hello students welcome to the next lecture on the differential equation in this lecture we will see what is the integrating factor myself dr gerk working in the school of mathematics staffer institute you can simply follow this link for finding the various video in the last lecture we have seen that any of the first order or first degree differential equation of the form mdx plus ndy is exact when this condition will be satisfied and when this condition will satisfy it then you can easily find the solution through this equation but sometime the equation is not being exact for example if you consider this differential equation what is the value of the m and n if you find them you can see that this is not the exact so once this is not exact then you are unable to find the solution through this approach then how you can find the solution of this problem for that we will need to discuss an integrating factor so that we are able to find the solution of this problem what is the integrating factor is so if the differential equation of this form is not exact then we have to make them exact by multiplying some function f that is we when if this is not the exact function exact differential equation we have to multiply this differential equation by something and that function is called as the integrating factor so whatever you have to multiply that is called as the integrating factor make sure that you have to multiply that factor which make them uh exactly for example if i revisit the example again here if i try to sum multiply this by 1 by x square then this will be here again you can compare this with the help of the mdx plus ndy then your new m will be y by x square and n will be my if i separate them will be here then you can check whether this will be the exact or not it is nothing but my here and if you find this value it is my here you can see that both are same it means this is my exact now this will exact with the help of only 1 by x square so therefore this update this 1 by x square is called as integrating factor so the term 1 by x square makes the differential equation to the exact so we called as 1 by x square is exact integrating factor but the rule is that how you can find the integrating factor this is by heating trial i have used as a 1 by x square now we will see in this lecture how you can find the integrating factors so first of all we can simply remark for you of differential equation mdx plus and dy always has the integrating factor but there is no general method for finding them the number of the integrating factor is always with the infinite it may be that 1 by x square is one of them if you multiply some another terms it can be again exactly there are the two ways you can find them integrating factors in this lecture we will see how you can find by inspection in the next lecture we will see some formula based integrating factors what is that if you think if you look about these terms if you try to solve any differential equation if if you find the term which is corresponding to this part i can write this term in instead of here as this you can easily see i can solve them here and so on similarly apart from them if you find some another terms like of x square plus y square it will be of the tan inverse r how you can remember that i just tell you a simple rule whatever the problems are there you have to look about the two terms are there y d x x d y or x d x y d y whatever the signs are there if you find this term which are present the differential equation then it may be of either this form y d x minus x x d y or it can be of y d x minus x d y provided divided by y square or x y and so on if the term is my x d x y d x then it can be of the x square plus y square if it is a denominator side then it's the sum trans if the equations are here now let's see some examples are there would be some say seven to eight examples so that you can easily understand very simple rule i will tell you don't focus about this you have to think about that you have to see that this part is y d x m d y you have to find that on the right hand side there is a term of the d y it means on the right hand side you need only terms which consists of the y it means i don't need the x so what you can do is i can divide x on both side what will happen here so now is there any of the form no because if it is a divided by x so if you term here it will be the x square so it will not be x square it means i can't divide it by x i can divide it by x y if i take as x y can it be the form of the any anything else yes it is nothing but my here so i can divide it by x now you can see the right hand side is only occupied what do you think about that since it is divided by here so it can be either of the tan inverse it can be either the log of this or it can be of any of are there but it's a y d x so it means either r of here x upon y or y upon x so can you find the derivative of the log here this is nothing but my x upon y and y square of this can be cancel out it means it will be y x upon here now you can integrate on the both side what will happen this is nothing but here this is nothing but y cube here is the right answer so you have to think about these terms are there look at the second part are there now here is the x d y can you find the terms of the y d x yes so you can firstly simplify them you can write like this way minus of the y d x and the other part is my x 5 plus x cube y square of dx now since on the right hand side there is a term of the x on dx so i don't want a y here i can take y on this part i can take here as x cube common x square plus y square of dx so which term so i need only for the x part it means this component is the extra term i can take this part as of divided side like of here now you can see that right hand side only of the x do you think which one is there think about that which term will consist of the x square plus y square r here either it can be of the tan inverse or it can be of the log of here but if you consider here what is the derivative of this it will be x square plus y square 2x dx plus 2 y d y but we need a minus it means it of the tan inverse formula so you can write has tan inverse x upon y y is x because x appears first time if appear y here then you can write y by x now you can integrate them you will get the right answer look at the another term are there here is the x d y appear can you find the y d x also yes this is appear so you can pair them the rest you can taken on the left right hand side this is the terms of the x over here so this is also of the x there is no problem so do you think that which one is there there is no term up because if it is a plus then it will be my x y but since it's a negative so we have to divide something here so since this is the x are there so we have to divide it both sides by x square what will happen are there do you find this term is there so once it will be x square so it means this is nothing but my here can you find the derivative of this x square x d y minus y d x you can see that they are same so what is the meaning of that this part is nothing but my y upon x this part is nothing but here now you can integrate them left hand side will be nothing but y by x right hand side is my 2 by x and off is the right answer look at the another part is there this is the x d y this is the y dx that's fine so i can take this part on the right hand side so this part is my here minus of 2 x square plus y square of dx now you can see this is the part of the x so i want x only on the right hand side so this is the part we need to divide on both sides so i can write this part up here do you think that now you can see this is a plus sign so it means this is nothing but my log of here so if you take the derivative of this what will happen this is nothing but x square plus y square you say 2x dx plus 2y d y so what is that i can i can multiply and divide it by 2 on both side on here so what will happen here this part is nothing but my what is that this is nothing but my half ln of x square plus y square is nothing but my here now you can integrate them on both sides you will get as half of this is there look at one more example are there now this is the x d y you so you need a y d x r is it there yes so you can open this bracket and you can simplify that here this is the term of the x here so i want y's to be on the left hand side i can divide it both sides by y square do you think here this is the y are there so it means this is nothing but my y is here so x upon y you can check that quickly this is the y square y d x minus of x d y you can see this is the same so it means this is nothing but my here you can integrate them you will get the right answer here look at this part again so you can see this is there is no ydx or xdy form you can see that there is no xdy and the ydx form so then how you can solve them you can pair them how you can pair them it can be written like this way x y i can pair them here now this is the x square and this is here but do you think that any of the term there is no term argon it means this inspection method is not applicable here so we can't find the inspection method because neither of the term is in the form of x d y plus minus of here it means this method is not applicable for this problem so for that we will try to look some another methods that is the formula based are there that is we will try to find the partial derivative derivatives and use either of them are there how you can use them that we will see in our next class because till then you can simply practice all those problem which is the inspection method so we will see our next class how you can find the integrating factor based on these two formula till then you can simply follow this link for finding the various videos best of luck students happy learning