Formal definition of implicit solution of a first order ordinary differential equation

Question

My book formally defines implicit solution of first order ordinary differential equation $M(t,y)dt + N(t,y)dy = 0$ in the following way:

Assume that $\Phi(t,y)$ is defined on some $Q \subset \mathbb{R}^2$ domain and is continuously differentiable in the $U \subset Q$ neighborhood of some $(t_0,y_0)$, also $\Phi(t_0,y_0) = 0$, $(\Phi_t'(t_0,y_0))^2 + (\Phi_y'(t_0,y_0))^2 \neq 0$.Then we will say that the implicit solution $t=t(y)$ (if $\Phi_t'(t_0,y_0) \neq 0$) or the implicit solution $y = y(t)$ (if $\Phi_y'(t_0,y_0) \neq 0$) is an implicit solution of the differential equation if $M(t,y)\dfrac{\partial \Phi(t,y)}{\partial y} - N(t,y)\dfrac{\partial \Phi(t,y)}{\partial t} \equiv 0 (*)$ in $U$.

This isn't a direct citation but a translation of the definition from the original language of the source. First of all I checked this on an example and it is true. But I can't make sense of it. For example, if we are controlling the case of $y = y(t)$ then by definition $y$ is a solution if $y' = -\dfrac{M(t,y)}{N(t,y)}$. Now $\Phi'_t(t,y(t)) = \Phi'_y(t,y)y'$ and we can write $(*)$ as $-\dfrac{M(t,y)}{N(t,y)} = -y'$ which is a contradiction. Could you explain how this definition works and where my mistake is?

Answer 1

The sign is wrong in your next to last line. Your solution is implicitly defined by $\Phi(t,y)=0$. Along that curve you have $d\Phi=0$, that is $$ \Phi_tdt+\Phi_ydy=0 $$ or $$ \frac{dy}{dt}=-\frac{\Phi_t}{\Phi_y} $$