# Where are the mistakes in the following reasoning?

Where is the flaw in my reasoning?

$$y''=y \iff y'=\frac{y^2}{2} \iff y=\frac{y^3}{6}$$

• Remember that you are dealing with $$\frac{d^2y(x)}{dx^2} = y(x)$$ Commented Feb 2, 2014 at 3:18
• Even if your approach were right (see answers) you would be missing constants of integration. Claude's solution shows how you get two constants to be determined from boundary conditions. Commented Feb 2, 2014 at 6:33

when you write $y''$ you mean differentiating with respect to some variable $x$.

However $\int y dy = y^2/2$

So when you go from $y''$ to $y'$ you are integrating with respect to $x$

So when you go from $y$ to $y^2/2$ you are integrating with respect to $y$

That is the flaw in your logic.

If you write out $$\frac{d y^2}{d x^2} = y$$ then $$\frac{d y}{d x} = \int y ~dx$$

For this reason alone, I am not a big fan of $y'$, $y''$ etc.

• Leibniz notation is far from perfect. Why not just write $y'(x)$ instead?
– user71641
Commented Feb 2, 2014 at 3:50
• The problem is, it does not say what the independent variable is. For example, how will you interpret $y'(0)=2$? While I agree that Leibniz's notation also has ambiguities it at least shows what the independent varible is Commented Feb 2, 2014 at 3:56
• If you're working with a differential equation, then you'll be writing $y'(x)$ which makes it clear what the independent variable is. If you're working with expressions like $y'(3)$, then what the independent variable is should have been specified, but even if it hasn't it shouldn't matter since it's just a constant.
– user71641
Commented Feb 2, 2014 at 4:13
• no, only $\dot{y}$ is correct, both $\frac{dy}{dx}$ and $y'$ are wrong ;) Commented Feb 2, 2014 at 4:56
• I still teach using a black board and I am not known to clean the black board all that well, so I don't like either $\dot{y}$ or $y'$ ;) Commented Feb 2, 2014 at 5:26

As said in previous answers, the problem is much clearer if you rewrite $$y''=y$$ as $$\frac{d^2 y}{d x^2} = y$$ This really shows the dependencies between $y$ and $x$.

So, you face a second order differential equation the characteristic equation being $r^2=1$ which has two roots $r=1$ and $r=-1$; so its general solution write $$y=c_1 e^x+c_2 e^{-x}$$

• why not $\frac{d^2 y}{d x^2} = y$? Commented Feb 2, 2014 at 6:43
• @dwarandae. Thanks for reporting the typo. Fixed now. Commented Feb 2, 2014 at 7:15