# Thread: Implicit Differentiation and then Euler's method

1. ## Implicit Differentiation and then Euler's method

In my studies, I encountered a question which I did not know how to do. The answer referred to implicit differentiation but I did not understand the steps undertaken, so could someone explain to me how to do the following question?

$\frac{dy}{dx}=y(1-y)$, show that $\frac{d^2y}{dx^2} = (1-2y)y(1-y)$.
The answer states that $\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx}) = \frac{d(y-y^2)}{dy} \cdot y(1-y) = (1-2y)y(1-y)$. Where does the $y - y^2$ come from?

Then I had problem with Euler's method with the question:
Given that $y=2$ when $x=0$, use Euler's method with a step-size of $\frac{1}{4}$ to estimate the value of $y$ when $x = \frac{1}{2}$
How do I use Euler's method if all I have if $f(y)$?

2. Originally Posted by Enedrox
In my studies, I encountered a question which I did not know how to do. The answer referred to implicit differentiation but I did not understand the steps undertaken, so could someone explain to me how to do the following question?

The answer states that $\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx}) = \frac{d(y-y^2)}{dy} \cdot y(1-y) = (1-2y)y(1-y)$. Where does the $y - y^2$ come from?

$\frac{dy}{dx} = y(1 - y)$

$\frac{d^2y}{dx^2} = \frac{d}{dx}[y(1 - y)]$

$= y\frac{d}{dx}(1 - y) + (1 - y)\frac{d}{dx}(y)$, from use of the product rule...

$= y\left(-\frac{dy}{dx}\right) + (1 - y)\frac{dy}{dx}$

$= \frac{dy}{dx}(-y + 1 - y)$

$= \frac{dy}{dx}(1 - 2y)$

$= y(1 - y)(1 - 2y)$.

3. Originally Posted by Enedrox
In my studies, I encountered a question which I did not know how to do. The answer referred to implicit differentiation but I did not understand the steps undertaken, so could someone explain to me how to do the following question?

The answer states that $\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx}) = \frac{d(y-y^2)}{dy} \cdot y(1-y) = (1-2y)y(1-y)$. Where does the $y - y^2$ come from?

Then I had problem with Euler's method with the question:

How do I use Euler's method if all I have if $f(y)$?
I don't know what you mean by that. The problem appears to be that you don't have f(x)! You don't say how the problems are arranged on the page. Could they be referring to the previous differential equation?

4. lol, the solution is so simple now that I actually understand it Thank you.

Although I would have done it slightly differently (now that I understand what's going on):
$\frac{d^2y}{dx^2} = \frac{d}{dx}[y(1 - y)]=\frac{d}{dx}[y-y^2)]$
$=1 \cdot \frac{dy}{dx} - 2y\frac{dy}{dx}$
$=\frac{dy}{dx}(1-2y)$
$=y(1-y)(1-2y)$

As for Euler's method, it is question 4 e (section 2). from the 2006 VCE specialist math exam

5. So the problem is to solve $\frac{dy}{dx}= y(1-y)$ y(0)= 2, using Euler's method with a step of 1/4, from 0 to 1/2. Did you expect us to guess that before?

Well, that's easy enough, you only have to do two steps! Since y(0)= 2, at 0, $\frac{dy}{dx}$, at x= 0, is 2(1-2)= -2. The line with slope -2, starting at (0,2) is y= -2x+ 2 and at x= 1/4, y(1/4)= -2(1/4)+ 2= -1/2+ 2= 3/2.

Now do the same thing again. If y= 3/2, what is dy/dx= y(1- y)? What is the equation of the line through (1/4, 3/2) with that slope? What is y on that line when x= 1/2?