Some basics about differential equatons.

These notes are set so that you get to prove the main results by solving smaller problems that when put together give the big result. The answers to the problems are in the videos. You will get the most out of these notes if you do (or try) the problems before looking at the videos. We will first study differential equations of the form $$ y' = f(x,y). $$ To be more explicit $f(x,y)$ is a function of two real variables $x$ and $y$ and we will generally assume that the first partial derivatives of $f$ exists. Then a function $y(x)$ defined on the interval $(a,b)$, then it is a solution to the differential if and only if for each $x\in (a,b)$ the value of the derivative $y'(x)$ is given by $$ y'(x) = f(x,y'(x)). $$ We will also often use the Leibniz notation $$ \frac{dy}{dx} = f(x,y). $$
For example if the equation is $$ y' = 1 + 2 x y $$ and we have that $y(1) = 3$ then we can compute the derivative $y'(1)$ by $$ y'(1) = 1 + (1)(y(1)) = 1 + (1)(3) = 4. $$

Problem: For the equation $$ \frac{dy}{dx} = \frac{1+x}{1+y^4} $$ if $y(-3) = -2$ find $y'(-2)$.

Solution.

To get a bit more of a feel for what a differential equation tell us, let us look at the special case of equations of the form $$ \frac{dy}{dx} = f(y) $$ That is the equation has no explicit dependence on the variable $x$. Here is a variant of one we did in class.

Problem: For the differential equation $$ y' = 10 - y $$ find all the constant solutions. Then make rough sketches of the solutions with $y(0) = 5$ and $y(0) = 15$. Finally for the solution with $ y(0) = 12$ estimate $y(100)$.

Solution.

Problem: Here is a slightly more complicated example. For the equation $$ \frac{dy}{dx} = .1 y \left( 10 - y\right) $$ do the following.
(a) Find the constant solutions.
(b) Graph the constant solutions along with rough sketches of the graphs of the solutions with $y(0)=5$ and $y(0)=15$. (c) For the solution with $y(0)=8$ estimate $y(50)$.

Solution.