 In this video, I explain the solution of ordinary linear differential equation by Laplace transform. Learning outcomes at the end of this session, the student will be able to solve the ordinary linear differential equation by Laplace transform method. Introduction, ordinary linear differential equation with a constant coefficient having a set of initial condition can be easily solved by the Laplace transform method without finding the general solution and arbitrary constants. Laplace transform of derivative of f of t statement. If L inverse of f of t is equal to f of s, then L of f dash of t is equal to s into L of f of t minus f of 0. Proof by definition of Laplace transform, the L of f dash of t is equal to integration 0 to infinity e to the power of minus s t f dash of t into d t. On integrating by parts L of f dash of t is equal to that is e to the power of minus s t into f of t between limit 0 to infinity minus the integration 0 to infinity derivative of e to the power of minus s t minus s into e to the power of minus s t and integration of f dash of t is f of t into the d t. Now, substituting the upper minus lower limit we know that the e to the power of minus s t f of t is equal to 0 when t tends to 0. Therefore, when substituting the upper limit that becomes 0 minus e to the power 0 is equal to 1 into f of 0 that becomes minus f of 0. Then the value to integrate with respect to t s is treated as a constant take this minus s as outside the sun sign that is minus into minus plus s integration 0 to infinity e to the power of minus s t f of t into d t. Therefore, L of f dash of t is equal to s into L of f of t minus f of 0. Similarly, L of f double dash of t is equal to s square into L of f of t minus s into f of 0 minus f dash of 0 and L of f triple dash of t is equal to s cube into L of f of t minus s square into f of 0 minus s into f dash of 0 minus f double dash of 0. In general, the Laplace of derivative of order n is given by L of f 2 over n of t is equal to s to the power n into L of f of t minus s to the power n minus 1 into f of 0 minus s to the power n minus 2 to f dash of 0 minus d dash minus f 2 over n minus 1 of 0. Working procedure for problems, step 1 the given differential equation is expressed as in the notation y dash of t y double dash of t y triple dash of t divided as for the derivative and step 2 take Laplace transform on both sides of the given equation. Step 3 use the expansion of L of y dash of t L of y double dash of t divided as and step 4 substitute the given initial condition and simplify to obtain L of y of t as a function of s and step 5 find the inverse Laplace transform to obtain y of t. Pause the video and write the formula for f to the power 4 dash of t I hope all of you have written the answer solution by the Laplace transform derivative L of f to the power 4 dash of t is equal to s to the power 4 into L of f of t minus s cube into f of 0 minus s cube into f dash of 0 minus s cube into f double dash of 0 minus f triple dash of 0. Come to an example solve d 2 y upon d t square plus 5 into d y by d t plus 6 y is equal to 5 into e to the power 2 t given y of 0 is equal to 2 and y dash of 0 is equal to 1 by using the Laplace transform method solution the given differential equation can be written as y double dash of t plus 5 into y dash of t plus 6 into y of t is equal to 5 into e to the power 2 t the call it is equation number 1. The given initial conditions are y of 0 is equal to 2 and y dash of 0 is equal to 1 call it is equation 2. Taking the Laplace transform on both sides to equation 1 we get L of y double dash of t plus 5 into y dash of t plus 6 into y of t is equal to L of 5 into e to the power 2 t. Now by the linear property the above equation can be written as L of y double dash of t plus 5 into L of y dash of t plus 6 into L of y of t is equal to 5 into L of e to the power 2 t by the Laplace transform of d 2 we know that L of y double dash of t is equal to s square into L of y of t minus s into y of 0 minus y dash of 0 L of y dash of t is equal to s into L of y of t minus y of 0 call it is equation number 4. Sub 2 to the equation 4 in equation 3 we get s square into L of y of t minus s into y of 0 minus y dash of 0 plus 5 into s into L of y of t minus y of 0 plus 6 into L of y of t is equal to 5 upon s minus 2 because Laplace transform e to the power 2 t is equal to 1 by s minus 2. Using the equation 2 that is the initial condition in the above equation and simplifying we get s square plus 5 s plus 6 into L of y of t minus 2 s minus 11 is equal to 5 upon s minus 2 that is the s square plus 5 s plus 6 into L of y of t is equal to 5 upon s minus 2 that the left hand term that is minus 2 s minus 11 can be taken to represent that becomes the 2 s plus 11. Therefore coefficient of L of y of t that is s square plus 5 s plus 6 can be written in the its linear factors as s plus 2 into s plus 3 into L of y of t is equal to 5 plus 2 s plus 11 into s minus 2 of s minus 2 upon s minus 2 by taking LCM here. Therefore the L of y of t is equal to on simplify the numerator it become the 2 s square plus 7 s minus 17 upon s plus 2 into s plus 3 into s minus 2 taking the inverse Laplace transform on both side we get that is L inverse of L is equal to that is 1 that is y of t is equal to L inverse of 2 s square plus 7 s minus 17 upon s plus 2 into s plus 3 into s minus 2 call it is equation number 5. Now consider the 2 s square plus 7 s minus 17 upon s plus 2 into s plus 3 into s minus 2 can be the 3 linear factors can be separated by using the partial fraction that is the a upon s plus 2 plus b upon s plus 3 plus c upon s minus 2 call it is equation 6 where a, b, c are constant to be determined on taking the LCM on the right hand side and simplifying we get that 2 s square plus 7 s minus 17 is equal to a into s plus 3 into s minus 2 plus b into s plus 2 into s minus 2 plus c into s plus 2 into s plus 3 call it is equation number 7. Now to find the value of a, b, c suppose we have to find the value of a, we have to choose the value of s such that the coefficient of the b and coefficient of c become 0. How to choose the value? Now what is the common factors in the both that is as coefficient of b and coefficient of c which is the common factor that is s plus 2 I am right here. Now put s plus 2 is equal to 0 it means s is equal to minus 2. Now to find the value of a, we have to put s is equal to minus 2 in equation 7 we get a to minus 14 minus 17 is equal to minus 4 a which implies a is equal to 23 by 4 and similarly the put s is equal to minus 3 in equation 7 we get 18 minus 21 minus 17 is equal to b into minus 1 into minus 5 which implies b is equal to minus 20 by 5 that is 5 1s are 5 4s are that is minus 4. Put s is equal to 2 in equation 7 we get 8 plus 14 minus 17 is equal to c into 4 into 5 that is c is equal to 5 upon that 20. Therefore the c is equal to 1 by 4 substitute the value of a, b, c in equation 6 we get 2s squared plus 7s minus 17 upon s plus 2 into s plus 3 into s minus 2 is equal to 23 by 4 into s plus 2 minus 4 upon s plus 3 plus 1 by 4 into s minus 2. Using the above result in equation 5 becomes the y of t is equal to L inverse of 23 by 4 into s plus 2 minus 4 upon s plus 3 plus 1 by 4 into s minus 2 that is y of t is equal to using the generative property that can be written as 23 by 4 into L inverse of 1 upon s plus 2 minus 4 into L inverse of 1 upon s plus 3 plus 1 by 4 into L inverse of 1 upon s minus 2 that is y of t is equal to 23 by 4 into L inverse of 1 upon s plus 2 is equal to e to the power of minus 2 t minus 4 into L inverse of 1 upon s plus 3 is equal to e to the power of minus 3 t plus 1 by 4 into L inverse of 1 upon s minus 2 is equal to e to the power 2 t. Thus y of t is equal to 23 by 4 into e to the power of minus 2 t plus e to the power 2 t by 4 minus 4 into e to the power 3 t references. Thank you.