# Understanding this differential equation $y' = |y|^{\alpha}$

I'm studying Numerical Analysis and the first example given in Suli and Mayers book on the section on initial value problems is the following -

Consider the differential equation $y' = |y|^{\alpha}$, subject to the initial condition $y(0) = 0$, where $\alpha$ is a fixed real number, $\alpha \in (0,1)$.

It is a simple matter to verify that, for any nonnegative real number $c$,

$y_c(x) = (1 - \alpha)^{\frac{1}{1 - \alpha}}(x - c)^{\frac{1}{1 - > \alpha}}$ when $c<=x< \infty$

$y_c(x) = 0$ when $0<=x< c$

is a solution to the initial value problem on the interval $[0, \infty)$.

I have not studied differential equations before so it is not a 'simple matter' for me to verify the solution to this equation. Can someone explain to me what is going on and how that $y_c(x)$ is a solution to the initial value problem? I don't see where the $x$ is 'coming from' for example...I see no $x$ in $y' = |y|^{\alpha}$.

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You plug it into the equation....... – Euler....IS_ALIVE Oct 18 '12 at 4:50

The solution to a differential equation is a function - in this case, you want to find the function $y(x)$ such that $$\frac{dy}{dx}=|y(x)|^{\alpha}, \ \text{and} \ y(0)=0$$

Even when a function $y$ is written without explicit reference to the variable it is acting on, there's still such a variable. It's just often omitted to make the notation simpler, and shouldn't cause any problems once you're used to seeing ODEs (as you probably should be if you're doing numerical analysis).

The nice thing about differential equations (both ordinary and partial) is that, while finding a solution is in general extremely difficult, once you think have a solution, it's extremely easy to check that it is indeed a solution: just calculate as many derivatives as the equation requires, and see if they satisfy the equation.

In this case, $y(x)=(1-\alpha)^{\frac{1}{1-\alpha}}(x-c)^{\frac{1}{1-\alpha}}$ (and by using $y(0)=0$, you could find what $c$ is). Then you get

$$\frac{dy}{dx}=(1-\alpha)^{\frac{1}{1-\alpha}}\cdot\frac{1}{1-\alpha}(x-c)^{\frac{1}{1-\alpha}-1}=(1-\alpha)^{\frac{1}{1-\alpha}-1}(x-c)^{\frac{1}{1-\alpha}-1}\\=(1-\alpha)^{\frac{\alpha}{1-\alpha}}(x-c)^{\frac{\alpha}{1-\alpha}}.$$

We just have to check that this is equal to $|y(x)|^{\alpha}$, which is obvious.

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There is a technique to solve first order differential equations called separation of variables. For example, if one has the differential equation $y'(x)=y(x)^2$, then the solution is

$$y'(x)=y(x)^2 \implies \frac{y'(x)}{y(x)^2} = 1 \implies \int y(x)^{-2}y'(x)\,dx = \int 1\,dx$$ $$\implies -y(x)^{-1} = x + c \implies \frac{1}{y(x)} = c-x \implies y(x) = \frac{1}{c-x}\,.$$

$c$ is a constant which can be determined exploiting the initial condition.

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