Solving the given differential equation which was supposedly a simple first order differential equation

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.

• solve for x' instead of y'. It's easier – LostInSpace Dec 1 '19 at 16:47

2 Answers

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

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

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