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 preceding 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 simplicity, 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 \(y\).