Solving Second Order Ordinary Differential Equations Exactly: Distinct Roots

In this segment, we'll solve a second order ordinary differential equation exactly. I will take the example with the distinct roots of the characteristic equation.

So, this is the second order ordinary differential equations we're going to solve, and it has this as the forcing function, while these are the two initial conditions which we'll need in order to solve this particular problem. So, let's go ahead and get started. Your full solution will be the homogenous part plus the particular part. So, if that is the case, then what's going to happen is that let's concentrate on finding the homogeneous part. In order to find the homogenous part what we're going to do is we have to look at the characteristic equation. So, it will be 1 times m squared because the coefficient here is 1 and this is the second derivative, so m squared plus 5 m raised power 1, because it’s the first derivative, plus 6 m to the power 0 because it’s the 0-derivative equal to 0 is the characteristic equation. Which gives us m square plus 5 m plus 6 is equal to 0, and I can factor it as m plus 3 times m plus 2 equal to 0. And that gives me the two roots of this character expression. You can very well solve this problem by using the quadratic equation and you will get the same two roots of the of the equation m squared plus 5 m plus 6 equal.

So, keep in mind that this particular lesson is for those who have already taken the ordinary differential equation course. So, the homogeneous part will be k1 e to the power of m1 x plus k2 e to the power of m2 x because we have two roots of the character equation. So, we get k1 e to the power of minus 3x plus k2 e to the power -2x. So, that's what we get as the homogeneous part of the solution.

Let's go and see what happens to the particular part of the solution. Let's go and recall the differential equation which we had we had this as our differential equation. The homogeneous part of the solution, which we just found out, is of the form as follows. Let's look at what the particular part will be; the particular part will be the forcing function and all its possible derivatives. All possible derivatives e to the power A*x are just e to the power A*x. So, what's going to happen is that we're going to get a*e to the power -4x as our form of the solution and it is not in conflict with any part of the homogeneous part of the solution, so we are okay there. So, the question now arises that hey how do we find the value of a in our particular part of the solution. So, we get the second derivative of A*e to the power -4x plus 5*the first derivative of a to the power -4x plus 6*a to the power -4x will be equal to the forcing function 7e to the power -4x. That's how we're going to find the value of A. So, the second derivative of this is 16A*e to the power -4x the first derivative of this is -4A*e to the power -4x. So, having said that, if we simplify the left side, we get 16 minus 20 plus 6 gives us 2A*e to the power minus 4x equal to 7e to the power -4x and tells us that A is equal to 3.5.

So, the particular part of the solution will be just A*e to the power -4x which is when we found A just now will be 3.5e to the power -4x. So, we have found the particular part of the solution exactly, we have found the homogeneous part of solution, but we still have to figure out what k1 and k2 would be so that we'll do in the next slide.

We found out that the homogenous part of the solution is k1 e to the power -3x plus k2 e to the power -2x the particular part of the solution we just found out to be 3.5e to the power -4x so the complete solution is the addition of the homogenous part and the particular part. For that we will get this. So, the question arises how do we find k1 and k2? We'll find k1 and k2 by using our initial conditions. We know that y of 0 is equal to 13, that allows us 13 by substituting 0 in our complete solution. So, that gives us k1 plus k2 plus 3.5. And that simply tells us that k1 plus k2 is 9.5, and that's our first equation. By taking 3.5 to the left side, that's what we get as one of the equations. The second equation will be found by using the initial condition on the first derivative of the function.

So, let's go and see how that happens. So, the next condition which we're going to apply is that the initial condition is on the derivative of the at zero is given at 17. So, if that is the case then let's go and see whatever y is. Y in the expression which we have found out so we have this as our y, let's go and look at what dy/dx is dy/dx would be -3*k1*e to the power -3x minus 2*k2*e to the power -2x and then minus 14 e to the power -4x. This is just based on the formula that the derivative of e to the power a*x is a*e to the power a*x. So, that being said, let's go and apply the condition that at 0 it is 17. So, we get 17 equal to minus 3*k1 minus 2*k2 and minus 14 here and if we rewrite this equation by taking these two unknowns to the left-hand side I get 3*k1 plus 2*k2 equal to -17 minus 14 minus 31. That is the second equation.

So now what we have is two equations, two unknowns in k1 and k2 and we should be able to solve them for those two unknowns. So, let's go and see how we're going to do that. So, we have 3*k1 plus 2*k2 equal to -31. That was our second equation and the previous equation which we had obtained from the first initial condition was as follows. So, what I can do is I can multiply this by 3, so I get 3*k1 plus 3*k2 equal to 28.5. Let me call it equation 3. What I’ll do is equation 2 subtract equation 3 from there, so if I do that I will get -k2 is equal to -59.5 and that allows me to say that k2 is 59.5. Now, if I substitute k2 in equation 1, what do I get? K1 plus k2 is 9.5, and we just found out k2 is 59.5. That allows me to say that k1 is -50. So, since I have now the values of k1 and k2 the complete solution is available to me; y is as follows. And just substituting the values of k1 and k2 gives me:

And so, that is the complete solution to the problem which finds the values of k1 and k2 as well for the homogeneous part of the solution. And that's the end of the segment