Finding the Homogenous Solution to a Variable Coefficients 2nd order ODE

In summary, the conversation discusses finding the homogenous solution for the equation x y'' + (x + 1) y' = 2 x, with the coefficients being a function of x. The person is unsure of where to start and asks for help in finding a solution. They mention using the variation of parameters method but are unsure of how to proceed. They also ask about using ansatz and how to choose one.
  • #1
joelio36
22
1
x y'' + (x + 1) y' = 2 x

Solve for y(x).

Due to the coefficients being a function of x, I have no idea where to start to find the homogenous solution (Complementary Function). I know how to proceed after this part with the variation of parameters method.

I just have no idea where to begin to find a solution of this equation, which seems to be the pre-requisite for all solution methods! Do you just used ansatz? If so how would I pick an ansatz?

I'm tearing my hair out here, any help would be greatly appreciated.
 
Physics news on Phys.org
  • #2
The homogenous equation x y'' + (x + 1) y' = 0 can be solved by separation of variables.
 
  • #3
y' = z
then separation of variables in a first order ODE.
 

1. What is a homogenous solution to a variable coefficients 2nd order ODE?

A homogenous solution to a variable coefficients 2nd order ODE is a solution that satisfies the differential equation when the right-hand side of the equation is equal to zero. In other words, it is a solution to the differential equation without any external forcing or input.

2. Why is finding a homogenous solution important?

Finding a homogenous solution is important because it allows us to understand the behavior of the system without any external influences. This can help us make predictions and analyze the system’s stability and response to different initial conditions.

3. How do you find the homogenous solution to a 2nd order ODE with variable coefficients?

To find the homogenous solution, we first need to solve the characteristic equation of the differential equation. This involves finding the roots of the characteristic polynomial, which can be done using the quadratic formula. Once we have the roots, we can use them to form the general solution by plugging them into the exponential function.

4. What is the role of initial conditions in finding the homogenous solution?

The initial conditions are used to determine the specific values of the constants in the general solution. These constants are then combined with the homogenous solution to form the particular solution, which satisfies both the differential equation and the initial conditions.

5. Can a homogenous solution also be a particular solution?

Yes, in some cases, the homogenous solution can also satisfy the initial conditions and therefore be the particular solution. This occurs when the initial conditions are such that the constants in the general solution are equal to zero, making the homogenous solution the only solution to the differential equation.

Similar threads

  • Calculus and Beyond Homework Help
Replies
7
Views
471
  • Calculus and Beyond Homework Help
Replies
1
Views
228
  • Calculus and Beyond Homework Help
Replies
3
Views
283
  • Calculus and Beyond Homework Help
Replies
16
Views
540
  • Calculus and Beyond Homework Help
Replies
3
Views
803
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
682
  • Calculus and Beyond Homework Help
Replies
3
Views
542
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
Back
Top