#Maths # ODEs - An Introduction ## What are ODEs? Ever wondered how we could model the rate of change in temperature of a hot drink as a function of its original temperature? Alright, maybe not with *that* wording, but it's a great, practical example of a *differential equation* - where we model the rates of changes of functions (their derivatives) against the original functions! There are actually different types of differential equation - we'll be going over some of the basics in this example. Listed below: - **Generic Differential Equation**: $\frac {dy}{dx} = f(x)$ - **Linear Differential Equation: $\frac{dy}{dx} + P(x)y = Q(x)$** - **Separable differential equation**: $\frac {dy}{dx} = f(x)g(y)$ - **Homogeneous differential equation**: $\frac{dy}{dx} = f \left( \frac{y}{x} \right)$ For the multivariable **Partial Differential Equation**, feel free to first take a look at[[Partial Derivatives (Maths)| Partial Derivatives]]. >[!Abstract]- The concept of "Order" in DEs >Look at the **highest-order** derivative in the equation. That is the **order** of the entire differential equation. > >For example, take the differential equation: $\frac {d^{2}y}{dx^2} = 2xy + y^2$ >This would be a <u>**second**</u> order differential equation, as the second order of the DE is the prevailing order in the equation. # Solving Simple ODEs ## An ODE in the form $\frac {dy}{dx} = f(x)$ Oh dear! Another differential equation lies before us, this time more daunting than ever! Nope, just integrate by the variable in question on both sides. That should show it! >[!Abstract]- Example Question 1) >Let us define $f(x)$ as a simple linear function $f(x) = 2x + 3$ >$\int_ \, dy = \int 2x+3\,dx$ >This thus equals to: >$y = x^2 + 3x + C$ >Therefore given linear values C can also be found through substitution. ## An ODE in the form $\frac {dy}{dx} = f(x)g(y)$ This is a **separable differential equation**, where there is a multivariable composite function. We'd typically integrate both sides as functions of $x$, but we evidently can't do that this time... This is difficult to visualise, so we can create a simple composite differential equation as a visual guide for us: $\frac{dy}{dx} = xy$ We can divide both sides by the composite $y$ function, allowing us to get something like: $\int \frac{1}{y}\frac{dy}{dx} \, dx = \int x \, dx $ This makes the equation easy to solve! All we have to do is solve the simple integrals to get our solution: $\ln y = \frac{x^2}{2} + c$ $y = e^{\frac{x^2}{2} + c}$ >[!Abstract]- Example Question 1) i) - HL AAI >Question: Solve the differential equation: >$\frac{dy}{dx} = \frac{y+2}{x+1}$ >**Solution:** > >We can first rewrite the equation (so that we don't get lost, pretty common) >$\frac{dy}{dx} = \frac{1}{x+1}y+2$ >We can then divide both sides by $y+2$. For the purpose of skipping steps, I'm going to directly integrate both sides by $dx$, giving us: >$\int \frac{1}{y+2} \frac{dy}{dx} \, dx = \int \frac{1}{x+1} \, dx $ >$\int \frac{1}{y+2} \, dy = \ln(x+1) + c$ >$\ln (y+2) = \ln(x+1) + c$ >$y = x + 3 + c$ ## Homogenous Differential Equations A homogeneous equation is one where a multivariable function $f(x,y)$ can be written as a composite with the function: $\frac{dy}{dx} = f\left( \frac{y}{x} \right)$ Where the $y$ values and $x$ values are constants to be found. An example of a homogeneous DE is shown below: $\frac{dy}{dx} = \frac{{y + x}}{x}$ Simplifying gives us a better idea of why it is homogeneous: $\frac{dy}{dx} = \frac{y}{x} + 1$ Now it's time for us to solve it! >[!Success]- Solving the Equation >To solve this, we can use the substitution $y = vx$. where $v$ is an arbitrary constant that can be expressed in terms of $x$. We therefore get the equation: >$\frac{dy}{dx} = \frac{{vx + x}}{x}$ >To continue solving this, let's try differentiating the function $y = vx$. Note that as $v$ can be written in terms of $x$, this lets us create the following expression for $\frac{dv}{dx}$ using the product rule: >$\frac{d}{dx}(vx) = v + x \frac{dv}{dx}$ >This serves as a substitution for $\frac{dy}{dx}$ as we have performed the equivalent of an implicit differentiation for $y$. This lets us continue to simplify: >$x\frac{dv}{dx} + v = \frac{vx + x}{x} = v + 1$ >$x \frac{dv}{dx} = 1$ >$\frac{dv}{dx} = \frac{1}{x}$ >Integrate on both sides! >$\int dv = \int \frac{1}{x} \, dx$ >$v = \ln \mid x\mid + c$ >We can therefore get a general formula of: >$\frac{y}{x} = \ln \mid x\mid + c$ >$y = x\ln \mid x \mid + \ cx$ # ODEs - Methods of Solving ## Euler's Iterative Method This method is a really, **really** slow method of approximation which can be used to solve ODEs if an initial numerical value (the **start**), an ending value, the **endpoint**, and the incrementation of the function, the **stepsize.** With a stepsize given by h with limits to ℎ=0 an approximation can be created of the rates of change. >[!Success]- Intuition >In each question that lets you use this method, there'll typically be a table and a range for you to fill in. For example, let's take the differential equation $\frac{dy}{dx} = x-y+7$ as an example. It is given that the range of the approximations we want to find are in the range $0 < x < 1$. The stepsize is defined as $h = 0.2$. >Remember that the $y$ value changes with any change in the $x$ value, giving us a formula for the $y$ value after each step (using the function to denote sensitivity): >$y_i = y_{i-1} + hf(x_{i-1}, y_{i-1})$ >Where $i-1$ denotes the values of the variables at the stepsize before. >You can imagine this as an extension of the fundamental theorem of calculus - the $y$ value increments with respect to $\frac{dy}{dx}$, similar to: >$y_{i} = \Delta x \left( \frac{dy}{dx} \right)$ >This is why the formula for $\Delta x$ using Euler's method is: >$x_{i} = x_{i-1} + h$ >We can then substitute the stepsize we have on $x$, $0.2$, in place of $\Delta x$. We can substitute both values into the original equation as shown given that the original $x$ value is 0 and $y(0)$ is 0.5: >$y_{i} = 0.5 + (0 - 0.5 + 7)$ >Getting us a new value for $y_i$ of $1.8$. We can use these new values for $x$ and $y$ to plug back into the function and get another $y$ value for the step. ## An ODE in the form $\frac{dy}{dx} + P(x)y = Q(x)$:(Integrating Factors) An integrating factor is a tool we use to solver *linear differential equations* - or DEs where the variables are arranged as shown above. Given functions $P(x)$ and $Q(x)$ in the first-order differential equation, we can write the integrating factor as shown: $\mu(x) = e^{\int P(x) \, dx }$ >[!Abstract]- Why $e$? > Let's look at an example to familiarise ourselves with LDEs: $\frac{dy}{dx} + xy = x^2$ This is a imple example of a linear differential equation. Note that our $P(x)$ can therefore be interpreted as $x$, meaning that our integrating factor will be: $\mu(x) = e^{\int x \, dx } = e^{\frac{1}{2}x^2}$ Multiplying both sides by the integrating factor yields: $e^{\frac{1}{2}x^2}\left( \frac{dy}{dx} \right) + xe^{\frac{1}{2}x^2}y = x^2$ Note that this is a product rule derivative. Let's rewrite this by integrating by parts: $\frac{d}{dx}(ye^{\frac{1}{2}x^2}) = x^2$ Now math the expression to get our answer! $ ye^{\frac{1}{2}x^2} = \frac{x^3}{3} + c$ $\frac{1}{2}x^2y = \ln\left( \frac{x^3}{3} + c\right)$ $y = \frac{2\ln\left( \frac{x^3}{3} + c\right)}{x^2}$ We get our general formula for the differential equation! The expression's pretty complicated though... Oh well, an answer's an answer! # Second-Order Differential Equations very difficult topic :( By this point, we're out of the HL AA range! Hurray! We'll be using the differential equation book by Zills and Cullen moving forward, giving us a better view and intuition of topics both new and old. Where the west is wild... We take a second-order differential equation as a differential equation with a highest order of 2 - these also form the backbone for our [[Boundary Value Problems (Maths)|boundary value problems]], which require a second-order derivative to find a solution for. ## Pursuit Curves # Example Problems >[!Example]- 11A Question 28) >Consider this system of Coupled Differential Equations: >$\frac{dx}{dt} = x+2y$ >$\frac{dy}{dt} = x-y$ >When $t = 0$, $x = 1$ and $y = 2$. >Use Euler's Method with a step length of 0.1 to estimate the values of $x$ and $y$ when $t = 1$. > >**Solution:** > >[!Example]- 11B Question 41) >a) Show that the substitution $u = y + \frac{1}{x}$ transforms the differential equation $x^2 \frac{dy}{dx} = xy - x + 2$ into $x \frac{du}{dx} = u - 1$. >b) Hence find the general solution of the equation $x^2\frac{dy}{dx} = xy-x+2$. > >**Solution a)** >Note that taking the derivative of $u$ with respect to $x$ gives us the following: >$\frac{du}{dx} = \frac{dy}{dx} - \frac{1}{x^2}$ >We can therefore perform a substitution with the $\frac{dy}{dx}$ in the main differential equation: >$x^2 (\frac{du}{dx} + \frac{1}{x^2}) = xy-x+2$ > >**Solution b)** ><u>*Method 2*</u>: >Just use our simpler differential equation and treat it like a $\frac{dy}{dx} = f(x)$ relationship: ><u>*Method 1*</u>: >Use the integrating factor method! >$x^2\frac{dy}{dx} - xy = 2-x$ >$\frac{dy}{dx} - \frac{y}{x} = \frac{2}{x^2} - \frac{1}{x}$ >The integrating factor for this equation is therefore given by $\mu_{x} = e^{\int \frac{1}{x} \, dx } = e^{1/x}$ >Multiply both sides by the integrating factor: >$\frac{1}{x} \frac{dy}{dx} - \frac{y}{x^2} = \frac{2}{x^3} - \frac{1}{x^2}$ >Isolate the $y$ and exterminate - integrate both sides! >$\int \frac{1}{x} \, dy - \frac{y}{x^2} = \int \frac{2}{x^3} - \frac{1}{x^2} \, dx $ >$\frac{y}{x} = \frac{1}{x} - \frac{1}{x^2} + c$ >This gives us our final answer: >$y = 1 - \frac{1}{x} + cx$ >[!Example]- 11B Question 45) >Variables x and y satisfy the differential equation $\sqrt{ 1 − x^2}\ \frac{dy}{dx}$ = $\sqrt{ 1-y^2 }$ when y = 1/2 and x = $\frac{\sqrt{3}}{2}$ . Show that this particular solution can be written in the form y = $x\sqrt{ A } + B \sqrt{ 1-x^2 }$, stating the values of the constants A and B. > >**Solution:** >Note that we can rewrite this as a separable differential equation with the formula $\frac{dy}{dx} = f(x) g(y)$: >$\frac{dy}{dx} = \frac{\sqrt{ 1-y^2 }}{\sqrt{ 1-x^2 }}$ >Let's simplify this using the general separable differential equation formula! >$\int \frac{1}{\sqrt{ 1-y^2 }} \frac{dy}{dx} \, dx = \int \frac{1}{\sqrt{ 1-x^2 }} \, dx $ >This nets us the following identity: >$\arcsin y = \arcsin x + c$ >Where $c$ is an arbitrary constant consisting of the constants from both integrals. >Substitute our numerical values to get a value for $c$: >$\arcsin \frac{1}{2} = \arcsin \frac{\sqrt{ 3 }}{2} + c$ >$c = -\frac{\pi}{6}$ >We can bring out the $\arcsin$ on both sides by applying the $\sin$ operator to both sides: >$y = \sin \left( \left( \arcsin x-\frac{\pi}{6} \right) \right)$ >Note that this is a compound angle. We can use the [[Trig Identities - Derivations DLC Ver. (Maths)|compound angle identity for sine]] to expand this: >$y = \sin \arcsin x\cos \frac{\pi}{6} - \cos \arcsin x \sin \frac{\pi}{6}$ >$y = x\cos c - \cos \arcsin x\sin \frac{\pi}{6}$ >This means that $\arcsin x$ must be an angle for $\cos x$ to be valid. Note that we can make a right triangle with hypotenuse length 1 and a height with length x. This means it's base is given by the length $\sqrt{ 1-x^2 }$ thanks to Pythagoras' Theorem. We can therefore take the $\sin$ of the right triangle as $\frac{o}{h} = \frac{x}{1}$, meaning that our angle to the base $\theta$ is equal to $\arcsin x$. Remember that $\cos \theta$ is given by the formula $\frac{a}{h}$, giving us: >$\cos \arcsin x = \sqrt{ 1-x^2 }$ >To prove that we can write the differential equation in the form specified, we must substitute the values from $c$ into the equation as shown: >$y = x\frac{\sqrt{ 3 }}{2} - \sqrt{ 1-x^2 } \frac{1}{2}$ >This gives us a value for the constants $A$ and $B$: >$A = \frac{x}{2} = \frac{\sqrt{ 3 }}{4}$ >$B = \frac{1}{2}$ >[!Example]- 11C Question 19) >Find the particular solution of the differential equation $\frac{dy}{dx} + \frac{2xy}{x^2+1} = 2x$ with the initial condition $y = 0$ when $x = 0$. >[!Example]- 11C Question 22) >A lazy accountant is given a bunch of differential equations to do. Unfortunately, he's since forgotten the maths he's done in high school in favour of learning how to ride a unicycle and really needs a way to simplify the equation. Show that the general solution of the differential equation $\frac{dy}{dx} - y\tan x = \cos x$ can be written int$y = A \sin x + (Bx + c)\sec x$. > >**Solution a)** >Find the integrating factor. >$\mu(x) = e^{\int -\tan x \, dx } = e^{\ln \cos x} = \cos x$ >Multiply it across the entire expression to get: >$\frac{d}{dx}(y\cos x) = \cos^2x$ >Integrate both sides to get: >$y\cos x = \frac{1}{2}(x + \sin x\cos x) + c$ >Now divide by $\cos x$! Make sure the algebra's always right - bad algebra is a question-killer for physics and maths. >$y = \frac{x}{2\cos x} + \frac{1}{2}\sin x + \frac{c}{\cos x}$ >Rewrite all the $\cos x$ terms into $\sec x$ to get your answer. We've proven it! >$y = \frac{x}{2} \sec x + \frac{1}{2}\sin x + c \sec x$ >$y = \frac{1}{2}\sin x + \left( \frac{x}{2} + c \right) \sec x$ >$y = A\sin x + (Bx + c)\sec x$ # What's Next?