 Alright students, now we will be learning how to solve differential equations, but let me tell you in this video we are only limiting ourselves to solving first order, first degree differential equations. Higher order, higher degree differential equations is not covered under this particular course, right? Now before I start talking about the methods to solve differential equations, let me tell you the biggest prerequisite for it is your integration. Your indefinite integration should be very very good, that means you should have a very good hand in integrating, right? If you are not, then please revisit that chapter and then come back to this solution of differential equation, right? The very first method that I am going to talk about is the variable separable method, right? The variable separable method. Now as the name itself suggests, right? Here we should be separating out the variables. Now let me show you how. Let us say that we have a differential equation which looks something like this. So dy by dx is a function of x and y, right? Such that you can write this equation as some function let me call as g of y times dy equal to some let's say kx times dx, okay? So this differential equation is of such a nature that you are able to separate out functions of y along with dy and functions of x along with dx. So separation of the two variables x and y is possible in that differential equation, right? To give you an example, let us say I want to solve the differential equation cx square x tan y dx plus cx square y tan x dy equal to 0, okay? So this is a differential equation which I will be solving by using the variable separable method. So variable separable method. So what we see here is that right now the dx contains both types of terms. Term containing x, term containing y. Similarly the dy term is along with terms containing x and terms containing y. We need to separate out these variables. That means y should be in a separate group along with all the functions of y and dx should be in a separate group along with all the functions of x, right? How do we do that? So basically I am going to rewrite this expression as cx square x dx by tan x equal to minus cx square y dy by tan y, right? Now see carefully what I have done. I have separated out all the terms which contain x along with dx, right? And all the terms which contain y along with dy, right? Now here should be taken that dx and dy should both be in the numerator. Never write a dx in the denominator or a dy in the denominator. So dx and dy should always be on the numerator, right? So what great have I achieved here? Nothing great but I have separated out the variables and by doing that I have made a ground for me to start integration, right? So I can integrate both sides now and try to solve this differential equation, right? Now somebody can ask me, I could have integrated here itself, right? Why did I separate out the variables, right? What was the need to separate out the variables? Now please note if you start integrating at this stage, right? You have two functions. Both are dependent on each other. Y and x here are not separate from each other. Y is dependent on x, correct? So while integrating c square tan y with respect to dx, don't assume that tan y will be like a constant and I can simply integrate c can square x. No, that doesn't happen that way. Y and x are interrelated to each other. So I cannot treat y as a constant with respect to x or y-serversa. So this is, this gives us the need to separate out the variables before we start the process of integration, correct? Now let's integrate this. Now to integrate this as I told you, you should be very, very good with your integration, especially indefinite integrals. So what is the integral of c can square by tan x? So if you assume your tan x to be t, c can square x dx will become dt, correct? So 1 by t dt integral will be nothing but ln tan x. If you are not able to find the integral for such functions, please, please, please revisit the indefinite integral chapter. Similarly, this one will become minus ln tan y. Now it's a normal practice that instead of c, I can write ln c also. Why? Why did not I write just a c over here? See, because all the terms involving here are in terms of log. So it's advisable that you write this c as ln c. This facilitates us to club all the terms, correct? So if I bring this on the other side, I'll have ln of mod tan x plus ln of mod tan y equal to ln of c, okay? We all know from the log properties, log a plus log b will become log of a b. So I can write this term as ln mod tan x tan y equal to ln c, all right? Now you can get rid of log from both the sides. It can give you, it will give you tan x tan y equal to c. So this is that curve or this is the family of curves which satisfies this differential equation, correct? So this is a solution for this differential equation. And how did I achieve this solution? By a method called the variable separable method. So I separated the variables y with y, x with x and I integrated both the sides and hence this answer. Hope you have made an honest attempt to solve the question displayed on the screen. So now let us look into its solution. So we want to find the general solution for this differential equation. Now the very first thing that I would try to identify is that can I separate out the variables, right? So can I separate out x's and y's terms in this particular question, okay? So looking at this question, in the present format it seems that x and y are linked to each other, right? So let us try to separate out the variables here. So I can write this as e to the power x into e to the power minus y. So by using the exponent law, so I can separate it out as e to the power x into e to the power minus y. This term also can be separated out as x square into e to the power minus y, right? We can take e to the power minus y common and the remaining term will be e to the power x plus x square. Now I can clearly see, I can clearly get a feeling here that x terms and the y terms are getting separated, right? So our attempt should be to separate out the x terms and the y terms, right? Once we do that, we can take the x and the y term along with the dx and the dy terms. So basically what I am trying to do is, I will bring this e to the power y down. So I will write it like dy by e to the power minus y, right? And I will take this dx term on the numerator side. So this term will become this, okay? Please take care that dy and dx terms should never be in the denominator, right? So dy dx terms should always be on the numerator while you are separating out the variables, okay? So once we have done this, we can say I have got the two variables separated out along with their differentials. Now this is the perfect ground for me to start my integration process because the variables have already been separated. So let me integrate both the sides, right? We all know integral of e to the power y with respect to dy will be e to the power y, correct? Integral of e to the power x is e to the power x, right? Don't ask me why. If you are asking me why, then you need to revisit the indefinite integrals chapter. What is the integral of x square? x cube by 3. And you can place an arbitrary constant C on any one of the sides. Please do not place arbitrary constants on both the sides. Though integration is happening on both left-hand side and right-hand side, the constant of integration or the arbitrary constant has to be only written on one side, right? So basically this particular equation or this particular curve is the general solution for this differential equation, correct? Now sometimes you may be given some initial value also. For example, if the question says find the particular solution for this differential equation under the condition that when y is equal to 1, x is equal to 0, right? So they have mentioned an initial condition by which we can find something called the particular solution, right? How do we do that? Once you have found the general solution, use this condition. So when you put y as 1, you get e to the power 1 which is e. When you put x as 0, this becomes 1, this becomes 0 and c is c, correct? So this gives you c as e minus 1. All I need to do is place this c back over here. So I need to just place this c back over here. So I get an equation like e to the power y is equal to e to the power x plus x cube by 3 plus e minus 1, right? So this becomes the particular solution. So this becomes the particular solution for this differential equation. Whereas this becomes the general solution for this differential equation, right? So the step of finding the particular solution goes via general solution. So once you have found the general solution, use the initial conditions given to you. This is the initial condition. So using this condition, I have got the value for the arbitrary constant which leads to the particular solution. Hope this is clear to you.