[Music] so [Music] [Applause] [Music] hello friends so welcome to lecture series on multi variable calculus so in this lecture we will deal with euler's theorem for homogeneous functions so what homogeneous function are first now suppose you have a function of two variable x and y now this function is said to be homogeneous function of degree n if f of alpha x alpha y is equals to alpha k raised to power n f of x y ok so then this if this property hold for a function of two variable ah f x y then we say that is a homogeneous function of degree n ah suppose ah suppose f is x square plus y square it is clearly homogeneous because if you take alpha x alpha y so it is alpha square times x square plus y square you simply replace x by alpha x and y by alpha y so alpha square will come out and it is alpha square times f of x y so we can say that it is a homogeneous function of degree two now now say f is sine inverse x upon y it is also homogeneous because if you replace x by alpha x y by alpha y so it is sin inverse of alpha x upon alpha y which is sine inverse x upon y and it is same as f x y so it is also homogeneous function of degree zero because this can be written as alpha raise to power zero ok so it is also homogeneous function of degree zero say you have this function x raised to the power four plus y raised to power four upon ah upon under root x plus and root y ok now this is also homogeneous function because if you replace x by alpha x and y by alpha y so from the numerator you are you will be getting alpha raise to power 4 and from the denominator you are getting alpha raise to power half so it is 4 minus half that is 4 minus half that is seven by two so this is homogeneous function of degree seven by two ok now there are some function which are not homogeneous like you are having say x cube plus x y it is not a homogeneous function because when you replace x by alpha x and y by alpha y from here we are getting alpha cube and here and from here we are getting alpha square so the entire alpha is not coming out so that means this function is not an homogeneous function similarly suppose you have this function sine inverse of x square upon y it is also not a homogeneous function ok so this is homogeneous function so and homogeneous function is also expressed as you see if we are saying that f is a homogeneous function of degree n then f can be expressed as x raise to power n some g of y by x or or f can be expressed as some y raised to power n h of x upon y if the homogeneous function of degree n now here are some examples the first example is homogeneous function of degree two the second example is homogeneous of degree zero we can easily verify the third example is homogeneous function of degree minus two the third example is not homogeneous now here we have a homogeneous function of n variables the same definition is applicable for homogeneous function for n variables also you simply replace x 1 by lambda x 1 x 2 by lambda x 2 and so on if you are getting lambda s to power and n times the same function that means the function is homogeneous of degree n now comes to euler's theorem it states that if f is a homogeneous function of degree n in x and y and has continuous first and second order partial derivatives then these two result hold so this is euler's theorem so let us try to prove this theorem first so f is a homogeneous function it is given the statement so f can be written as x raise to power n g of y by x so we are assuming f as a homogeneous function of degree n ok now what will be del f upon del x it will be x raise to power n first as it is derivative of second plus second as it is derivative of first similarly what will be del f upon del y it will be x raise to power n g dash y by x and this again which is one by x now what will be x f x plus y f y you simply multiply this by x and this by y and add them what we obtain we obtain it is x raised to the power n minus one because it is x square minus two into x it is x raised to n minus one with negative sign times y g dash of y upon x plus n x raise to power n g of y by x plus here when you multiply by y it is y x is to power n minus one g dash of y by x and these two terms cancel out it is n times and this is nothing but f only so it is nf so we can say that x f x plus y f y is n f ok it is true only for homogeneous functions ok now the second result now we have obtained that x f x plus y f y is equals to n f ok if f is a homogeneous function of degree n now let us differentiate both the side partially with respect to x when you differentiate both side partial respect to x what we obtain it is first as it is there to second plus second derivative first plus y into f y of x is equals to n of f x which is or it is x f x x plus f x plus y f y x is equals to n f so this is a second equation now again you differentiate equation one partial respect to y both sides so what you will obtain it is x f x y plus y f y y plus f y is equals to n fy now ah you multiply equation two by x and equation three by y and add them ok so what you will obtain it is x square f x x from here it is ah y square f y y plus it is x y times f y x plus it is ah when you multiply this by y it is xy times xy plus ah it is xfx plus f y that we can put on the right hand side so what we will obtain it is equals to n minus one times x f x plus y f y ok now since ah since in the statement it is given to us that it has continuous first and second order partial derivatives so we can say that f x y is same as f y x both are equal because it is having first and second order continuous partial derivatives ok so from here we can say that it is x square f x x plus y square f y y plus two x y we can call it f x y or f y x because both are same and it is equals to n minus one and this value is n f so it is n into f so in this way we obtain the second part of the theorem now say we are the first problem first problem is u equal to under root y square minus x square sine inverse x by y now sine inverse x by y is a homogeneous function of degree zero ok and here it is also homogeneous function of degree one because ah when you take when you replace x by alpha x and y by alpha y so alpha square will come out from the from this inside bracket and under root we are having so it will be alpha alpha will come out so basically the degree of this ah problem this function is one so by the euler's theorem we can easily say that n u x sorry x u x plus y u y is equals to n u and n is one so it is equal to u directly by the euler's zero so the first problem is over now come to the second problem now for second problem second problem is also homogeneous you see u equal to y cube minus x cube whole divided by y square plus x square the degree of this function is one so x u x plus y f u i will be equal to n u and n is one so it is u and for a second part x square u x x plus two x y u x y plus y square u y y it is equals to n n minus 1 into f and n is 1 so when you substitute n equal to 1 so this term will be 0 and since the second part hence the right hand side is 0 okay so this problem is also over directly by euler's theorem now let us come to these problems ok now the first problem is u is equal to ln x raised to power 4 plus y raised to power 4 upon s plus y now this function this function is not homogeneous function is not a homogeneous function because you can simply obtained because when you replace x by alpha x and y by alpha y so alpha is not coming out so it is not an homogeneous function ok and suppose you want to find out x u x plus y u y or x square u x x plus two x y u x y plus y square u y y how can we obtain that now this implies e raise to power u is equals to x raise to power four plus y raise to power four upon x plus y now now this is homogeneous ok say if this function is say f if if e raised to power u is f ok now it is an homogeneous function so for homogeneous function euler theorem is applicable so we can simply say that ah x f x plus y f y will be equals to n f and n is three so it is three f and what is f f is e raised to power u so what will be f x it is del by del x of e raise to power u which is del by del u of e raised to power u into del u by del x and which is e raise to power u into u x and similarly f y will be e raise to power u into u y so when you substitute these values here so what you will obtain it is x okay e raised to power u will come out from both the terms so e raise to power u will come out it is x u x plus y u y is equals to 3 and f is e raised to power u e raised to the power u cancels out so this implies x u x plus y u y is equals to 3 so that is how we obtain the first part of the problem now second part can also be obtained you differentiate partial respect to as both the sides what you will obtain x u x x plus u x plus y u y x is equals to zero ok now we differentiate pass respect to y both the sides so it is x u x y plus y u y y plus u y is equal to zero multiply this by x and this by y add them assuming these two are equal u x y and u y x are equal ok so this implies x square u x x plus y square u y y plus 2 x y u x y will be equals to it is x time u x plus y time u y that will go to right hand side which is minus x u x minus y u y which is minus three because x u x plus y u is three so hence we can obtain that this value is minus three now ah come to second problem for second problem also if you directly see the problem directly the problem is not a homogeneous one you see what is what is the function here function is u is equal to cos inverse x plus y upon it is under root x and root y for x y lying between 0 and 1 okay now this function is this function as a whole is not a homogeneous function but if you take f is equals to cos of u which is equals to x plus y upon under root x plus under root y now f is homogeneous function f is equal to this and it is an homogeneous function of degree one by two so euler's theorem is applicable for f for the function f so what will be by euler's theorem it is x f x plus y f y is equals to n f and n is one by two that is one by two times f now what is f f is cos u so what will be f x again it is del by del x of cos u which is del by del u of cos u into del u by del x and that will be minus sin u into u x again what is fy fy is del by del y of cos u which is del by del u of cos u into del u by del y which is again minus sin u into y ok now when you substitute these two values in this equation what you will obtain this is x f x is minus sin u so minus sin u can come out it is x u x plus y u y will be equals to one by time one by two times f and f is cos u so we obtain that x u x plus y u y will be equals to minus one by two ah cot u ok so that is how we obtain the first part of the problem so what i want to say basically sometimes ah the problems are not homogeneous but we can make it homogeneous by substituting f as some function of u we can apply euler's theorem for this f because this f is equal to this which is homogeneous and later on we find that we find f x and f y in terms of u and then we can simply find the values of x u x plus y u y again again suppose you want to find out x square u x x plus y square u y y plus two x y u x y for this problem so we differentiate both sides respect to x partially so what you will obtain x u x x plus u x plus ah y u u y x is equals to minus one by two now derivative of ah derivative of cot is minus cos x square u and u again that is u x ok now do differentiate again this equation both sides respect to y partially so what you will obtain x u x y plus y u y y plus u y is equals to minus one by two into minus cos x square u into u one multiply first equation by x second equation by y multiply this equation by x this equation by y and add them so what you will obtain we obtain x square u x x plus two x y u x y plus y square u y y will be equal to it is one by two cos of square u will come out it is x u it is x u x plus y u y which is minus one by two cot u and this u x x u x and y u y will go to the right hand side which is negative of minus one by two coty so what we have to show we have to show this thing so you can take one by four common when you take one by four common it will be ah from this side we got ah minus cos sega square u into cot u and it is plus two cot u and it is equals to one by four cos x square is one by sin square cot u is ah cos upon sine plus two cos upon sine now you take the lcm and simplify it is one by four you take the lcm sine cube u it is minus cos u plus two cos u times sin square u so what you have to show cos u into cos two u so cos you can come out it is 2 sine square u minus 1 and this is negative of cos two u so it is minus one by four cos u cos two u upon sin cube u so that is how we can prove this part of the problem now similarly we can we can go for the third problem here we can assume 10u as fu so that f which is x cube plus y cube upon x minus y becomes an homogeneous function we can apply euler's theorem for f ok and later on we can take f u as 10 u find f x and f y substitute in these equations so that we can obtain the values of a and b part of this problem now we have one result for euler's theorem let us discuss this this result suppose function f is a sum of two functions g and h this is a homogeneous function of say degree n ok this is also an homogeneous function of some other degree say m [Applause] ok what is given o m a m g is m and this is n so suppose this is m and this is n so suppose f is a function which is a sum of two functions g is a homogeneous function of degree m h is a homogeneous function of degree n now is f homogeneous it can be homogeneous if m is equal to n if m is not equal to n so f will not be a homogeneous function so can we apply euler's theorem for such type of problems you see g is an homogeneous function of degree m so for g euler is applicable so we can say that x g x plus y g y is equals to m g h is a homogeneous function of degree n so for h euler's theorem is applicable so that will be x h x plus y h y is equals to n h you add the two equations when you add the two equations it is x g x plus h x plus y g y plus h y and this equals to m g plus n h now if f f is equals to g plus h so what is f x it is g x plus h x and what is f y it is g y plus h y so we can say that it is x f x plus y f y which is mg plus h so this is a basically a consequence of euler's theorem suppose a function can be expressed as sum of two homogeneous functions of different degree then x f x plus y g y y f y is equals to m g plus n h where degree of ah f is g is m and degree of h is n similarly similarly this is an homogeneous function so we can say that x square g x x plus two x y g x y plus y square g y y is m m minus 1 into g and again this is a homogeneous function of degree n so h x square h x x plus 2 x y h x y plus y square h y y will be equals to n n minus 1 into h now you again add the two equations so what you will obtain it is x square g x x plus h x x plus two x y g x y plus h x y plus y square g y y plus h y y will be equals to m m minus 1 into g plus n n minus 1 into h now g x x plus h x x from here is f x x plus two x y it is f x y plus y square it is f y y which is equals to m m minus one into g plus n n minus 1 into h so hence we have shown the second consequence of euler's theorem okay now suppose we have the first problem okay so in the first problem u is equals to this term plus this term now this term is a homogeneous function of degree two the first term the second term is also an homogeneous function of degree one so both are homogeneous function having ah different degree it's a homogeneous function of degree two it's a homogeneous function of degree one so if you want to compute x u x plus y u y so we can directly apply the first consequence of euler's theorem that is mg plus nh where what is m here m is 2 and g is g is the first term plus n n is 1 into second term so the answer of the first part will be so the answer the first part will be ah first is x u x plus y u y will be the first term has a degree two so it is two times g g is the first term is it ok plus the second term has a degree one that is n into h n is one and h is x sine inverse y by x so this is the value of the ah this expression now the value of second expression which is x square u x x plus y square u y y plus two x y u x y will be equal to m m minus 1 into g plus n n minus 1 into h that is simply 2 x raised to the power 4 plus y raised to the power 4 upon x square plus y square so that is how we can solve this problem now suppose you want to solve second problem ok in the second problem this is not homogeneous however this is homogeneous ok the second is u is equals to e raised to power x square plus y square plus it is x square y square upon x plus y suppose suppose it is g plus h h is clearly homogeneous of degree three but g is not an homogeneous function so g is e raised to the power x square y square so you can make ln of g is equals to x square y square now this capital g is an homogeneous function of degree two ok so now we can apply the consequence of euler's theorem on capital g on capital g and h ok so we can say that x g x plus y g y will be equals to ah two g ok and what will be g x from here it is ah one by g times one by g can come out x g x plus y g y will be equals to two lng so this implies x g x plus y g y will be two g lng and for h it is x h x plus y h y will be equals to three h now you add the two equations it is two it is one you add the two equation so what you will obtain it is x times g x plus h x which is u x ok y times g y plus h y which is u y is equal to this plus this and hence x u x plus y u y which we which is equal to two times x ah g is e raised to the power x square by y square log j is x square y square this side plus three h h's so that is how we can find out the value of this expression so thank you very much [Music] 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