Differential Equations in a Nutshell
Definition
A differential equation is an equation involving one or more derivatives of an unknown function.
Here, the unknown function will be y = y(x) unless it arises from a problem involving time: y = y(t)
examples
$${\frac{\mathbb{d}y}{\mathbb{d}x} = 3y}$$ $${\frac{\mathbb{d^2}y}{\mathbb{d}x^2} - 6 \frac{\mathbb{d}y}{\mathbb{d}x} + 8y = 0}$$ $${\frac{\mathbb{d^3}y}{\mathbb{d}t^3} + t \frac{\mathbb{d}y}{\mathbb{d}t} + (t^2 - 1)y = \mathbb{e}^t}$$ $${y' - y = \mathbb{e}^{2x}}$$ $${y'' + y' = \cos{t}}$$
A function $y = y(t)$ is a solution of a differential equation on a given interval if the equation is satisfied for every x in that interval when y and its derivatives are substituted into the equation.
example 1
$y = \mathbb{e}^{2x}$ is the solution of the differential equation
$${\frac{\mathbb{d}y}{\mathbb{d}x} - y = \mathbb{e}^{2x}}$$
on the interval $(-\infty, \infty)$.
Check whether the solution satisfies the differential equation.
TODO solution
On a given interval a solution of a differential equation from which all solutions on that interval can be derived by substituting arbitary solutions is called the general solution.
The graph of a solution is called an integral curve. The general solution leads to a family of integral curves.
In applied problems leading to differential equations there are usually conditions that determine the arbitary constants.
Generally, we need n conditions to determine n constants.
Euler’s Numerical Method
Separable Differential Equations
$$\frac{dy}{dx} = f(x)g(y)$$
$$\frac{1}{g(y)}\frac{dy}{dx} = f(x)$$
Integrate with respect to x.
$$\int{\frac{1}{g(y)}\frac{dy}{dx}\mathbb{d}x} = \int{f(x)\mathbb{d}x}$$
$$\int{\frac{\mathbb{d}y}{g(y)}} = \int{f(x)\mathbb{d}x}$$
Homogeneous Differential Equations
$$\frac{dy}{dx} = f(\frac{y}{x})$$
We need to creat a new variable $v=\frac{y}{x}$ which gives us $y = vx$
Differentiate $$\frac{\mathbb{d}y}{dx} = \frac{\mathbb{d}[vx]}{dx}= v + x \frac{\mathbb{d}v}{dx}$$
Example
Solve
$$\frac{dy}{dx} = \frac{x^2 + y^2}{xy}$$
Solution
$$\frac{dy}{dx} = \frac{x^2 }{xy} + \frac{ y^2}{xy}$$
Simplify
$$\frac{dy}{dx} = \frac{x }{y} + \frac{ y}{x}$$
$$\frac{dy}{dx} = \left(\frac{y}{x}\right)^{-1} + \frac{ y}{x}$$
Substitute $v=\frac{y}{x}$.
$$ v + x \frac{\mathbb{d}v}{dx} = v^{-1} + v$$
Subtract v.
$$ x \frac{\mathbb{d}v}{dx} = v^{-1} $$
Separate variables
$$ v \mathbb{d}v = \frac{\mathbb{d}x}{x} $$
Integrate
$$ \frac{v^2}{2} = \ln|x| + C $$
Let C = ln(k)
$$ \frac{v^2}{2} = \ln|x| + \ln(k) $$
$$ \frac{v^2}{2} = \ln(kx) $$
$$ v = \pm \sqrt{2\ln(kx)} $$
Substitute back.
$$ \frac{y}{x} = \pm \sqrt{2\ln(kx)} $$
$$ y = \pm x\sqrt{2\ln(kx)} $$
Linear Differential Equations
An equation of the form
$$\frac{dy}{dx} + P(x)y = Q(x)$$
is called a first order linear differential equation.
The method to find the solution of the ODE is to find a factor called the integrating factor so that the left hand side is the result of the product rule:
$$I(x)\frac{dy}{dx} + I'(x)y = I(x)Q(x)$$ where $I'(x) = I(x)P(x)$
intergrating factor
How do we find I(x)?
Rearrange $I'(x) = I(x)P(x)$ to get
$$\frac{I'(x)}{I(x)} = P(x)$$
Next integrate both sides with respect to x.
$$\ln{I(x)} + C = \int{P(x)}dx$$
Apply the exponential function on both sides.
$$e^{\ln{I(x) + C}} = e^{\int{P(x)}dx}$$
$$I(x) \cdot e^C = e^{\int{P(x)}dx}$$
$$I(x) = Ae^{\int{P(x)}dx}$$ where $A = e^{-C}$ By convention we use A = 1.
product rule revisited
Now, if $I(x)P(x)y = I'(x)y$ we can rewrite the LHS as
$$\frac{\mathbb{d}[I(x)y]}{\mathbb{d}x} = I(x)Q(x)$$ which is the product rule reversed.
Intergrate with respect to x.
$$I(x)y = \int{I(x)Q(x)\mathbb{d}x} + C$$
$$y =\frac{1}{I(x)} \int{I(x)Q(x)\mathbb{d}x} + C$$
Use the initial conditions to find C.