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

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?

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 $$\frac{dy}{dt}=-\frac{\Phi_t}{\Phi_y}$$
• Oh I can see that now. But why did my book define $\Phi(t,y)$ with those conditions instead of defining it as $d \Phi = 0$ and $\Phi_t = M$, $\Phi_y = N$? Commented Jan 10, 2023 at 17:30
• The conditions ensure that $\Phi=0$ indeed defines a function in the neighborhood of $(t_0,y_0)$. Depending on which partial derivative of $\Phi$ is not zero, you have $y=y(t)$ or $t=t(y)$. Commented Jan 10, 2023 at 17:43