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Suppose we have a differential equation of the form

$$x'(t)= f \left( \frac{c_1 x + c_2 t+c_3}{d_1 x + d_2 t + d_3}\right), $$

where $c_1 d_2 - d_1 c_2 =0$.

Why is it sufficient to solve the differential equation

$$x'(t)=\tilde{f}(c_1x+c_2 t)? $$

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If $c_1d_2-c_2d_1=0$, then $c_1d_2=c_2d_1$ and $c_1=\frac{c_2d_1}{d_2}$. Substitute the right side for $c_1$. Then,

$$f\left(\frac{c_1x+c_2t+c_3}{d_1x+d_2t+d_3}\right)=f\left(\frac{\frac{c_2d_1}{d_2}x+c_2t+c_3}{d_1x+d_2t+d_3}\right)$$

$$=f\left(\frac{c_2}{d_2} \left(\frac{d_1x+d_2t+\frac{c_3d_2}{c_2}}{d_1x+d_2t+d_3}\right)\right)$$

which is clearly a function of $d_1x+d_2t$ alone.

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  • $\begingroup$ I think I wanted a function of $c_1 x + c_2 t$ but of course this can be done in the same way. $\endgroup$ – user144921 Mar 13 '15 at 23:45
  • $\begingroup$ I think the "$c_3$" in the last line must be wrong. $\endgroup$ – user144921 Mar 14 '15 at 9:18
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You must assume in addition that $c_1$ and $c_2$ are not both zero. If they are, then the second differential equation takes the form $x'(t)=K$, but solving this obviously does not suffice, in general, to solve the first equation.

Given this, it's because the second form can be made to include the first form by choosing $\bar f$ appropriately: assuming that $c_1 d_2-d_1 c_2=0$ and that $c_1$ and $c_2$ are not both zero, $$\tilde{f}(c_1x+c_2 t)=f \left( \frac{c_1 x + c_2 t+c_3}{d_1 x + d_2 t + d_3}\right)$$ if $$\tilde{f}(u)=f\left({u+c_3\over \beta u+d_3}\right),$$ where $\beta=d_1/c_1$ if $c_1\ne 0$ and otherwise $\beta=d_2/c_2$.

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