Differential Equations Practical 2
Recap
Definition-of-a-Differential-Equation
Definition-of-Order-in-Differential-Equations
Definition-of-a-Partial-Differential-Equation
Definition-of-a-Linear-Ordinary-Differential-Equation
You can also verify solutions to these equations by substituting the solution into the original equation and testing if the left-hand side (LHS) is equal to the right-hand side (RHS) over a given interval ( if not given).
When we’re thinking about intervals of validity, then we need to study solution curve for any discontinuities or other points which aren’t differentiable.
What are the types of solution to a differential equation?
There are two types:
Definition-of-an-Explicit-Solution-in-Differential-Equations
Definition-of-an-Implicit-Solution-in-Differential-Equations
The latter of which is usually more common.
EXAMPLE
To show that is an implicit solution to the differential equation on the interval , we just implicitly differentiate the solution: \begin{align*}\frac{d}{dx}(x^{2}) + \frac{d}{dx}(y(x^{2})) = \frac{d}{dx}(25)&\implies2x+2y\frac{dy}{dx}=0\\&\implies \frac{dy}{dx}=- \frac{x}{y}\end{align*} To show its validity, you can demonstrate the two solutions based on the positive/negative nature of the square root, because the equation is an equation of a circle.
How do we handle families of solutions?
A note on validity
We can test validity just by substituting as we did before, but with both equations, like so:
\begin{align*} \begin{cases} \frac{dx}{dt}=x+3y,\ \frac{dy}{dt}=5x+3y \end{cases}:\begin{cases} x=e^{-2t}+3e^{6t},\ y=-e^{-2t}+5e^{6t} \end{cases}&\implies\begin{cases} (-2e^{-2t}+18e^{6t})=(e^{-2t}+3e^{6t})+3(-e^{-2t}+3e^{6t})\ (2e^{-2t}+30e^{6t})=5(e^{-2t}+3e^{6t})+2(-e^{-2t}+5e^{6t}) \end{cases}\ &\implies \begin{cases} RHS=-2e^{-2t}+18e^{6t}=LHS\ RHS=2e^{-2t}+30e^{6t}=LHS \end{cases}\ &\therefore \text{a valid pair of solutions} \end{align*}
When solving differential equations, we often implicitly integrate, leaving us with a single constant $c$. This is a **one parameter family of solutions**. If we have higher order, $n$, differential equations, then we have **$n$-parameter family of solutions**. In the case where there are no parameters, we name it a **particular solution**. We can handle these in two ways (plus just leaving them as a family of solutions): - Use multiple differential equations (**a system of differential equations**) - similar to solving simultaneous equations, you end up with a pair of differentiable functions as solutions defined on interval $I$, without any constant parameters. - Use a contraint (**initial value problems**) - these contraints, often the initial value of a function, allow us to solve for the parameter in a solution and give us its particular solution. ### More details on initial value problems ![[Definition of an Initial Value Problem]] >[!EXAMPLE] > >Considering the ordinary differential equation $\frac{dy}{dx}=y$ with general solution $y=ce^{x}$, we can impose the initial condition $y(x=0)=3$ such that: $y(x=0)=ce^{x=0}=c=3$, which gives us the particular solution $y=3e^{x}$. >[!EXAMPLE] > >Considering the ordinary differential equation $\frac{dy}{dx}=y$ with general solution $y=ce^{x}$, we can impose the initial condition $y(x=1)=-2$ such that: $y(x=1)=ce^{x=1}=-2:c=-2e^{-1}$, which gives us the particular solution $y=-2e^{x-1}$. Naturally, if our differential equation is a higher order, let's say an order $n$, then we'd need to integrate it $n$ times to find a solution, hence need to impose $n$ conditions to solve for $n$ constants of integration. ## How can we tell that a solutions is unique, or even exists? In other words, how can we check that any of a differential equation's solutions pass through $(x_{0},y_{0})$ and does only one solution curve pass through this point? Sometimes, no, but there's a theorem which help us discover this quicker: ![[Definition of the Theorem of Unique Solution Existence in Differential Equations]] Basically, if $f(x,y)$ and $\frac{\partial x}{\partial y}$ are continuous in some region containing the origin, then there will be a unique solution - this is easy to check for, as we just look for discontinuities first. >[!EXAMPLE] > >We can show that $y=\tan(x+c)$ is a one-parameter family of solutions to the ordinary differential equation $y^{\prime}=1+y^{2}$ by substitution and then showing that $\sec^{2}(x)=1+\tan^{2}(x)$. >Then, we can verify if there exists a unique solution for the initial condition $y(0)=0$ by using the aforementioned theorem - so, if there are no discontinuities in the curve around the origin, then there is a unique solution, and vice versa. In this case, the particular solution is $y=\tan(x)$, which passes through the origin and is valid in the interval $(- \frac{\pi}{2}, \frac{\pi}{2})$ and hence has a unique solution. >Finally, we can conclude that the solution is not valid in the region $(-2,2)$ because it contradicts the aforementioned interval of validity (constrained by the asymptotes at $\pm \frac{\pi}{2}$).