 Namaste. Myself, Dr. Basaraj Emberadhar, Assistant Professor, Department of Communities and Sciences, Valchain Institute of Technology, Solapur. In this video, I explain how to find the inverse Laplace transform by Carnolation Theorem. Learning Outcomes. At the end of this session, student will be able to find the inverse Laplace transform using Carnolation Theorem. Watch the video, write down the statement of Carnolation Theorem. I hope all of you return the statement solution. If L inverse of f of s is equal to f of t and L inverse of g of s is equal to g of t, then the Carnolation Theorem is given as L inverse of f of s into g of s is equal to integration u is equal to 0 to t f of u into g of t minus u into d o. Come to example, using Carnolation Theorem, find inverse Laplace transform of 1 upon s into s squared plus 9 solution. Let the given function 1 by s into s squared plus 9 can split as the product of the two functions that is 1 by s into 1 upon s squared plus 9. Now, let f of s is equal to 1 by s and g of s is equal to 1 upon s squared plus 9. Now, further find the f of t and g of t that is f of t is equal to L inverse of f of s which is equal to L inverse of 1 by s which is equal to 1 and the g of t is equal to L inverse of g of s. The L inverse of 1 upon s squared plus 9 is equal to 1 by 3 into sin 3 t. Now, by the Carnolation Theorem L inverse of f of s into g of s is equal to integration u is equal to 0 to t f of u into g of t minus u into d o. It means you have to replace in the f of t t by u that remains same and g of t t by t minus u that is g of t minus u is equal to 1 by 3 into sin of 3 into t minus u. Therefore, L inverse of 1 by s into s squared plus 9 is equal to 1 by 3 into integration u is equal to 0 to t into sin of 3 t minus 3 u into d o. Now, the integrate the sin of 3 t minus 3 u with respect to u treating t as a constant that t is equal to 1 by 3 into integration of sin that is minus cos 3 t minus 3 u by minus 3 between the limits 0 to t when substituting the upper and the lower limit and simplifying we get the 1 minus cos 3 t by 9. Come to another example using the Carnolation Theorem what is the inverse Laplace transform of s upon s squared plus e squared whole squared solution. Let s upon s squared plus e squared whole squared can split as the product of the two functions that is s upon s squared plus e squared into 1 upon s squared plus s squared. Let f of s is equal to s upon s squared plus e squared and g of s is equal to 1 upon s squared plus a squared. Now, further we find f of t and g of t that is f of t is equal to L inverse of f of which is equal to L inverse of s upon s squared plus a squared is equal to cos a t and g of t is equal to L inverse of g of s which is equal to L inverse of 1 upon s squared plus a squared which is equal to 1 by e into sin a t. Then by the correlation theorem L inverse of f of s into g of s is equal to integration u is equal to 0 to t f of u into g of t minus u into d o. It means replacing f of t by u that is f of u is equal to cos a u and in g of t t minus u that is g of t minus u is equal to 1 by e into sin of a into t minus u. Then the correlation theorem becomes integration u is equal to 0 to t cos a u into 1 by e into sin a t minus u into d o. Now L inverse of s upon s squared plus a squared whole squared is equal to 1 by e integration u is equal to 0 to t cos a u into sin a t minus a u into d o which is equal to 1 by a integration 0 to t. Now here the integrand is the product of the two trigonometric functions using the basic trigonometric formulae or to express it into sum of difference that is of the form that cos e into sin b. But by the basic trigonometric formulae the cos a into sin b which is equal to 1 by 2 into sin of a plus b minus sin a minus b that becomes 1 by a integration 0 to t 1 by 2 into sin of a u plus a t minus a u minus sin of a u minus a t plus a u into d o which is equal to 1 by 2 a integration 0 to t plus a u minus a u get cancelled it becomes sin a t and plus a u plus a u that is becomes 2 a u that is minus sin 2 a u minus a t into d o. Now 1 by 2 a integrate with respect to u that sin a t is the constant integration of d u is u minus the integration of sin is minus cos minus into minus it become plus the cos 2 a u minus a t by 2 a bit limit 0 to t which is equal to 1 by 2 a substitute in the upper and lower limit that is a t into sin a t plus cos a t by 2 a minus 0 plus cos a t by 2 a now after simplifying plus cos a t by 2 a and plus into minus minus cos a t by 2 a get cancelled which is equal to t into sin a t by 2 a come to another example find L inverse of 1 upon s to the power 4 into s squared plus 4 by Carnullation theorem solution consider 1 upon s to the power 4 into s squared plus 4 s split as the product of the 2 function that is 1 by s to the power 4 into 1 upon s squared plus 4 let f of s is equal to 1 upon s to the power 4 and g of s is equal to 1 upon s squared plus 4. Now find the f of t and g of t that is f of t is equal to L inverse of f of s which is equal to L inverse of 1 by s to the power 4 is equal to T cube upon 3 factorial and g of t is equal to L inverse of G of s, which is equal to L inverse of 1 upon s card plus 4, which is equal to sign 2 t by 2. Then by the Carnot-Reyson theorem, L inverse of 1 upon s to the power 4 into s card plus 4 is equal to integration u is equal to 0 to t f of u into G of t minus u. It means we have to replace in the f of t t by u and G of t t by t minus u. It becomes the integration u is equal to 0 to t u cube upon 6 into sign 2 into t minus u by 2 into du that is 6 to the power 12, which is the constant for integral to take outside that is equal to 1 by 12 integration 0 to t u cube into sign 2 t minus 2 u into du. Now, integrate by Bernoulli's integration by parts, which is equal to 1 by 12 and choose the first and second function by using the Euler rule. It means u cube is the first function and sign 2 t minus 2 u is a second function is equal to 1 by 12 u cube first function as it is integration of second function integration of sign is minus cos 2 t minus 2 u by minus 2 minus derivative of first term in the preceding derivative of u cube is equal to 3 u squared into integration of second in the preceding that is minus sign 2 t minus 2 u by minus 2 into minus 2 plus the derivative of first in the preceding derivative of 3 u squared is 6 u into integration of sign is again minus minus into minus it become plus cos 2 t minus 2 u by minus 2 into minus 2 into minus 2 minus derivative of the first in the preceding that is 6 u is 6 and integration of second in the preceding that is the sign 2 t minus 2 u by minus 2 minus 2 minus 2 minus 2 bit limits 0 to t. Now, substitute in the upper and minus lower limit that becomes 1 by 12 into t cube into cos 0 by 2 plus 60 into cos 0 by minus it minus 0 minus 0 plus 0 minus 6 into sign 2 t by 16 which is equal to 1 by 12 into t cube by 2 minus 3 t by 4 plus 3 by 8 into sign 2 t. Thank you.