Absolute value => Directly integrable, separable and/or linear?

Question

I have previously supplied this question, however it went unanswered: Is the following directly integrable, separable and/or linear. EDIT: This is no longer homework as of 10 minutes after this edit.

$\frac{da}{db} = |a|,a(0) = 0$

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Direct integration gives: $|a|\cdot b +c$, $\therefore$ $c= 0$, and I believe it is therefore directly integrable. I am not entirely sure however due to the fact that depending on y, it has two derivatives(?)

Also, it takes the form $\frac{da}{db} = f(a)$, whereas I believe the form required for direct integration is $\frac{da}{db} = f(b)$.

$$\frac{1}{|a|}\cdot da = 1\cdot db$$ $$\ln a + c \text{ Note this is undefined(ln0)}$$

I don't believe it is separable, since using initial values, it shouldn't be defined.

However in this case it takes the form $\frac{da}{db} = f(a)f(b)$, where $f(b) = 1$ which seems correct for separation!

Finally, I know it isn't linear as this proves.

Hints and/or justification are preferred over simply answers! Any help is greatly appreciated!

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Answer 1

This IVP has no solution as the Cauchy-Peano Theorem states: Where $\frac{dy}{dx} = f(x,y), y(x_0) = y_0$

If $f(x,y)$ is continuous in some rectangle, $R =${$(x,y)| |x-x_0| \lt a, |y - y_0| \lt b$} then the initial value problem has at least one solution. ($x_0,y_0 \in \mathbb{R}$)

Since it is not continuous, it has no solution.

It is not directly integrable as it has form $\frac{da}{db} = f(a)$. It is required that the IVP take the form $\frac{da}{db} = f(b)$ for it to be directly integrable, however: $\frac{da}{db} = f(a)f(b)$ is a form that is separable.

$$\frac{da}{db} = |a|$$ $$\frac{1}{|a|} \cdot da = 1 \cdot db$$ $$\ln |a| = x + c \text{ This intermediate step has ln|0| which is underfined(given IV)}$$ $$|a| = C\cdot e^x \text{ Where C is a constant e^c}$$

Since it has an intermediate step which is undefined, it cannot be integrated via separation.

Showing that it is not linear can be done via the method outlined in this similar question: