 Namaste, Myself, Mr. Berardar Bala Saheb, Assistant Professor, Department of Humanities and Sciences, Walchen Institute of Technology, Solapur. In this video lecture, we will discuss Laplace transform using property effect of multiplication by T. Learning outcome, at the end of this session, students will be able to apply effect of multiplication by T property to find Laplace transform of U1 function. Let us pause the video for a while and write the answer to the question. Find the Laplace transform of cos of 3T. Come back, I hope you have written answer to the question. Here I will going to give the solution for this question. Question is, find the Laplace transform of cos of 3T. We know that Laplace of cos of 8 equal to S upon S square plus A square. This is from Laplace transform table. In this, replace A equal to 3 so that Laplace of cos of 3T equal to S upon S square plus 3 square. That is Laplace of cos of 3T equal to S upon S square plus 9. Now, let us start with property effect of multiplication by T. It states that if Laplace of f of T equal to f of S, then Laplace of T raise to N into f of T is equal to minus 1 raise to N into nth order derivative of f of S with respect to 2S, where N is positive integer. Let us see the proof of this statement. By definition of Laplace transform, we have Laplace of f of T equal to f of S which is equal to integration of e raise to minus S T into f of T dt with the limit T equal to 0 to infinity. Now, differentiating both this side with respect to 2S using differentiation under integral sign rule, we get d by dS of f of S equal to d by dS of integration of e raise to minus S T into f of T dt where T equal to 0 to infinity. By DUIS rule, when we take this ordinary derivative inside the integration, it converts to partial derivative with respect to 2S. Therefore, d by dS of f of S equal to integration 0 to infinity dou by dou S of e raise to minus S T into f of T dt. Now, differentiate this integrand with respect to S partially keeping T constant, we get which is equal to integration 0 to infinity derivative of e raise to minus S T with respect to 2S is minus T into e raise to minus S T into f of T as it is dt which is equal to minus integration e raise to minus S T in bracket T into f of T dt where T equal to 0 to infinity. Now, this definite integration is nothing but Laplace transform of T into f of T. Therefore, d by dS of f of S equal to minus into Laplace of T into f of T. From this result, we can write Laplace of T into f of T equal to minus 1 into d by dS of f of S. Similarly, Laplace transform of T square into f of T equal to minus 1 raise to 2 into second order derivative of f of S with respect to 2S and Laplace transform of T cube into f of T equal to minus 1 raise to cube into third order derivative of f of S with respect to 2S. And so on in general Laplace transform of t raise to n into f of T equal to minus 1 raise to n into n third order derivative of f of S with respect to 2S where n is the positive integer, this is the problem. Now, let us consider example, example 1 find Laplace transform of t into sin of a t, here f of t nothing but sin a t, first we find it is Laplace transform, therefore Laplace of sin a t equal to a upon s square plus a square denoted by f of s. Now, by property Laplace transform of t into f of t is equal to minus 1 first order derivative of f of s with respect to 2 s. In this f of t is sin a t and f of s means Laplace of sin a t. After substituting we get Laplace of t into sin a t equal to minus 1 into d by d s of a upon s square plus a square. Now, differentiate this right hand side a upon s square plus a square with respect to t s using cosine true we get which is equal to minus 1 as it is keeping denominator s square plus a square in numerator into derivative of numerator a is 0 minus a into derivative of denominator with respect to t s is 2 s whole divided by square of s square plus a square which is equal to minus 1 into numerator minus 2 s upon in denominator s square plus a square bracket square. Therefore, Laplace transform of t into sin a t equal to 2 a s upon s square plus a square bracket square this is the required answer. Now, consider example 2 find the Laplace transform of t square into cos of 2 t. Here f of t means cos of 2 t and now we have to find its Laplace transform. Laplace transform of cos 2 t equal to s upon s square plus 2 square which is s upon s square plus 4 and denoted by f of s. This is due to formula Laplace of cos a t equal to s upon s square plus a square. Now, by property of effect of multiplication by t that is Laplace of t square into f of t equal to minus 1 bracket square into second order derivative of f of s with respect to t s. Therefore, by this property Laplace of t square into cos of 2 t is equal to minus 1 bracket square into d square upon d s square of s upon s square plus 4 that is equal to minus 1 bracket square is 1 into d square upon d s square you can write d by d s into d by d s of s upon s square plus 4. Now, keeping one d by d s as it is and first find first order derivative of s upon s square plus 4 using cos n true we get which is equal to d by d s as it is and in numerator write denominator s square plus 4 and into derivative of numerator s with respect to t s is 1 minus s into derivative of denominator is 2 s upon s square plus 4 bracket square that is equal to d by d s in bracket after simplification numerator becomes 4 minus s square and denominator s square plus 4 bracket square. Again differentiate this RHS with respect to t s using cos n true we get in bracket keeping s square plus 4 bracket square in numerator as it is into derivative of numerator with respect to t s is minus 2 s minus keeping numerator 4 minus s square as it is and derivative of this denominator with respect to t s is 2 in bracket s square plus 4 bracket close into 2 s whole divided by square of that denominator gives s square plus 4 bracket rest to 4 which is equal to. Now, taking s square plus 4 common factor in numerator in bracket minus 2 s into bracket s square plus 4 minus 4 into s in bracket 4 minus s square bracket close and curly bracket close whole divided by s square plus 4 bracket rest to 4. Now, one factor s square plus 4 gets cancelled from numerator and denominator and it is equal to we get after opening the bracket in numerator minus 2 s cube minus 8 s minus 16 s plus 4 into s cube whole divided by s square plus 4 bracket cube. Therefore, law plus transform of t square into cos 2 t is equal to now in numerator minus 2 s cube plus 4 s cube use plus 2 s cube minus 8 s minus 16 s use minus 24 s whole divided by s square plus 4 bracket cube this is the required answer. Now, let us consider example 3 find the law plus transform of t into under root 1 plus sin of 2 t here f of t equal to under root 1 plus sin 2 t now find it is law plus transform before that we have to simplify to remove square root sin. Therefore, root of 1 plus sin 2 t is equal to under root of 1 you can write as sin square t plus cos square t from the trigonometric function and plus sin 2 t you can write 2 into sin t into cos t which is equal to under root sin square t plus cos square t plus 2 into sin t cos t you can write as sin t plus cos t bracket square and now in right side square root and bracket square gets cancelled it gives sin t plus cos t. Now, taking law plus operator on both the side so that law plus transform of under root 1 plus sin 2 t equal to law plus of sin t plus cos t we know that law plus transform of sin t is 1 upon s square plus 1 and plus law plus of cos of t is s upon s square plus 1. Therefore, law plus of under root 1 plus sin 2 t equal to 1 upon s square plus 1 plus s upon s square plus 1 which is equal to now taking L c m in the right hand side we get numerator 1 plus s upon s square plus 1 which is equal to f of s. Now, by effect of multiplication by t property law plus of t into f of t equal to minus 1 into d by d s of f of s. Therefore, law plus of t into under root 1 plus sin 2 t equal to minus 1 into d by d s of f of s means 1 plus s upon s square plus 1. Now, by differentiating with respect to t s using quotient rule we get which is equal to minus 1 into numerator keeping denominator s square plus 1 as it is into derivative of numerator with respect to t s is 1 minus numerator 1 plus s as it is and derivative of denominator is 2 s whole divided by s square plus 1 bracket square which is equal to minus 1 into when we open the bracket in numerator we get s square plus 1 minus 2 s minus 2 s square upon s square plus 1 bracket square that is equal to minus 1 into numerator 1 minus 2 s minus s square upon s square plus 1 bracket square. Hence, law plus of t into under root 1 plus sin 2 t equal to now s square plus 2 s minus 1 upon s square plus 1 bracket square this is the required answer. To prepare this video lecture I refer these two books as references. Thank you.