Implicit Differentiation of $x+y= \arctan(y)$?

Question

I'm trying to verify that $x+y = \arctan(y)$ satisfies this differential equation: $1+(y^2)+(y^2)y'= 0$. To do so, I tried to differentiate $x+y = \arctan(y)$ to get $y'$, but only got so far: $$1 = \left(\frac{1}{1+y^2}y' -1 \right) y'.$$ Now I'm not sure how I can isolate $y'$, or whether I am even taking the right first steps to solving the problem. Any hints/help about what direction I should take?

Answer 1

Applying implicit differentiation to $x+y=\arctan(y)$ gives $$1+\frac{dy}{dx}=\frac{1}{1+y^2}\frac{dy}{dx}.$$ Multiplying both sides by $1+y^2>0$ we get $$(1+y^2)(1+\frac{dy}{dx})=\frac{dy}{dx}.$$ Expanding the left-hand side, $$1+y^2+y^2\frac{dy}{dx}+\frac{dy}{dx}=\frac{dy}{dx}.$$ Subtracting $dy/dx$ from both sides, $$1+y^2+y^2\frac{dy}{dx}=0.$$ So we are done. The point is, it's not always necessary to isolate $y'$ first before verifying the equation.

Answer 2

You had differentiated correctly, but made an error when you re-arranged your equation:

$$ [ \ x \ + \ y \ ] \ ' \ \ = \ \ [ \ \arctan(y) \ ] \ ' \ \ \Rightarrow \ \ 1 \ + \ y' \ \ = \ \ \frac{1}{1 \ + \ y^2} · y' $$

$$ \Rightarrow \ \ 1 \ \ = \ \ \frac{1}{1 \ + \ y^2} · y' \ - \ y' \ \ \Rightarrow \ \ 1 \ \ = \ \ \left(\frac{1}{1+y^2} \ - \ 1 \right) · y' \ \ . $$

rather than $ \ \left(\frac{1}{1+y^2}y' -1 \right) y' \ \ . $ You will indeed verify the solution of the differential equation once you have the correct expression:

$$ 1 \ \ = \ \ \left(\frac{1}{1+y^2} \ - \ 1 \right) · y' \ \ \Rightarrow \ \ 1 \ - \ \left(\frac{1}{1+y^2} \ - \ 1 \right) · y' \ \ = \ \ 1 \ - \ \left(\frac{-y^2}{1+y^2} \right) · y' \ \ = \ \ 0 $$

[since $ \ 1 + y^2 \ \neq \ 0 \ \ , $ it is "safe" to multiply the equation through by it]

$$ \Rightarrow \ \ 1·(1+y^2) \ - \ (-y^2) · y' \ \ = \ \ 1 \ + \ y^2 \ + \ (y^2) · y'\ \ = \ \ 0 \ \ . $$