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How would you solve the next ODE?

$$\frac{dy}{dt} = \frac{at + by + m} { ct + dy + n},$$

where $a, b, c, d, m, n$ are constants and $ad = bc$.

Corrected.

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    $\begingroup$ there is a mix of $t$ and $x$. $\endgroup$
    – Maesumi
    Commented Feb 28, 2013 at 0:00
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    $\begingroup$ And if $t$ and $x$ are the same, then this equation is not, in general, separable. $\endgroup$ Commented Feb 28, 2013 at 0:04
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    $\begingroup$ @GerryMyerson I believe if we let $u=ax+by$ then it becomes separable. $\endgroup$
    – Maesumi
    Commented Feb 28, 2013 at 0:08

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Let $u=at+by$ then $ct+dy=Ku$ for some constant $K$ (as $ad=bc$). Now ${du \over dt}=a+b {dy \over dt}$ and ${dy \over dt}=-{a \over b}+{1 \over b} {du \over dt}$. Next we have ${dy \over dt}={{u+m} \over {Ku+n}}$ so ${ -{a \over b}+{1 \over b} {du \over dt}} = {{u+m} \over {Ku+n}}$. And this can be solved by separation as it is autonomous.

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  • $\begingroup$ Perhaps it should be explicitly noted that $K$ is a constant. $\endgroup$ Commented Feb 28, 2013 at 0:20

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