Answer:
StTo find the general solution of the first-order linear differential equation:
(x + 8)y' + 2y = 2(x + 8)^2, x > -8
We can start by rearranging the equation in the standard form for a linear differential equation, which is:
y' + P(x)y = Q(x)
In this case, we have:
y' + (2/(x + 8))y = 2(x + 8)
Now, we can use an integrating factor to solve this equation. The integrating factor (IF) is given by:
IF = e^(∫P(x)dx)
In this case, P(x) = 2/(x + 8), so:
IF = e^(∫(2/(x + 8))dx)
Now, let's find the integral:
∫(2/(x + 8))dx = 2∫(1/(x + 8))dx
This is a straightforward integral, and it equals:
2ln|x + 8| + C1, where C1 is the constant of integration.
Now, we can find the integrating factor:
IF = e^(2ln|x + 8| + C1)
IF = e^(ln(x + 8)^2 + C1)
IF = (x + 8)^2 * e^(C1)
Now, we can rewrite our differential equation using the integrating factor:
(x + 8)^2 * e^(C1) * y' + 2(x + 8)^2 * e^(C1) * y = 2(x + 8)^2
We can simplify this equation by dividing both sides by the common factor:
y' + 2y = 2
Now, we have a much simpler first-order linear differential equation. The solution to this equation can be found using the method of integrating factors or separation of variables.
Let's use the integrating factor method. The integrating factor in this case is simply 1, so we don't need to multiply the entire equation by it. The equation becomes:
y' + 2y = 2
Now, we can solve for y:
y' + 2y = 2
First, we find the solution to the homogeneous equation (setting the right-hand side to 0):
y' + 2y = 0
This is a separable differential equation:
dy/dx + 2y = 0
dy/y = -2dx
Integrate both sides:
ln|y| = -2x + C2
Now, exponentiate both sides:
|y| = e^(C2) * e^(-2x)
Since e^(C2) is a positive constant, we can write:
|y| = Ce^(-2x)
Where C is a positive constant.
Now, we can consider the non-homogeneous part:
y' + 2y = 2
We will look for a particular solution of the form y_p = A, where A is a constant.
y_p' = 0
Substituting into the equation:
0 + 2A = 2
2A = 2
A = 1
So, the particular solution is y_p = 1.
Now, the general solution is the sum of the homogeneous and particular solutions:
y = Ce^(-2x) + 1
This is the general solution to the given first-order linear differential equation.ep-by-step
Step-by-step explanation: