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# Linear Differential Equations (Lecture) | First Order LDE | Solution Of Differential Equations | Differential Equations

We've encountered a lot of time with Linear Differential Equations from class 11th and have also used the concept of Integrating factor. But, we never tried to analyze the concept behind these Linear Differential Equations. So, here's a detailed video by Jeff Chasnov Sir to show the concepts of LDE, Watch and enjoy.

## Linear Differential Equations

We've encountered a lot of time with Linear Differential Equations from class 11th and have also used the concept of Integrating factor. But, we never tried to analyze the concept behind these Linear Differential Equations.
So, here's a detailed video by Jeff Chasnov Sir to show the concepts of LDE, Watch and enjoy.

A linear differential equation or a system of linear equations such that the associated homogeneous equations have constant coefficients may be solved by quadrature, which means that the solutions may be expressed in terms of integrals. This is also true for a linear equation of order one, with non-constant coefficients. An equation of order two or higher with non-constant coefficients cannot, in general, be solved by quadrature.

Subtitles:

So, let's try to solve another type first-order differential equation. This is called a linear first-order. Linear means linear in the y-variable. So we have dy, dx is now multiplied anywhere by, its not multiplied by y. We have a y by itself multiplied by any function of x. Then we have just a function of x on the right-hand side. So this is the standard form of linear first-order ODE. It may not be separable, right? If you pull the px times y on the right-hand side, you may not be able to factor out the y and separate it. So some first-order linear ODEs are separable but the general one is not. Luckily, there's analytical method to solve all linear first-order ODEs and that's what I want to tell you about today. So, the idea is that you take this equation and you multiply it by something that's called an integrating factor. So we multiply by integrating factor which is what I'm going to call Mu, the Greek letter of Mu. So we multiply both sides of this equation by Mu and what we get is Mu which is a function of x times dy, dx plus p of xy. That's going to be equal to Mu of x times g of x. This integrating factor is going to help us to integrate the differential equation. So, how does this help us to integrate the differential equation? Well, we have to find this integrating factor and if we find the correct integrating factor, we can write this left-hand side, Mu of x times this dy, dx plus p of xy. We can write that left-hand side as d, dx of something. Okay? If we can write that left-hand side as d, dx of something, then we can integrate with respect to x and use the fundamental theorem of calculus to integrate. You can integrate a function f prime of x dx which just becomes f of x. So, what is this something here? This Mu will turn out to be independent of y and will depend only on this p. This something here turns out to be Mu of x times y of x. Okay? The goal is to find the integrating factor Mu such that this left hand side, we can write as d, dx of Mu times y. Okay? Then we can integrate. So I'll show you how to integrate in a moment. Let's figure out what that Mu is. So that Mu is a solution of this equation here. Mu times dy, dx. Sorry the Mu is a solution of this second equation Mu times dy, dx plus py equals d, dx Mu y. If we take that second equation. Let me write it here. If we take the second equation and multiply it out. We have a Mu times dy, dx plus Mu times p times y. Then we can use the product rule for the derivative. So, that's equal to Mu times dy, dx plus dMu, dx times y. Okay? We want to solve this equation for Mu, solve this equation for the integrating factor. If we look at this equation, we see a nice cancellation. This Mu, dy, dx cancels Mu, dy, dx. Then we have Mup times y equals dMu, dx times y. So both of these terms are multiplied by y. So, this y cancels this y. Then we end up with the differential equation dMu, equal to p of x times Mu.Okay? That's another differential equation. So, our method of looking for an integrating factor results in a differential equation for the integrating factor. Luckily, that differential equation is separable. So, since that differential equation is separable, we can separate it and integrate. So here, we end up with dMu divided by Mu and then we have p of x, dx. Okay? So we've separated, then we can integrate. So, we integrate from the x naught to x. Then we have Mu of x naught to Mu. We don't know what Mu of x naught is. In the end, it turns out it doesn't matter what Mu of x naught is. Any value will give the same result. So let's just choose Mu of x naught equal to one. So if we do that and we integrate, we have the integral here from x naught to x and the integral here then from Mu of x naught which we choose to be one to Mu. The left-hand side is just for log. So it's log Mu minus log one which is just log Mu, natural logarithm. The right hand side is this integral x naught to x p of x dx and then we exponentiate. So, we get our integrating factor then is equal to e to the integral from x naught to x p of x ex. Okay? We've determined the integrating factor. This integrating factor is e to the integral of x naught to x p of x dx is exactly the function that we need to multiply the differential equation by to convert it to d, dx of Mu times y. Okay? So using this integrating factor then we tried to solve the differential equation. So, now the differential equation is d, dx of Mu times y equals Mu times g. So, that's the left-hand side is d, dx Mu times y. The right hand side is Mu times g. Now we integrate this equation. We use the fundamental theorem of calculus. So, the integral of the derivative of a function is just the function. Then we have to evaluated at the upper and lower limit. So here we're integrating from x naught to x. So we end up with a x. So we have a Mu of x times y minus the value at the lower limit which is Mu of x naught times y naught. We chose Mu of x naught equal to one. So, that's minus y naught. What we've done is we've integrated here from x naught to x of Mu of x times g of x, dx. Okay? So, the fact that Mu allowed us to write the left-hand side as d, dx of something gave us the ability to integrate the differential equation. Then finally, we just need to solve for y. So we solve this equation for y. We add y naught to both sides. We divide through by Mu of x and we get y of x then is equal to one over Mu of x. Then we have y naught plus the integral from x naught to x, Mu of x, g of x, dx. That's the analytical solution to the linear first-order equation. Okay, so let me summarize, it's a little bit complicated for the first time you see it, but it's actually not all that difficult but it will look a little bit messy. So, let me summarize. You have a first-order equation. You always have to put it in standard form, because eventually you're basically just applying some formula for integrating a first-order equation. So, you put it in standard form, dy, dx plus some function of x, p of x times y equals some other function of x which is g of x together with an initial condition, y of x naught equals y naught. The method of solving a linear first order equation, is to have an integrating factor. We multiply the whole equation by Mu of x is unknown function which is an integrating factor. This integrating factor allows us to write the left-hand side as d, dx times Mu times y. Then this one will be equal to Mu times g. Then we can integrate that equation to get a solution. So, we can determine what the integrating factor is. So, if you're given this differential equation, the integrating factor is given by Mu of x equal to the exponential of the integral from x naught to x of p of x, dx. That's the integrating factor. When you solve this differential equation you find this function. This integrating factor by doing this integral. Then once you have the integrating factor, then you can integrate the differential equation. This is then the integration of the differential equation. So, y of x is one divided by this integrating factor times the initial value of y naught plus the integral from x naught to x of Mu of x, the integrating factor, times g of x the right hand side of the differential equation dx. So, these two yellow circled the equations is essentially the formula that one remembers to solve this first-order linear differential equation. I'm Jeffrey Chasnof. Thanks for watching and I'll see you in the next video.

*Self Typed
*Source- Internet, Books and Self-Analysis
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