 Namaste. Myself, Dr. Basaraj Ambiradhar, Assistant Professor, Department of Communities and Sciences, Walchain Institute of Technology, Svalapuram. Now, in this video, I explain 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. Carnolation theorem statement, if L inverse of f of x is equal to f of t and L inverse of g of x is equal to g of t, then L inverse of f of x into g of x is equal to integration u is equal to 0 to t, f of u into g of t minus u into d o. Working procedure for problems. Step one. The given equation is expressed as the product of the two functions f of s and g of s. In a step two, we find 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 and in step three, we apply the Carnolation theorem that is 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 and in step four, we evaluate the integral to obtain the required inverse Laplace transform. Pause the video and answer the question. What is the inverse Laplace transform of number one s upon s squared plus 4 and number two 1 upon s minus 3? I hope all of you have written the answer. Answer. First one. L inverse of s upon s squared plus 4 is equal to cos 2 t because L inverse of s upon s squared plus e squared is equal to cos 80 and second one is L inverse of 1 upon s minus 3 is equal to e to the power 3 t because L inverse of 1 upon s minus a is equal to e to the power 80. Here a is equal to 3. Come to an example. Use the Carnolation theorem. Find the inverse Laplace transform of 1 upon s squared into s minus a solution. Let 1 upon s squared into s minus a can be split as the product of the two functions that is 1 by s squared into 1 upon s minus a. Let f of s is equal to 1 by s squared and g of s is equal to 1 upon s minus a. Now further to find f of t and g of t that is the f of t is equal to L inverse of f of s which is equal to L inverse of 1 by s squared is equal to t and g of t is equal to L inverse of g of s which is equal to L inverse of 1 upon s minus a which is equal to e to the power 80. Now by the Carnolation theorem we know that 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 du. Therefore L inverse of 1 upon s squared into s minus a is equal to integration u is equal to 0 to t and replacing in the f of t t by u that is it becomes u and in g of t t by t minus a u that becomes e to the power a into t minus u that is u into e to the power a into t minus u into du. Therefore L inverse of 1 upon s squared into s minus a is equal to integration u is equal to 0 to t by replacing in the f of t t by u and g of t by t minus a u it becomes u into e to the power a into t minus u into du. Therefore the L inverse of 1 upon s squared into s minus a is equal to integration u is equal to 0 to t u into e to the power 80 into e to the power of minus a u by using the law of indices here. Now when you are integrating with respect to u means t is treated as a Carnolation take this e to the power 80 as a outside the integral sign that is equal to e to the power 80 into integration u is equal to 0 to t into u into e to the power of minus a u into du. On integrating the Bernoulli integration by parts that is L inverse of 1 upon s squared into s minus a is equal to e to the power 80 into. Now here the first function u as the first function and e to the power minus a u as a second function. Now according to Bernoulli integration by part the first function as it is that is u as it is and integration of e to the power of minus a u is e to the power of minus a u by minus a minus derivative of first in the preceding the derivative of u with respect to u is 1 and integration of second term in the preceding that is e to the power of minus a u by minus a minus a into minus a that is plus a square between the limits 0 to t. Now substituting the lower now substituting the upper minus lower limit and that is equal to e to the power 80 into substituting upper limit it becomes minus t into e to the power of minus a t by a minus a minus e to the power of minus a t by a square plus when substituting the lower limit the first term becomes 0 and minus and minus it become plus e to the power 0 is equal to 1 1 upon a square that therefore L inverse of 1 upon a square into s minus a is equal to e to the power 80 by a square after multiplying this e to the power 80 that is multiply the entire term by e to the power 80 and it is simplifying the L inverse of 1 upon s square into s minus a is equal to e to the power 80 by a square minus t by a minus 1 upon a square. Now come to another example find the inverse Laplace transform of s upon s plus 2 into s square plus 9 by Carnot Lation theorem solution s upon s plus 2 into s square plus 9 can be stood as a product of the two function that is 1 upon s plus 2 into s upon s square plus 9 that f of s is equal to 1 upon s plus 2 and g of s is equal to s upon s square plus 9. 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 upon s plus 2 is equal to e to the power of minus 2 t and g of t is equal to L inverse of g of s is equal to L inverse of s upon s square plus 9 is equal to cos 3 t m. Now by the Carnot Lation theorem we know that 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 u that is L inverse of s upon s plus 2 into s square plus 9 is equal to integration u is equal to 0 to t and replacing the f of t t by u and g of t t by t minus u then it becomes the e to the power of minus 2 u into cos 3 into t minus u into d u and that is equal to integration u is equal to 0 to t e to the power of minus 2 u into cos of 3 t minus 3 u into d u. For the integration purpose we know that the integration of e to the power of x into cos v x into d x is equal to e to the power of x upon a square plus b square into a cos v x plus b sin b x here. Now using that formula the integration becomes e to the power of minus 2 u upon minus 2 square plus bracket minus 3 whole square into minus 2 into cos 3 t minus 3 u minus 3 into sin 3 t minus 3 u between the limits 0 to t m. Hence of 2 to the upper and lower limit it becomes 1 upon 13 e to the power of minus 2 t into minus 2 into cos 0 minus 3 into sin 0 minus of e to the power of 0 minus 2 into cos 3 t minus 3 into sin 3 t which is equal to 1 by 13 into e to the multiplied this by e to the power of minus 2 t that becomes the minus 2 into e to the power of minus 2 t because cos 0 is equal to 1 plus sin 0 is equal to 0 is equal to 0 and then e to the power is equal to 1 minus into minus it become plus 2 into cos 3 t plus 3 into sin 3 t. Therefore, n inverse of s upon s plus 2 into s square plus 9 is equal to 1 by 13 into 2 cos 3 t plus 3 sin 3 t minus 2 e to the power minus 2 t m reference. Thank you.