 Hello students, I am Deepak Maslaker working as an assistant professor in the Department of Mechanical Engineering at WIT, Solapur. In this session we are going to look at the solution of differential equation by weighted residual method. Here are the learning outcomes of this session. So at the end of this session students will understand what is weighted residual method, what are the steps involved in it and we will try to solve one example problem by using weighted residual method. So weighted residual method is a method, weighted residual method is a numerical method which is used to find out the approximate solution of a boundary value problem. Here is a problem statement. So here you can see the problem, the differential equation is given d square y by dx square plus y plus x equal to 0 and the domain is 0 to 1 and these are the boundary conditions y of 0 equal to y of 1 equal to 0. So we are going to find out the approximate solution of this differential equation by using Galerkin's weighted residual method. The first step involved in the solution is assumption of trial solution. So let us assume a trial solution. Here is our trial solution y of x equal to a0 plus a1x plus a2x square. So the thumb rule is that number of constants involved in the trial solution should be more in number than the order of differential equation. So we are trying to solve second order differential equation and I have chosen three constants here a0 plus a1 a0 a1 and a2. Let us apply the boundary condition over this trial solution. So first boundary condition is at x equal to 0 y is 0 and if I put these values in equation number 1 I get the value of a0 a0 equal to 0. So please remember I have assumed trial solution and this trial solution contains number of constants which are one more in number than the order of differential equation. That is the order of differential equation is 2. So the trial solution contains three constants. If the order of differential equation is 4 then you have to incorporate five constants in the trial solution. After applying first boundary condition let us apply second boundary condition and the second boundary condition is at x equal to 1 y is 0. If I put these values in equation number 1 I will get a1 plus a2 equal to 0 because here a0 is a0 that already we have determined and I can simplify this as a1 equal to minus a2. So till this moment we got two values a0 equal to 0 and a1 equal to minus a2. Now putting the values of a0 and a1 in equation number 1. So after substituting values of a0 and a1 in equation number 1 what we get y of x equal to a0 that is 0 plus a1x but a1 equal to minus a2 minus a2x plus a2x square. So after taking a2 common I will get y of x equal to x square minus x equation number 2. Please remember equation number 2 is the approximate solution of differential equation and this is called one parameter solution because this equation contains only one unknown that is a2 therefore this solution is called one parameter solution. Now in the second step I am defining weighting function and this weighting function or weight function is w of x equal to dou y by dou a2 that is partial derivative of y with respect to a2. So if I differentiate equation 2 partially with respect to a2 I will get w of x that is equal to x square minus x. So in the second step I have calculated weighting function or weight function and how to calculate it just by differentiating the trial solution with respect to the unknown parameter. Once we get the weighting function we have tried to find out domain residual and this domain residual is nothing but the left hand side of the differential equation that is rd of x equal to d square y by dx square plus y plus x. This is our left hand side of differential equation. Now substituting the value of y from equation 2. So from equation 2 we have y equal to a2 into x square minus x plus y and here is the value of y y equal to a2 into x square minus x plus x. So after taking the second derivative of this quantity I get 2 into a2 plus a2 into x square minus x plus x. So here is the value of domain residual and the fourth step is minimization of domain residual. So in order to minimize the domain residual in overall sense I integrate the domain residual multiplied by weighting function over the entire domain and this integral should be zero that means in overall sense the error should be zero over the domain. So substituting the values of wx and rdx here is the value of wx here is the value of domain residual. Now expanding these brackets I have expanded these brackets x square minus x into first term plus a2 into x square minus x into second term it becomes x square minus x whole squared plus x into x square minus x dx. So I expanded this bracket. Now after expansion let us carry out the integration and if I carry out the integration I will get like this 2 a2 into x cube by 3 minus x square by 2 plus a2 into x raise to 5 by 5 minus 2 x raise to 4 by 4 plus x cube by 3 between the limit 0 to 1 plus x raise to 4 by 4 minus x cube by 3 between the limit 0 to 1 equal to 0. Now imposing upper and lower limit x equal to 1 and x equal to 0. So if I substitute these values I will get 2 a2 into minus 5 by 6 plus a2 into 1 by 30 minus 1 by 12 equal to 0 and adding the terms of a2 a2 is taken common and if I add these 2 quantities minus 5 by 3 plus 1 by 30 I will get minus 49 by 30 into a2 minus 1 by 12 equal to 0. If I simplify this expression and extract the value of a2 I will get a2 equal to minus 1 by 12 into 30 by 49. After solving this after simplifying this I will get a2 equal to minus 5 divided by 2 into 49 and a2 equal to minus 5 by 98. Now putting the value of a2 in equation number 2 y of x equal to minus 5 by 98 into x square minus x or we can write y of x equal to 5 by 98 into x minus x square. This negative sign is absorbed inside the bracket. So in this way we can find out the approximate solution of a differential equation by using Galerkin's Wetter residual method. So please remember the steps and please try to recall the steps. The first step is assumption of trial solution and once we assume the trial solution we apply boundary conditions and find out the constants. The second step is finding the value of wetting function. Wetting function is the partial derivative of trial solution with respect to unknown parameter. In the third step we find out domain residual and in the fourth step we go for the minimization of domain residual. By operating these four steps we can obtain the approximate solution of a boundary value problem. Thank you.