# Multiplication by ‘t’ Property – Laplace Transform – Engineering Mathematics 3

Hello friends so today we are gonna learn a new theorem called as multiplication by T theorem of Laplace transform now what is the statement of multiplication by T Theorem so it is given like this if Laplace or function of T is some function of s then the Laplace of T raised to n into function of T is equal to minus 1 raise to n D by DS raised to n Phi of A that means nth derivative of Phi of S so your why it is called as multiplication by T so if you’ll observe then we have Laplace of P raise to N F of T means our function f of T is getting multiplied by T raise to n term or else A by P hence the name is given as multiplication by T theorem now we have to prove that this theorem is true for all natural numbers so now let’s start with the derivation of this theorem so since I want to prove that this theorem is true for all natural numbers n what I’ll do is Here I am going to use a method of mathematical induction because the method of mathematical induction is used to prove the theorem for all natural numbers or to prove any statement for all natural numbers so here and say that we will prove given state mein by method of mathematical induction now and what is the method of mathematical induction so it is a two-step process in which first step is called as base case and the second step is called as inductive step so in base case what we do that we prove that the given statement is true for first natural number so now the first natural number is 1 so we will prove that the given statement is true for n equal to 1 next in inductive step what we’ll do is we’ll assume that this statement is true for any natural number and we’ll prove that it is true for the next natural number so these are the two steps and after that we can infer that it is true for all the natural numbers so let’s start so in start with step number one which is called as base case where we prove that it is true for n equal to 1 now let’s prove for n equal to 1 so let’s start from here so it is given that let Phi of s is equal to Laplace of function of T now guys to prove that Laplace of function of TB use the definition of Laplace transform so what is the definition of Laplace transform failure I’d say it is integration from 0 to infinity e raised to minus s T into the function of T DT now as I want to find out or as I want to prove the theorem or the given statement for now first natural number what I’ll do is I take derivative on both sides and I’ll apply the rule of differentiation under the integral sign which is called as DUI s so here I say I’ll say this as equation number 1 and I’ll say differentiating both sides with respect to s and applying duis that is differentiation under integral sign so what is DUIs so in duis we get the differentiation of fire faces fire – of s and we applied differentiation under the integral sign so this is the integral sign so you and say zero to infinity as it is let us apply the differentiation so it is always a partial differentiation so it is dou by dou X of e raised to minus s T F of T DT so I write down the DT here now we have to find out the differentiation of this function with respect to X with respect to s partially so here F of T is constant which will come outside because it does not have s so we will get therefore Phi dash of s equal to integration 0 to infinity F of T outside next the derivative of e raised to minus s T with respect to s is e raised to minus s T into minus T so that is e raised to minus s T into minus T DT now we’ll rewrite this terms so it is integration 0 to infinity minus and outside a raised to minus s T into T f of T DT and now if I compare this term with the definition of Laplace transform then this integral is nothing but Laplace of T into F of T because here function of T is this term so we got the value of Phi dash of s so can I say that therefore this Laplace of P into function of T is equal to by sending this negative sign on the left hand side minus Phi dash of s and therefore I can say that Laplace of T into f of T is equal to minus off now Phi dash of S is nothing but derivative of Phi of s so it is minus D by D’s of Phi of s so guys if you’ll see carefully then I have derived or I have proved that the given statement is true for n equal to 1 where CR the power of T is 1 here we have minus 1 raised to 1 which is minus 1 and D raised to 1 D s raised to 1 so this was the original statement and if we’ll put n equal to 1 then we’ll get the exactly same result so here I’ll say that therefore the rule is true for and equal to 1 so in the base case we have proved that the given statement is true for first natural number that is n equal to 1 now let’s proceed to step number 2 that is inductive step so here and say that for step number two which is inductive step so an inductive step what we do is we consider that the given statement is true for any natural number and we prove that it is true for the next national number so here I say that now we assume that the rule is true for any natural number so let’s say that natural number is M so n is equal to M and we’ll prove that it is true for next natural number so the next natural number of M is M plus one so true for n equal to M plus one so let’s start with this so since n is equal to M so our statement so on statement Laplace of T raised to n will become T raised to n here we’ll get minus 1 raise to M D raise to M D s raise to M Phi of s so you will get Laplace of T raise to M f of T is equal to minus 1 raise to M D raise to M be s raise to M Phi of s so we have assumed that role is true for n equal to M hence I am substituting n equal to M everywhere now on left hand side we can apply the definition of Laplace transform so therefore on left hand side will get integration 0 to infinity e raise to minus s T into function of T so your function of T is T raise to M F of T DT which is equal to minus 1 raise to M D raise to M D s raise to M Phi of s now as we want to prove that this statement is true for n equal to M plus 1 we have to get M plus 1 everywhere and for that I will differentiate on both sides with respect to s so that here we can get M plus 1 but if you lobster the left hand side then we have integration so your we have to differentiate with respect to s as well as we have to apply the D you is that a differentiation under the integral sign so you will say differentiating both sides with respect to s and applying bu is that is differentiation and the integral sign so on the right hand side if you will see then minus 1 raise to M is a constant so I’ll keep it outside now as we want to differentiate this term so right now we have differentiation or I’ll say the M order differentiation so M the order differentiation when I differentiate once again with respect to s that will become M plus 1 to order so because we are taking the next order of derivative so we are taking to you to one more time hence that M is becoming M plus 1 so we go to value on right hand side on left hand side as we are applying the U is what will do is inside the integral sign will differentiate partially with respect to s so that will become dou by dou s of a raised to minus s T into T raise to M F of T DT now to differentiate this function with respect to s this P raised to M is constant because it does not have s as well as F of T is also constant so I take these two functions outside and the derivative of e raised to minus s T is e raised to minus s T into minus T because I am differentiating with respect to s so therefore integration 0 to infinity T raise to M F of T outside and I raise to minus s T into minus T DT is equal to right hand side as it is which is minus 1 raise to M D raised to M plus 1 upon D s raised to M plus 1 or M plus 1 derivative of Phi of s now you’re now here I can take minus sign outside so that will become minus in aggression 0 to infinity next arrays to minus s T as it is and T raised to M into T raised to 1 will become T raised to M plus 1 and f of T as it is DT is equal to minus 1 raise to M D raised to M plus 1 upon D s raised to n plus 1 into Phi of s so now on left hand side we have the definition of Laplace transform so erased 2 minus s T into this is my function of T and integration is with respect to 0 to infinity so I can say minus n as it is that it is Laplace of T raised to M plus 1 into f of T which is equal to minus 1 raise to M d raised to M plus 1 upon D s raised to M plus 1 into Phi of s so therefore we can say that Laplace of T raised to M plus 1 into f of T equal to now I’ll send this minus sign on other side so when it will go there that will become minus 1 raised to M plus 1 and here it is d raised to M plus 1 upon D s raise to n plus 1 or the M plus 1 derivative of Phi of s so there is now if you observe then we again got the same statement with n equal to M plus 1 so here I can say that therefore if this statement or the property is true for n equal to M then it is true for and equal to M plus one so you have seen here so we have completed the two steps of method of mathematical induction so from these two steps now will infer something so now I am going to step number three or the final step so here I say that since the property is true for n equal to one as we have seen in step number one we can say that by step number two we can say that it is true for n equal to M plus 1 where m is 1 that is 1 plus 1 that is 2 it means it is also true for n equal to 2 because in step number 2 we have proved that if the property is true for n equal to M then it is true for next natural number that is M plus 1 so since it is true for n equal to 1 from step number 2 we can say that it is true for n equal to 2 as well now I will repeat the concept once again for n equal to 2 so now I will say since the property is true for n equal to 2 then by step number 2 we can say it is true for the next natural number that is it is true for n equal to 2 plus 1 that is 3 and if I repeat this process again then you will find that the above property is true for all natural numbers so therefore you are and say that it is true for any value of n so here we accrued that multiplication by T theorem and we will use that result to solve the numerical based on multiplication by T of Laplace transform thank you

Good one πππ

can you please teach the multiplication and division property of inverse laplace transform ?

please make video on second shiftingtheorem

Thank you, it was pretty useful for me π

Plz make video on vector

it was pretty helpful for me

Where the minus sign gone at last step while proving result is true for m+1??

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