Math 555: Differential Equations I |
A first order differential equation (FODE) is called separable
if it can be written in the form
$$
M(x)\,dx + N(y)\, dy = 0. \tag{1}
$$
In a real sense, this is the simplest possible form for a FODE to
take. The solution curves are obtained as implicit functions $y = y(x)$
by merely integrating equation $(1)$. We obtain
$$
\int M(x)\,dx + \int N(y)\, dy = C. \tag{2}
$$
When possible, we shall usually want to solve for $y$ as an honest
function of $x$. In case the differential equation comes with an initial
condition, then it will sometimes require careful attention in determining
the proper form of the solution function and its domain.
We look at some examples below.
Example 1. Consider the differential equation
$$
\frac{dy}{dx} = \frac{xy}{x^2 + 1}. \tag{3}
$$
This equation separates to become
$$
-\frac{x}{x^2 + 1}\,dx + \frac{1}{y}\,dy = 0. \tag{4}
$$
Integrating equation $(4)$ yields
$$
-\tfrac{1}{2}\ln(x^2 + 1) + \ln y = C. \tag{5}
$$
Solving equation $(5)$ for $y$ gives
$$
y(x) = C\sqrt{x^2 + 1}. \tag{6}
$$
We plot the slope field and some integral curves below.
Example 2. Consider the initial value problem (IVP)
$$
\begin{cases}
\dfrac{dy}{dt} = \dfrac{6t^2 + 4t}{2y - 4}, \\[0.5 ex]
y(0) = -3.
\end{cases} \tag{7}
$$
This can be separated to become
$$
(2y - 4)\,dy = (6t^2 + 4t)\, dt, \tag{8}
$$
and the Fundamental Theorem of Calculus may be applied with the
initial values to give
$$
\int_{-3}^y 2u - 4 \, du = \int_{0}^t 6s^2 + 4s\, ds. \tag{9}
$$
This is easy to integrate, but let's use Sage to do it
for us. The left-hand integral is
The right-hand integral is
Therefore, the implicit solution is
And solving for $y$ explicitly yields
Notice that this IVP has two "solutions": one with a positive square
root and one with a negative square root. Here "solutions" is in quotes
because in reality there can only be one solution. The subtle issue
here is that we are looking for a solution curve by trying to
find a solution function. But what if the curve is not given by
the graph of a function? Then you arrive at a conundrum like our current one.
We will interpret this situation as follows. Of the two functions that
we've found, only one of them is acceptable as the solution function:
the one that obeys the initial condition. However, we will regard the
concatenation of the two graphs as a single solution curve to the IVP.
As always in math, context and perspective matter. Be mindful of whether
you want a solution function or a solution curve when completing these
problems in class. In practical applications, both points-of-view can
be fruitful.
We check the initial values.
We see that only the negative square root obeys the initial condition. Therefore the solution is
We plot the slope field and both functions. The solution function is blue, while the entire curve may be regarded as the solution curve. ${\tt Sage}$ has trouble plotting the portion of the curve that has close-to-vertical tangent lines. You should regard this graph as connected in your mind.
Another option is to use ${\tt Sage}$ to find the domain of the solution function(s), and not let the $t$-values extend outside the domain.
Notice that the imaginary part of the middle solution is very small.
Indeed, this is again a numerical artifact. A cubic polynomial with
integer coefficients cannot have three imaginary solutions.
We use the real part of this solution as the left-hand bound for the
$t$-values and re-plot the solution curves.
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