Hello. In this series, we will revisit the classical exact solution techniques used to solve second-order ordinary differential equations. The nominal form for a second-order differential equation has this form. You can see that the highest derivative here is the second-order derivative, and that's why we call this a second-order linear ODE. K₀, K₁, K₂ are the coefficients of this differential equation, and f(t) is the right-hand side. t is the independent variable, y is the dependent variable, and our mission is to find an exact solution or expression for y(t).

Similar to first-order ODEs, the solution has two parts: a homogeneous part and a particular part. Let's see how to solve for each of them.

For the homogeneous part, this is the part of the solution that makes the differential equation equal to zero. If we reflect on this equation, you see that you have a function yₕ, when you add it to a scaled version of its derivative and add to a scaled version of the second derivative, you're going to zero. So this suggests that y, the derivative of y, and the second derivative of y should have the same function anatomy such that when we add them together, we can get rid of them. This property happens in exponentials because derivatives of exponentials are also exponentials of the same exponent, and then we can add them together using some scaling factor to get a result of zero. So this suggests that the homogeneous solution will be in the form of exponential functions.

In order for us to generate that homogeneous solution, we construct what we call the characteristic polynomial. To construct the characteristic polynomial, we replace the derivative operator by some variable s. So the second derivative here will be replaced by s², the first derivative here would be replaced by s, and if no derivative, it would be s⁰, which is equal to one. Okay, this gives us what we have here, what we call the characteristic polynomial.

The roots of this characteristic polynomial determine the possible solution for the homogeneous solution. So we have the characteristic polynomial, which for a second-order differential equation is a second-order polynomial. A second-order polynomial has two roots that can be obtained using this nominal equation.

Now, since we have two roots, we have multiple possibilities on the shape of these two roots.

Option number one: We might have two unique real roots. If this is the case, then the homogeneous solution would be two exponentials: one whose exponent is the first root, the second is an exponential whose exponent is the second root. K₁ and K₂ are some constants that can be obtained using the initial conditions of the differential equation.

The second possibility of the roots is that we get two identical roots, meaning that α₁ and α₂ are both equal. The homogeneous solution will still be an exponential function whose exponent is α (because α₁ is equal to α₂, which is equal to some real number α), but now we have this extra t here.

The third and last possibility for our two roots is that they are both unique but complex conjugates. Okay, so the two roots are complex conjugates with the real part α and imaginary part β. In this situation, the homogeneous solution would be a sinusoidal function with an exponential envelope. The exponential envelope has an exponent α, which is the real part, and the sinusoids are oscillating at a frequency β, which is the imaginary part of our roots.

So, knowing the roots of the characteristic polynomial would guide us to one of three possibilities for the solution of the homogeneous part. Each possibility has constants K₁ and K₂, which can be obtained using the initial conditions, as we are going to see in the next video.

The second part of the solution is the particular solution. The particular solution is the solution that makes the differential equation give us the right-hand side, which is equal to f(t). Similar to the first-order ODE, we usually have the particular solution in a similar form to f(t). So we have the same table from last time. Given the nature of f(t), we can hypothesize a possible solution for the particular solution. You can see that the particular solution has some unknowns. We can use that proposed solution—substituted into the differential equation—in order for us to solve for those unknowns.

So, if the differential equation is given as K₂y'' + K₁y' + K₀y = f(t), all I need to do to solve for those unknowns in the proposed solution is to plug in that expression into this differential equation, and this will give me equations to solve for those unknowns.

We are going to solve an example in the next video to reinforce the methodology to compute those parameters.

So, in conclusion, we have seen the nominal form for second-order differential equations. The highest derivative in the differential equation is the second derivative, and the solution is also composed of two parts: homogeneous part and particular part. The homogeneous part is one of three options, depending on the roots of the characteristic polynomial.

Thank you.