In this lecture, we summarized a method for solving second-order differential equations with constant driving functions and show how to use initial conditions to solve for the natural response of the system. We also explained the situations that lead to over damped, critically, damped and under damped solutions. Well, here's a second-order differential equation with a constant driving function a. And the coefficients of the equation are the lower case variables, a B and C, which multiply in the second. Derivative of the function we're, trying to find the first derivative of the function and the function itself, the general solution to this equation that is for the unknown function. Ax is made up of two components.
One component is called the forced response. And it represents that part of the solution that's caused by the constant driving function. A other component is called the natural response, and it's caused by the differential equation. And the particular values of the equation coefficients. For a differential equation like this with a constant driving function.
The forced response is the ratio of the driving function a to the coefficient C that multiplies X of T in the differential equation. It is useful incorrect to think of the forced response as the final value that the function X of T will approach. If we let time become very large, because of this, we sometimes refer to the forced response as the steady-state value of the function. And again, the steady-state value of the function is. The forcing function, a divided by the coefficient that multiplies X of T. Now, the natural response for the equation is the response when there is no driving function.
That is when the driving function. A is a zero now, whereas the forced response gives us the final value for the solution. The natural response gives us the transient values.
In other words, the natural response tells us how the solution moves from its initial conditions to its final value. The key to determining the natural response is to. Find the roots for something called the characteristic equation. The characteristic equation is a second-order polynomial in a variable S and its coefficients are the same as those for the original differential equation. The coefficient for s squared is the coefficient for the second derivative term.
The coefficient for s is the coefficient for the first derivative term and the coefficient for s raised to the 0 or for the constant term, that's, the coefficient for the zeroth derivative term, which is. Simply the function ax, the characteristic equation will in general have two roots. And those roots can be obtained through the quadratic equation. Now in general, there are three possible situations that can occur for these routes.
And these situations are determined by the relationship between B squared and 4ac, which make up the terms inside the square root in those three situations occur when B squared is greater than 4ac when B squared is equal to 4ac. And when B squared is less than 4ac. So if B. Squared is greater than 4ac. Then we'll be taking the square root of a positive number. And the roots for the characteristic equation will both be really valued as I've shown here. Now, for this situation, the natural response will be the sum of two exponential terms whose coefficients s1 and s2 are the roots of the characteristic equation.
And the total response will include both the natural response and the forced response. Now, the roots s1 and s2 along with the forced response, which is the ratio of a. Over C are all determined by the original differential equation, but the coefficients the constants k1 and k2 that multiply each of these exponential terms must be determined by, including two initial conditions for the system. Well, first we'll need to specify the value for the function X of T at some particular time.
Now, the most common time to specify, the function is T equals zero for which the function is then the sum of k1, plus k2, plus an over C, because the exponential terms get, the s-1 T and e. To the S to T when T is equal to zero will both be equal to one now, a and C are known that k1 and k2 are not this means that by specifying the initial condition for X of T we'll have one equation, but two unknowns so to determine k1 and k2 we'll need another initial condition. The other condition is typically specified through the value for the derivative of X of T at the initial time. Now, the derivative for X of T will be s, 1, K 1, e to the s 1, T, plus s, 2, K 2, e to the s 2, T, plus 0. So if we let T be.0 then the exponential terms will both be 1, and we'll be left with an expression for the derivative at T equals 0 is s, 1, K 1, plus s, 2, K 2. Now with these two initial conditions, then we have two equations and two unknowns, and we can solve for K 1 and K 2. And once we know, K 1 and K 2, we'll have a general expression for the solution to the differential equation. Well, it's possible.
The B squared will be equal to 4ac. And when this happens, both of the roots s1 and s2 be equal to some value. S, which is. Equal to negative B over 2a. Now, in this case, the natural response will be of the form K 1. E to the S T, plus K 2, T, e to the St, the two exponential coefficients S and s are the same equal to negative B over 2a.
But one of the terms is also multiplied by the time T now, it's important to remember that this factor of T must be included in one of the terms for the solution. Now, again, the exponential coefficient as the ratio a to C, which is the forced response are all determined by the original. Differential equation, but the constants k1 and k2 must be determined by the initial conditions. Well, if we specify X of T at some time and let's use T equals zero as one initial condition, then X of 0 will be k1. E to the 0 that'll, be k1, k2 times 0 and then an over C. So X of T, when T is equal to 0 is k1, plus an over C. And because a and C are known this initial condition is enough to determine k1, but we'll need another initial condition.
If we want to determine k2 well-to-do that. We typically again, Specify the initial condition for the derivative. Now, if we evaluate this derivative will have s times k1, e to the S T, plus k2, e to the S T, plus s times k2, T, e to the S T, plus 0. And if we evaluate this when T is equal to 0, then this condition provides an equation equal to s, k1, plus k2. And that since we'll know, K 1 from this equation allows us to solve for K 2. So again, the initial conditions. The two initial conditions for X of T at T equals 0 and the derivative of X of T at T equals 0 provide two.
Equations in the two unknowns, K 1 and K 2 where again a and C and the root s are all determined by the original specification of the differential equation. Well, finally, we could have a situation where B squared is less than 4ac. And in this case, the quantity inside the square root would be negative, and that will result in complex roots. So here J is the square root of -1, the complex number and inside the square root. Now we've put 4ac minus B squared, because we've pulled out the negative 1.
The. Square root of negative 1 as J. Now, if we define the variable Sigma equal to B over 2a and Omega as the square root of 4ac, minus b squared over hue, then the two roots will be negative, Sigma, plus J, Omega and negative Sigma, minus J Omega, the form for the natural response in this situation is e to the negative Sigma T times, k1, cosine, Omega, T, plus k2, sine, Omega, T and again, Sigma and Omega are determined by the roots of the characteristic equation. And then, of course, we also have then the forced. Response to make up the total response again, Sigma and Omega and a and C are all determined by the original differential equation. So we again need to determine the constants k1 and k2. And once again, we use initial conditions to do that.
If we look at X of T, when T equals 0 we'll have e to the 0 that's 1, k1 times, the cosine of 0 that will be k1 and k2 times the sine of 0 that would be 0 and then plus an over C. So the initial condition for X of T. When T is equal to 0 gives an expression equal to k1. Plus an over C accordingly, we get an equation for k1. Now, if we use the initial condition for the derivative, we get another equation that allows us to solve for K 2.
So what we do is look at the derivative for X of T, and then evaluate that when T equals 0, and the only terms that will be remaining will be negative Sigma times, k1, plus Omega times k2. And as before we all have two equations through our initial conditions that give us two equations for the two unknowns, K 1 and K 2. So in summary here are. The three situations we can have B squared greater than 4ac. We can have the coefficient b squared equal to 4a c and b squared less than 4ac.
Now when b squared is greater than 4ac, we refer to the solution, the natural response that tells us how we go from the initial conditions to our final value for X of T as an over damped system. When b squared is equal to 4ac. We call that critically damped. And when b squared is a less than 4ac, we call that under damped. And these expressions of damping over. Damped, critically, damped, an under damped refer to how the solution moves from those initial conditions to the final value. And the best way to illustrate that is by now, using the methods that we've just developed for solving second-order differential equations and apply them to circuits that contain resistors sources, inductors and capacitors.
So that the solution, the expressions for unknown, voltages and currents involve second-order differential equations will apply these methods solve for those. And take a look at some changes in voltages or currents and see if we can see why we would call them over damped, critically, damped and under damped.