CHAPTER 08.05: HIGHER ORDER AND COUPLED DIFF Eq Back: Part 2 of 2
Then I will get, for the second derivative of y with respect to x I will have to put du/dx and then plus 3(dy/dx) will be 3u, because dy/dx is now written as u, minus 5 first derivative of z with respect to x is w, plus 11y mius 7z, minus 7z is equal to zero. Not equal to zero, it is equal to e^-x. That's what's given to you. So you can already see that by making these substitutons I don't have any second derivative terms now, only first derivative terms. Same thing I will do to the second differential equation which is 13 and second derivative of z with respect to x that'll become first derivative of w with respect to x minus 29, I have first derivative of y with repect to x, that'll become u, plus 31 first derivative of z with respect to x, that'll become w, minus 37y plus 41z. These are the rest of the terms which are as is, is equal to sin(x). So again what you are finding out is that you have all these terms which are written there so this becomes my, this is my equation number 3 and this is my equation number 4 which I have here. So let's go ahead and see that now how I can write down these 4 , 4 differential equations which I have. So what I done is that I already, let me go back here, let me go back here. I've already shown that I have written my dy/dx, dy/dx, let me just focus on these two equations here. So I have already shown that dy/dx is equal to u and dz/dx is equal to w. So as I said that this will be my one first order differential equation, this will be another first order differenetial equation. Then I reduce the main two ones which are given to my to first order differential equations. So that gives me four first order ordinary differential equations. So since it gives me four first order differential equations that means that now I have simultaneous, four simultaneous first order ordinary differential equations. Now I can use the methods which I have learned for my numerical methods for solving ordinary differential equations to do that. So I am going to rewrite my equations now. So my first equation which was: dy/dx=u So my first differential equation is dy/dx=u and this is my first differential equation for which I need an initial condition and the initial condition would be needed on y. So that will be y(0)=13 which is given to me. And that becomes my first equation. Now if I look at the second equation it is: dz/dx=w and now I will need the initial condition on the value of z and z(0) is already given to me as 19 and that becomes my second differential equation. So you can see how I'm, how I am setting these up. Now, the third differential equation is going to come from here, so let me go back here and focus on these two equations here. So what you are going to find out is that you are going to get a differential equation in terms of u. So you are going to take everything which you have here to the right hand side and also divide it by 2. And same thing here, you are going to take everything which is here to the right hand side and divide by 13 to set up the equation number 4. So that's how this is going to pan out. So let me write this down now, so I get du/dx as my third differential equation would be equal to: (e^-x - 3u + 5w - 11y + 7z)/2 and I will need a corresponding initial condition on u. Now you got to understand that u is nothing but dy/dx. So that means that I will need the condition on whatever the dy/dx is given to me. So I need u(0). u(0) is nothing but dy/dx(0) and that value is turning out to be 17. And that's my third differential equation. Same thing, my last differential equation which I will need will be dw, written in terms of dw/dx and that will turn out to be: (sin(x) +29u - 31w + 37y - 41z)/13 and again the initial condition which I will need is on w, but w is not our original variable. w(0) I will need the initial condition, but w(0) is nothing but dz/dx(0) and that value here is 23. And that gives me equation number 4. So, I have, what I have now is that, if I go here, all the way from the first one. I have this first differential equation and this will be my first function, f1, which will be given in terms of x, y, z, u and w. All those, only is, u is the only term but it is actually the function of the independent variuable and the two dependent variables and the two which I introduced here. This will become, this is the second differential equation, this will become the function, f2, again a function of x, y, z, u and w. The independent variable, the two dependent variables and the two which I intoduced. And here will be function f3 now, which will be a function of x, y z, u and w. And here you are seeing instances of all of those in the expression there. And this one here will be the function f4, and the arguments being x, y, z, u and w. So now you have been able to set it up in the form which will allow you to solve them by using your numerical methods for solving ODE's. And that's the end of this segment. |