over du}=z=f\left({a+bz\over m+nz}\right)$$which is different from $\mathbf{(B)}$

Any help will be appreciated!

-- Jason

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    $\begingroup$ @Ana Galois: Hi Ana, thanks for helping me putting them into proper formula forms, appreciate it:) $\endgroup$ – Vol_Smile Sep 11 '13 at 15:15
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  • $au + bv = a + b \dfrac{v}{u} = a + b z$
  • $mu + nv = m + n \dfrac{v}{u} = m + n z$
  • You have the transformation $v = zu$, find the derivative and substitute for the final expression, so we have: $dv = z du + u dz$, so:

$$\dfrac{dv}{du} = z + u \dfrac{dz}{du} = f\left(\dfrac{a+bz}{m+nz}\right)$$

Simplifying yields:

$$\dfrac{dz}{f\left ( \dfrac{a+bz}{m + nz}\right)-z} = \dfrac{du}{u}$$

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  • $\begingroup$ Thanks for the detailed explanation Amzoti, I finally see what I did wrong. I was treating z as a constant without realizing it was in fact a function of u and I need to use the product rule when doing d(z*u)/du. Thanks again:) $\endgroup$ – Vol_Smile Sep 11 '13 at 15:13
  • $\begingroup$ @JasonYe: You are very welcome and glad I could help. Regards $\endgroup$ – Amzoti Sep 11 '13 at 15:27
  • $\begingroup$ Yay...an accept! How goes your day? Happy POETS Day tomorrow!! $\endgroup$ – amWhy Sep 13 '13 at 0:04

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