Class 34 - What is a differential equation? (11/28/2017, Tuesday)
Logistics
Final Exam- 12/14 8:00 AM
- due today
- 10.3 due 11/27 Mon
- 8.7 due 11/29 Wed
- 8.8 due 11/30 Thurs
Lecture
A differential equation is an equation involving derivatives whose solution is given by a function. For example, consider the following differential equation:
\[y' = y.\]
Any solution to the differential equation must be a function $y = f(x)$ such that if we differentiate, we get it back. What could be a possible candidate? We can plug in $y = e^{x},$ so we say the function $x \mapsto e^{x}$ is a solution to the above differential equation.
Exercise. Show that the constant function $y = 1$ is NOT a solution to the differential equation $y' = y.$
Exercise. Show that for any constant $c,$ the function $y = ce^{x}$ is a solution to the differential equation.
In fact, every solution of the differential equation $y' = y$ is of the form $y = ce^{x}.$ To see this, first note that having $y' = y$ is the same as having $y'(x) - y(x) = 0.$ This is the same as having $y'(x)e^{-x} - ye^{-x} = 0$ because $e^{-x} > 0$ for any $x.$ Then we have
\[\frac{d}{dx}(y(x)e^{-x}) = y'(x)e^{-x} - ye^{-x} = 0.\]
If there is such a differentiable function $y = y(x),$ by the 1st Fundamental Theorem of Calculus, we must have $y(x)e^{-x} = c$ for some constant $c.$ Multiplying $e^{x},$ this implies that
\[y(x) = ce^{x}.\]
Exercise. Find every function $y = y(x)$ that satisfies $y' = y$ and $y(0) = 1.$
You will realize that $y(0)$ is actually what determines $c$ in the previous discussion. We call the value $y(0)$ initial value.
Exercise. Show that $y = e^{3t}$ is not a solution to the differential equation $y'' = 4y.$ Can you find a solution to this differential equation? What if we do the same for $y'' = -4y$ instead?
If a differential equation only involves first derivative, we say such an equation is of first order (e.g., $y' = y$). If it involves second derivative and no higher derivatives, we say the equation is of second order (e.g., $y'' = 4y$).
Slope fields. Given a first order equation $y'(x) = f(x, y),$ the slope field is a picture consisting of $xy$-plane and for every point $(x, y),$ the slope $f(x, y)$ marked at the point. Please read Section 11.2 of your book to understand this.
Exercise. Draw the slope field of the equation:
\[\frac{dy}{dx} = -\frac{x}{y}.\]
Separation of variables. It is no coincidence that the slope field above looks like a collection of circles. Let us try to solve the equation.
Exercise. Show that $y = \sqrt{1 - x^{2}}$ is a solution to the equation
\[\frac{dy}{dx} = -\frac{x}{y}.\]
Can you find any other solutions?
Rewriting the equation above, we have $y'(x) = -x/y(x).$ This gives $y(x)y'(x) = - x,$ so
\[\int y(x)y'(x) dx = - \int x dx = - x^{2}/2 + C_{1}.\]
On the other hand, we have
\[\int y(x)y'(x) dx = \int y dy = y^{2}/2 + C_{2},\]
so we get
\[x^{2} + y^{2} = 2(C_{1} - C_{2}) = C.\]
Since the left-hand side is a sum of squares, it cannot be negative, so $C \geq 0.$ Hence, we can write $C = r^{2}$ where $r = \sqrt{C} \geq 0,$ so the above can be written as
\[x^{2} + y^{2} = r^{2}\]
for any $r \geq 0.$
Here, we get a trick called separation of variables. We notice that from the step
\[\frac{dy}{dx} = -\frac{x}{y},\]
we could deduce
\[\int y dy = \int -x dx,\]
so we imagine as if there is a step saying "$y dy = - x dx,$" although this does not make much sense in our course. This trick works in a more general context, and you can read about this in your book to see more examples.
Exercise*. Do #7 on Exam 2 Winter 2017.
Exercise*. Do #7 on Exam 3 Fall 2016.
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