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I recently encountered the following question on my exam-

$$(\tan^{-1}y-x)dy=(1+y^2)dx$$

I have only been taught how to solve linear and first order differential equations.

My attempts included the following-

  • Substitution using $y=\tan q$ and then trying to simplify as I observed the term in RHS would yield $\sec^2q $ post this substitution.

  • I also tried to use the integrating factor method but the expression wasn't being simplified into a form where it could be used.

After a while I thought maybe it was a mistake and couldn't be solved at my level of understanding. Kindly help me evaluate the solution for this equation.

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    $\begingroup$ solve for x' instead of y'. It's easier $\endgroup$ – LostInSpace Dec 1 '19 at 16:47
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Hint: $$(\tan^{-1}y-x)dy=(1+y^2)dx$$ $$\implies (1+y^2)x'+x=\tan^{-1}y$$ Solve for x

Or rewrite it as: $$(\tan^{-1}y-x)\frac {y'}{(1+y^2)}=1$$ $$(\tan^{-1}y-x)(\tan^{-1}y)'=1$$ $$(z-x)(z)'=1$$ $$x'=z-x$$

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The substitution $u = \tan^{-1}(y)-x$ and some simplification transforms this differential equation into $$ u \;du = (1 - u)\; dx $$ which is separable.

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  • $\begingroup$ How did this substitution come into your mind? Is it due to practice and experience? $\endgroup$ – DinoManPhyLab Dec 1 '19 at 16:51
  • $\begingroup$ $\tan^{-1}(y)-x$ occurs in the equation. First I tried $v = \tan^{-1}(y)$, then saw that $v-x$ would be better. $\endgroup$ – Robert Israel Dec 1 '19 at 18:00
  • $\begingroup$ Ok thanks for the insight $\endgroup$ – DinoManPhyLab Dec 1 '19 at 18:07

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