In calculus we learn how to “solve” two kinds of differential equations of the form

$$\frac{dy}{dx} = f(x, y) \tag{1}$$

Here are two solution methods that come up regularly when studying PDE.

First Order Separable Equations

By definition a separable differential equation is a diffeq of the form

$$y'(x) = F(x) G(y(x)),\quad\text{or}\quad \frac{d y}{d x} = F(x)G(y).$$

Thus the function $f(x,y)$ on the right hand side in (1) has the special form

$$f(x,y) = F(x)\, G(y). \tag{2}$$

For example, the differential equation

$$\frac{d y}{d x} = \sin(x) \bigl(1+y^2\bigr)$$

is separable, and one has $F(x) = \sin x$ and $G(y) = 1+y^2$. On the other hand, the differential equation

$$\frac{d y}{d x} = x+y$$

is not separable.

Solution method for separable equations.

To solve this equation divide by $G(y(x))$ to get

$$\frac{1}{ G(y(x))} \frac{d y}{d x} = F(x). \tag{3}$$

Next find a function $H(y)$ whose derivative with respect to $y$ is

$$H'(y) = \frac{1}{G(y)} \quad\left(\text{solution: } H(y) = \int {\frac{dy}{G(y)}}.\right)$$

Then the chain rule implies that the left hand side in (3) can be written as

$$\frac{1}{ G(y(x))} \frac{d y}{d x} = H'(y(x)) \frac{d y}{d x} = \frac{d H(y(x))}{d x}.$$

Thus (3) is equivalent with

$$\frac{d H(y(x))}{d x} = F(x).$$

In words: $H(y(x))$ is an antiderivative of $F(x)$, which means we can find $H(y(x))$ by integrating $F(x)$:

$$H(y(x)) = \int F(x) dx +C. \tag{3}$$

Once we have found the integral of $F(x)$ this gives us $y(x)$ in implicit form: the equation (3) gives us $y(x)$ as an implicit function of $x$. To get $y(x)$ itself we must solve the equation (3) for $y(x)$.

Determining the constant. The solution we get from the above procedure contains an arbitrary constant $C$. If the value of the solution is specified at some given $x_0$, i.e. if $y(x_0)$ is known then we can express $C$ in terms of $y(x_0)$ by using (3).

First Order Linear Equations

Important example: Exponential growth and decay.

The solution of

$$\frac{du}{dt} = ku$$

is

$$u(t) = u(0)e^{kt}.$$

General case.

Differential equations of the form equation

$$\frac{d y}{d x}+a(x)y = k(x) \tag{IL}$$

are called first order linear.

The Integrating Factor.

Linear equations can always be solved by multiplying both sides of the equation with a specially chosen function $m(x)$ called the integrating factor. It is defined by

$$m (x) = e^{A (x)},\text{ where } A (x) = \int a(x)\,d x. \tag{IF}$$

If we multiply the equation (IL) with the integrating factor $m(x)$ we get

$$m(x)\frac{d y}{d x}+a(x)m(x)y = m(x)k(x).$$

By the chain rule the integrating factor satisfies

$$\frac{d m(x)}{d x} = \frac{d\, e^{A(x)}} {d x} = \underbrace{A'(x)}_{=a(x)} ~\underbrace{e^{A(x)}_{}}_{=m(x)} = a(x)m(x).$$

Therefore one has

$$\begin{aligned} \frac{d\, m(x)y}{d x} &= m(x)\frac{d y}{d x}+a(x)m(x)y \\ &= m(x)\Bigl\{\frac{d y}{d x}+a(x)y \Bigr\}\\ &= m(x)k(x). \end{aligned}$$

Integrating and then dividing by the integrating factor gives the solution

$$y=\frac1{m(x)}\left(\int m(x)k(x)\,d x+C\right).$$

In this derivation we have to divide by $m(x)$, but since $m(x)=e^{A (x)}$ and since exponentials never vanish we know that $m (x)\neq0$, so we can always divide by $m (x)$.

Oscillators

The solution of

$$\frac{d^2 u}{dt^2} + \omega^2 u = 0$$

is

$$u(t) = u(0)\cos \omega t + \frac{u'(0)}{\omega}\sin \omega t$$