 Namaste. Myself, Dr. Basaraj Ambiradak, Assistant Professor, Department of Humanities and Sciences, Wachan Institute of Technology, Swalapur. In this video, I explain the solution of ordinary linear differential equation by Laplace transform method. Learning Outcomes At the end of this session, the student will be able to solve the ordinary linear differential equation by Laplace transform method. Pause the video and write the answer for the question that is, what is the partial fraction of x upon x plus 1 into x plus 5? I hope all of you have written the answer solution x upon x plus 1 into x plus 5. Now, here in the denominator, both are the linear factor can be separated by the partial fraction as a upon x plus 1 plus b upon x plus 5. Now, on taking the same and simplifying, which is equal to a into x plus 5 plus b into x plus 1 upon x plus 1 into x plus 5. On the cross multiplication, which becomes x is equal to a into x plus 5 plus b into x plus 1. Call it as equation 1. Now, find the value of a and b here. Now, to find the value of b, you have to choose the value of x such that the coefficient of b becomes 0. That is, the x plus 1 is equal to 0. It means the x is equal to minus 1. Put x is equal to minus 1. That minus 1 is equal to 4 a, which is a is equal to minus 1 by 4. Now, similarly, suppose you have to find the value of b, you have to choose the value of x such that the coefficient of a becomes 0. What is the coefficient of a? That is x plus 5. They make it as x plus 5 is equal to 0 implies x is equal to minus 5. That is, put x is equal to minus 5. That is equation 1 implies minus 5 is equal to minus 4 b, which implies b is equal to 5 by 4. The x upon x plus 1 to x plus 5, which is equal to minus 1 by 4 upon x plus 1 plus 5 by 4 upon x plus 5. Now, come to an example. Solve d cube y upon d t cube plus d 2 y upon d t squared minus d y by d t minus y is equal to 0. Given y of 0 is equal to 0, which is equal to y dash of 0 and y double dash of 0 is equal to 6 by using the Laplace transform method. Solution, the given differential equation can be written as y triple dash of t plus y double dash of t minus y dash of t minus y of t is equal to 0, call it as equation 1. The given initial conditions are y of 0 is equal to y dash of 0, which is equal to 0 and y double dash of 0 is equal to 6, call it as equation number 6. Taking the Laplace transform on both sides, two equation 1 we get, L of y triple dash of t plus y double dash of t minus y dash of t minus y of t is equal to 0. By linear property, the above equation can be written as L of y triple dash of t plus L of y double dash of t minus L of y dash of t minus L of y of t is equal to 0. By Laplace transform derivative, we know that L of y triple dash of t is equal to s cube into L of y of t minus s squared into y of 0 minus s into y dash of 0 minus y double dash of 0. L of y double dash of t is equal to s squared 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 as equation number 4. Subtruty the equation 4 in equation 3, we get s cube into L of y of t minus s squared into y of 0 minus s into y dash of 0 minus y double dash of 0 plus s squared into L of y of t minus s into y of 0 minus y dash of 0 minus s into L of y of t plus y of 0 minus L of y of t is equal to 0. using the equation 2 in the above equation and simplifying we get s cube plus s square minus s minus 1 into L of y of t is equal to 6 s cube plus s square minus s minus 1 into L of y of t is equal to 6. Now further for simplification the coefficient of L of y of t can be split as the it is linear factor for that s square into s plus 1 minus 1 into s plus 1 into L of y of t is equal to 6 that is s plus 1 is a further common that become s square minus 1 into s plus 1 into L of y of t is equal to 6. Therefore L of t is equal to 6 upon s minus 1 into s plus 1 whole square. Now taking the inverse Laplac transform on both sides we get y of t is equal to L inverse of 6 upon s minus 1 into s plus 1 whole square call it is equation 5. For taking the inverse of this function we have to separate these linear factors for the case consider 6 upon s minus 1 into s plus 1 whole square. Now these linear factor two linear factors can be separated by using the partial fraction in the denominator one is the non-repetitive linear factor and there is a repeated twice that because a upon s minus 1 plus b upon s plus 1 plus c upon s plus 1 whole square call it is equation 6 where a, b, c are constant to be determined that is now on taking the same on the retention on the cross multiplication it become 6 is equal to a into s plus 1 whole square plus b into s plus 1 into s minus 1 plus c into s minus 1 call it is equation number 7. Now a, b, c are the constant to determine. Now suppose you have to find the value of a to make choose the value of such that coefficient of b and c become 0. Now how to choose that one which the factor is a common in both that is s minus 1 in the both coefficient of b as well as coefficient of c. Now put s is equal to 1 in equation 7 we get a is equal to 6 by 4 is equal to 3 by 2 now similarly we have to put s is equal to minus 1 in equation 7 we get that is 6 is equal to c into minus 2 which is equal to c is equal to minus 6 by 2 is equal to minus 3. Now to find the value of the b we have to put s is equal to 0 in equation 7 we get 6 is equal to a minus b minus c this b is equal to a minus c minus 6 implies c is equal to minus 3 by 2 substitute the value of a, b, c in equation 6 we get 6 upon s minus 1 into s plus 1 whole square which is equal to 3 by 2 into s minus 1 minus 3 by 2 into s plus 1 minus 3 upon s plus 1 whole square using the above result in equation 5 becomes y of t is equal to L in workshop 3 by 2 into s minus 1 minus 3 by 2 into s plus 1 minus 3 into s plus 1 whole square that is on using the linearity property that is a retention completeness that is y of t is equal to 3 by 2 into L in workshop 1 upon s minus 1 minus 3 by 2 into L in workshop 1 upon s plus 1 minus 3 into L inverse of 1 upon s plus 1 whole square which is equal to 3 by 2 into but without that L in workshop 1 upon s minus 1 is equal to e to the power t minus 3 by 2 L in workshop 1 upon s plus 1 which becomes e to the power of minus t minus 3 into e to the power of minus 3 into L in workshop 1 upon s square L because by first shifting property thus y of t is equal to 3 is the common in the first two terms that is 3 into e to the power t minus e to the power of minus t by 2 minus 3 into e to the power of minus t into L inverse of 1 upon s square is equal to t thus y of t is equal to 3 into sin x t minus 3 into e to the power of minus t into t is the required solution once higher engineering mathematics by H. K. Das and engineering R. S. Verma. Thank you.