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My textbook says that solutions for the equation $y'=-y^2$ must always be 0 or decreasing. I don't understand—if we're solving for y', then wouldn't it be more accurate to say it must always be 0 or negative. Decreasing seems to imply that we're looking at a full graph, even though the book is talking about individual solutions. Can someone explain this?

Secondly, it gives that the family $y=\frac{1}{x+C}$ as solutions for the equation. It then tells me that 0 is a solution for y in the original equation that doesn't match the family, but I don't quite understand that. How can we know that y' will equal 0 if we're specifically looking outside of the family of solutions it gives?

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3 Answers 3

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If a function is differentiable and is decreasing, then the derivative is non-positive. In your case, $y^2 \geq 0 \Rightarrow y' = -y^2 \leq 0$. Hence, the solution you obtain must be decreasing.

Note that we obtain the family $y = \frac{1}{x+c}$ assuming $y \neq 0$. This can be seen from the way we solve the equation. We have $$\frac{dy}{dx} = -y^2$$ Assuming $y \neq 0$, we can divide by $y^2$ to get $\frac{dy}{y^2} = -dx$. Now we can integrate to get, $y = \frac{1}{x+\frac{1}{y_0}}$. Note that for each $y_0$, we get a different solution. So we have a one parameter family of solutions. Also, note that this family of $y$'s is nowhere $0$ and hence it is fine when we divided by $y^2$ in the original equation.

So the process of solving assumes that $y \neq 0$. We also observe that $y=0$ is a solution by plugging it into the differential equation.

Hence, the solutions to the differential equation are $$y=0 \text{ and } y=\frac{1}{x+\frac{1}{y_0}}, \text{ where } y_0 \in \mathbb{R}$$

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You're confusion can be resolved by clarifying one thing (I think). It seems you think that to "solve" a differential equation, you are looking for $y'$? Really, a solution is a function, maybe to decrease confusion we'll call it $f(x)$, with the property that $f'(x)=-[f(x)]^2$. That is what is meant by $y'=-y^2$.

So when we think of the graph of $y=f(x)$ we know that it has the property that $\frac{dy}{dx}=-[f(x)]^2$, or since a square is always positive, and then we take the negative, we know the derivative of $y=f(x)$ is negative and hence the function is decreasing.

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  1. Remember if $y'(x) > 0$, $y(x)$ is increasing; if $y'(x) < 0$, $y(x)$ is decreasing; if $y'(x) = 0$ then $y(x)$ is constant. In our case $y'(x) \le 0$ which means $y(x)$ is always constant or decreasing.

  2. You can verify yourself: if $y(x) = 0$ for all $x$, then $y'(x) = 0$ so it is true that $y' = -y^2$ therefore $y(x) = 0$ is a solution but it isn't in the form $\frac{1}{x + C}$.

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