Solving First Order Linear Differential Equation

In summary, the problem is to find the set of solutions of the given equation (E) for different intervals I, including (0,+∞), (-1,0), (-∞,-1), (-1,+∞), and ℝ. The equation is rearranged and an integrating factor is chosen to obtain a general solution in terms of a constant k. The solution works fine for intervals (0,+∞), (-1,0), and (-∞,-1), but there may be problems at points 0 and -1 for intervals (-1,+∞) and ℝ. Further clarification and help is needed for these intervals.
  • #1
the0
14
0

Homework Statement



Given (E): [itex](x+1)^{2}(xy'-y) = -(2x+1)[/itex]

Determine the set of applications from the interval [itex]I[/itex] to [itex]ℝ[/itex] which are solutions of (E) for:

a) [itex]I = (0,+∞)[/itex]
b) [itex]I = (-1,0)[/itex]
c) [itex]I = (-∞,-1)[/itex]
d) [itex]I = (-1,+∞)[/itex]
e) [itex]I = ℝ[/itex]


The attempt at a solution

I have rearranged (E) into the form:

[itex]y'-\frac{y}{x}[/itex] = [itex]\frac{-(2x+1)}{x(x+1)^{2}}[/itex]

and chosen integrating factor:

[itex]exp(∫\frac{-1}{x}dx) = \frac{1}{x}[/itex]

now multiplying by the integrating factor and integrating gives:

[itex]∫\frac{y'}{x}-\frac{y}{x^{2}}dx = ∫\frac{-(2x+1)}{x^{2}(x+1)^{2}}dx[/itex]

partial fractions now give:

[itex]∫\frac{y'}{x}-\frac{y}{x^{2}}dx = ∫\frac{-1}{x^{2}}+\frac{1}{(x+1)^{2}}dx[/itex]

After integrating and a tiny bit of algebra I get a general solution, for [itex]k\inℝ[/itex] as:

[itex]y=\frac{1}{x+1}+kx[/itex]

Now I think I am right in saying that this solution works fine for the intervals in a), b) and c)?

However for [itex]I = (-1,+∞)[/itex] there is (or potentially is) a problem with the point [itex]0[/itex] and for [itex]I = ℝ[/itex] there are (or potentially are) problems with the points [itex]0[/itex] and [itex]-1[/itex]?

Now I'm not sure how to proceed. Am I thinking along the right lines?

Any help/pointers would be very much appreciated, Thanks!
 
Physics news on Phys.org
  • #2
Can the general solution already posted simply extend to [itex]I = (-1,+∞)[/itex] ??
If so how can this be shown in detail?

Also I don't yet understand what happens when we are considering [itex]ℝ[/itex]

Thanks for any help!
 

1. What is a first order linear differential equation?

A first order linear differential equation is a mathematical equation that describes the relationship between a function and its derivative. It can be written in the form y' + p(x)y = g(x), where y' represents the derivative of the function y, p(x) is a function of x, and g(x) is a function of x.

2. How do you solve a first order linear differential equation?

To solve a first order linear differential equation, you can use the method of separation of variables or the method of integrating factors. In the method of separation of variables, you separate the variables on either side of the equation and then integrate both sides. In the method of integrating factors, you multiply both sides of the equation by an integrating factor to make it easier to solve.

3. What are the initial conditions in solving a first order linear differential equation?

The initial conditions refer to the values of the function and its derivative at a specific point. They are necessary to solve the differential equation and can be used to determine the particular solution.

4. What is the difference between a general solution and a particular solution?

A general solution is a solution that contains an arbitrary constant and can be used to represent all possible solutions to a differential equation. A particular solution is a specific solution that satisfies the initial conditions of the differential equation.

5. Why is solving first order linear differential equations important?

Solving first order linear differential equations is important because they are used to model various real-world phenomena in fields such as physics, engineering, and economics. They also provide a foundation for understanding and solving more complex differential equations.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
763
  • Calculus and Beyond Homework Help
Replies
21
Views
843
  • Calculus and Beyond Homework Help
Replies
19
Views
777
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
291
  • Calculus and Beyond Homework Help
Replies
5
Views
620
  • Calculus and Beyond Homework Help
Replies
2
Views
280
  • Calculus and Beyond Homework Help
Replies
6
Views
854
  • Calculus and Beyond Homework Help
Replies
10
Views
1K
Back
Top