My Coding >
Mathematics >
First order ordinary differential equations >
Linear first order ODE
Linear first order ODE (Page: 1)Go to Page:
A firstorder linear ordinary differential equation (ODE) is an equation that can be written in the form \(y\prime + p(x)y = q(x)\), where \(y\prime\) denotes the derivative of \(y\) with respect to \(x\), \(y\prime = \frac{\mathop{dy}}{\mathop{dx}}\). These equations are called "linear" because the dependent variable y and its derivative y' appear only in the first degree, and "ordinary" because they involve only one independent variable x. Firstorder linear ODEs can be solved using integrating factors, which are functions that transform the equation into an exact differential equation. The general solution to a firstorder linear ODE involves one arbitrary constant, which can be determined by applying initial or boundary conditions. Integration factorNow, I will show you, how to solve these linear first order ordinary differential equation with the help of integrational factor. But it is very important to understand, that the best practice is to derive this formula every time you are solving these equaions, rather than using ready made formula, to reduce number of errors. So, Once again, we need to solve differential equation: \[y\prime + P(x)y = Q(x)\] For simplicity, let's use the following abbreviations and meanings: \[y\prime \equiv \frac{\mathop{dy}}{\mathop{dx}}; P(x) \equiv P; Q(x) \equiv Q; y \equiv y(x)\] Now we will introduce integrational factor \(I\): \[I \equiv I(x)\] and multiply both parts of the equation by \(I\): \[I(y\prime + Py) = IQ\] \[Iy\prime + IPy = IQ\] Let's write the product rule for for \(Iy\): \[(Iy)\prime = Iy\prime + I\prime y\] Now if we compare the two last equations, we can make an interesting observation: if \(IPy = I\prime y\), or \(IP = I\prime\), then: \[(Iy)\prime = IQ\] Because we can create any integration factor, this will be our way of making it, and we can find it: \[I\prime = IP\] \[\frac{\mathop{dI}}{I} = P\mathop{dx}\] \[\int\frac{\mathop{dI}}{I} = \int P\mathop{dx}\] \[\ln{\leftI\right} = \int P\mathop{dx} + C\] \[e^{\ln{\leftI\right}} = e^{\int P\mathop{dx} + C}\] \[\leftI\right = e^{\int P\mathop{dx} + C} = e^{\int P\mathop{dx}}e^C = Ae^{\int P\mathop{dx}}\] Because this result is a solution to an equation \(I\prime = IP\), then constant \(A\) will cancel out and we can set it up into any arbitrary fixed value of our choice. I think the value \(A = 1\) is the most easiest and obvious way to do it. Therefore an Integration factor will be:: \[I = e^{\int P\mathop{dx}}\] Let's return to the original equation and use this substitution: \[(Iy)\prime = IQ\] \[\frac{\mathop{d}}{\mathop{dx}}[Iy] = IQ \Rightarrow Iy = \int IQ\mathop{dx} + C\] This is a very important to always keep in mind about constsnt of integration. And now we can express y: \[\boxed{y = \frac{1}{I}\int IQ\mathop{dx} + \frac{C}{I}; I = e^{\int P\mathop{dx}}}\] As for now, I can declare, that this equation is solved win Integration factor, and now we can consider few examples.: For more detailed explanation of all these equations, please watch our video:

Last 10 artitles
9 popular artitles


© 2020 MyCoding.uk My blog about coding and further learning. This blog was writen with pure Perl and frontend output was performed with TemplateToolkit. 