3.4 KiB
Section 2 - First Order Differential Equations
Section 2.1 - Linear Differential Equations
This section is from Paul's Online Math Notes.
Let the following first-order linear differential equation be given, with p(t) and g(t) continuos.
\frac{dy}{dt} + p(t)y = g(t)
Deriving the Solution
Next, we let \mu(t) be our integrating factor. Multiply both sides of the equation through by \mu(t).
\mu(t)\frac{dy}{dt} + \mu(t)p(t)y = \mu(t)g(t)
Now, define \mu(t) so that \mu(t)p(t) = \mu'(t). Thus, we can state the following:
\mu(t)\frac{dy}{dt} + \mu'(t)y = \mu(t)g(t)
The left of the preceeding equation is simply the product rule, so we can write (\mu(t)y(t))' = \mu(t)g(t). Take the integral of both sides.
\begin{align} \int (\mu(t)y(t))' dt &= \int \mu(t)g(t) \ \mu(t)y(t) + C &= \int \mu(t)g(t) dt \ y(t) &= \frac{\int \mu(t)g(t) dt - C}{\mu(t)} \end{align}
Let C absorb the negative sig, and we see the following.
y(t) = \frac{\int \mu(t)g(t) dt + C}{\mu(t)}
This is the general solution to the differential equation. However, it is incomplete, as we do not know \mu(t)
To derive the function, recall that we defined \mu(t)p(t) = \mu'(t). Thus, we can rewrite this equation.
\begin{align} \frac{\mu'(t)}{\mu(t)} &= p(t) \ (\ln \mu(t))' &= p(t) \ \end{align}
Integrate both sides.
\begin{align} \ln \mu(t) + k = \int p(t) dt \ \mu(t) &= e^{\int p(t) dt + k} \ &= e^k e^{\int p(t) dt} \end{align}
As k is an unknown constant, rewrite this as \mu(t) = k \exp(\int p(t) dt).
Summary
The following differential equation is given.
\frac{dy}{dt} + p(t)y = g(t)
To find a solution to this differential equation, construct the integrating factor \mu(t).
\mu(t) = k \exp(\int p(t) dt)
Thus, the solution to the differential equation can be written as the following.
y(t) = \frac{\int \mu(t)g(t) dt + C}{\mu(t)}
Section 2.2 - Separable Differential Equations
This section is from Paul's Online Math Notes.
Let the following differential equation of the following forms be given.
\begin{align} N(y) \frac{dy}{dx} &= M(x) \frac{dy}{dx} &= \frac{M(x)}{N(y)} \ \frac{dy}{dx} &= \frac{N(y)}{M(x)} \ \frac{dy}{dx} &= N(y)M(x) \ \end{align}.
For the sake of simplicty, select the following form:
N(y) \frac{dy}{dx} = M(x)
Thus, integrate both sides with respect to x.
\int N(y) \frac{dy}{dx} dx = \int M(x) dx
Since y is really y(x), we can make the following substitution:
u = y(x) \text{ and } du = y'(x)dx = \frac{dy}{dx}{dx}
This reduces the integral to the following:
\int N(u) du = \int M(x) dx
This is solvable from here.
Section 2.4 - Bernoulli Equations
This section is from Paul's Online Math Notes.
Let a differential equation of the following form be given, with n \in \mathbb{N}; n \geq 2
y' + p(x)y = q(x)y^n
This is a Bernoulli equation.
Divide by y^n.
y^{-n}y' + p(x)y^{1-n} = q(x)
Now, make the substitution v = y^{1-n}. Thus, the derivative is as follows.
v' = (1-n)y^{-n}y'
Substituting into the first equation yields the following.
\frac{1}{1-n}v' + p(x)v = q(x)
After solving, be sure to rewrite in terms of y.