Hello everyone. In this first part of the series, we will revisit the classical exact solution technique used to solve first-order differential equations. The nominal form for a first-order differential equation is given by this equation here. So, y as the dependent variable, t as the independent variable, K1 and K0 are some constant coefficients, and f(t) is a function of time. Our purpose is to find an exact solution for y(t) that satisfies this differential equation.

Another way we can write down this differential equation is by using this format: K1 y' + K0 y equals to f(t). So, this prime here is also equivalent to taking the first-order derivative. In general, the solution y has two main components: yH, which we call the homogenous solution, and yP. The total solution or the exact solution is the superposition between these two components. Let's visit each of these two components and see how to solve for them.

So, we start with the homogenous solution. The homogenous solution is the solution that makes the left-hand side of our differential equation equal to zero. With that differential equation equal to zero on the right-hand side, we can use simple calculus steps to generate a solution. So here, this differential equation can be written as K1 (dyH/dt) = - K0 * yH.

Now, let's separate variables and have everything that belongs to y on one side and everything that belongs to t on the other side, where t is our independent variable and y is our dependent variable. So, we can write this equation as follows: dyH/yH (so I divided both sides by this y) equals to - (K0/K1) dt.

Now, if I integrate both sides of this equation, on the left-hand side, I'm going to get the logarithm of yH, and on the right-hand side, to integrate this part, I'm going to get - (K0/K1)*t, and then we have this integration constant.

So, let's continue our solution. This is what we have for yH. We aim to solve for yH, so how about we take the exponential of this entire equation? If you take the exponential on the left-hand side, we're going to get yH. On the right-hand side, we're going to get e^(−K0/K1 t + C). Using the exponential property, we have an addition here. This can be written as e^(−K0/K1 t) * e^C, where C is our integration constant. Since C is a constant, e^C is also a constant, so this can be written as K e^(−K0/K1 t).

And this completes the formula that we have for the homogenous solution. So, the homogenous solution would be an exponential function whose exponent is proportional to the coefficient of our differential equation, and this constant here, we can solve for it using the initial conditions.

The second part of the solution of the differential equation is the particular solution. The particular solution is the part of the solution that gives f(t) when we substitute it into the differential equation. Usually, the particular solution has a similar anatomy to the function f(t).

So, this table here gives us the options that we might have for f(t). If f(t) is a constant, then the particular solution will also be a constant. If it is a polynomial, the particular solution will also be a polynomial. If it is exponential, it will be an exponential, and so on and so forth. You can see that the particular solution has coefficients b0, b1, b2 here, for example, or this constant A.

Our method to compute this is by substituting our proposed particular solution back into the differential equation. So, the differential equation has this form: K1 (dyP/dt) + K0 yP equals to f(t). We substitute this proposed yP into the differential equation and solve for those parameters (b0, b1, b2, or this big A) that validate this equation.

In the next video, we are going to see an example to reinforce the solution method for the first-order differential equation.

So, in conclusion, we have seen the nominal form of the first-order differential equation: the independent variable t, the dependent variable y. We have seen that the total solution is composed of two parts: the homogeneous solution, which is an exponential form whose exponent depends on the coefficient of our differential equation, and the forced solution would have one of some options depending on the shape of the right-hand side of our differential equation.

In the next video, we will follow up with a couple of examples to reinforce our method of solution. Thank you.