Hello. In the second video, we solve examples to reinforce our method to solve for the exact solution of a first-order differential equation. Specifically, we are going to learn how to get those constants associated with the homogeneous and particular solutions.

So, we start with this example. We have a first-order differential equation, and we have this initial condition. We can start by comparing that to the nominal form, which was:

K₁ dy/dt + K₀ y = F(t)

And we know that the complete solution y(t) has two components: a homogeneous part plus a particular part. In the previous video, we derived the solution for the homogeneous part yₕ(t) as K, a constant, multiplied by exp(-K₀/K₁ * t), where K₀ and K₁ are the coefficients of our differential equation.

By comparison, you can see that K₀ = 2 and K₁ = 3, and as a result, our homogeneous solution is equal to K * exp(-2/3 * t). Here, K is a constant that we need to solve for using the initial condition, as we're going to see later.

So now we have the proposed solution for the homogeneous part. For the particular part, the particular part takes the same form as F(t). For this problem, F(t) = exp(-t), so as a result, our proposed particular solution would be in this form:

yₚ(t) = A * exp(-t)

Now, we don't know this variable A—this constant A—and we need to solve for it. One way to solve for it is to plug in this proposed solution for yₚ(t) into the differential equation and solve for it.

So, our differential equation is:

3 * dyₚ/dt + 2 * yₚ = exp(-t)

We substitute our proposed solution for yₚ(t) into that equation and solve for A. So we have:

3 * (derivative of yₚ) = -A * exp(-t)
+ 2 * A * exp(-t)

And this has to equal the right-hand side, which is F(t) = exp(-t).

Now, you can see that we have exp(-t) on both sides, and the coefficient of exp(-t) on the left-hand side is equal to (-3A + 2), and on the right-hand side, the coefficient of exp(-t) is simply 1.

For this equation to hold, the coefficients of exp(-t) on both sides should be the same. So we have to set (-3A + 2) = 1 so that the differential equation holds using our proposed particular solution. From here, we can solve for A as -1 (A = -1).

So now we have solved for this constant associated with the particular solution, and our particular solution is equal to -exp(-t).

Now we have a proposal for the two parts: homogeneous and particular. So the complete solution y(t) is equal to the homogeneous part plus the particular part. You can see that we still have an unknown K, which is the constant associated with the homogeneous part. We solve for K using the initial condition.

The initial condition is y(0) = 5, so we plug t = 0 into our solution:

y(0) = K * exp(-0) - exp(-0)

And this has to equal 5. Since exp(-0) = 1, from here we have:

K - 1 = 5

And as a result, K = 6.

Now we have our solution for K using the initial condition. Now we can complete our solution:

y(t) = 6 * exp(-2/3 * t) - exp(-t)

And this completes the solution for this first-order differential equation.

Now, let's do another example. In this example, we have a first-order differential equation. Just to let you know, the independent variable can be any variable. In the previous example, the independent variable was t; here, it is x. No worries.

So now we are going to generate a solution y as a function of x, and you can see that the right-hand side is constant. Okay, comparing to the nominal form, we have:

K₁ * dy/dx + K₀ * y = F(x)

Where x is our independent variable. Okay, the solution y(x) would be of two components: a homogeneous component and a particular component.

The homogeneous component, as we derived in the previous video, would be a constant times an exponential function whose exponent is -K₀/K₁ times the independent variable, which in our case is x:

yₕ(x) = K * exp(-K₀/K₁ * x)

By comparison, you can see that K₀ = 2 and K₁ = 1, so the homogeneous solution is equal to:

K * exp(-2 * x)

Again, K is a constant, and we are going to solve for it using the initial conditions later, after we solve for the particular solution.

Now, we can propose the particular solution yₚ(x). yₚ(x) would be a scaled version of the function on the right-hand side of the differential equation, F(x), which was a constant 50. So, yₚ(x) would be simply B₀, a constant.

To solve for this B₀, we substitute our proposed solution into the differential equation. So the differential equation was:

dy/dx + 2 * y = 50

Now, we have a proposal for yₚ. The derivative of yₚ is equal to 0, plus 2 * yₚ (so yₚ = B₀), equals 50. So from here, we have:

B₀ = 25

And as a result, our particular solution yₚ(x) is equal to simply 25.

Now we have the two components: the particular solution and the homogeneous solution. We add them to get the exact solution. The exact solution y(x) is equal to the homogeneous part, K * exp(-2 * x), plus the particular part.

Okay, now we have this unknown K, and we solve it using the initial condition. So we have y(0) = 40. We plug x = 0 into that equation and equate it to 40.

Using our proposed complete solution:

y(0) = K * exp(-0) + 25

And this has to equal 40. Since exp(-0) = 1, from there you can see that:

K = 15 (K = 15)

And as a result, our complete solution is:

y(x) = 15 * exp(-2 * x) + 25

And this completes our solution. Thank you.