Solving a 2nd Order Linear ODE: Any Clever Tricks?

In summary, the conversation discusses solving the differential equation $$ay''-(2x+1)y'+2y=0$$ with initial conditions ##y(0)=1## and ##y(1)=0##, where ##a## is a non-zero constant. The suggested methods are using a change of variable and a power series solution.
  • #1
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Homework Statement


$$ay''-(2x+1)y'+2y=0$$ subject to ##y(0)=1## and ##y(1)=0## where ##a## is a non-zero constant.

Homework Equations


Not too sure

The Attempt at a Solution


I know an analytic solution exists since I solved with mathematica. My thoughts were to try a series expansion, but since the analytic solution is in closed form and is using the imaginary error function, I'd rather not waste a lot of time with a power series guess if someone knows (or sees) something insightful.

Any ideas or clever tricks?

Thanks!
 
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  • #2
I would use two methods. First, let u= 2x+ 1. Then du/dx= 2 so [itex]dy/dx= (dy/du)(du/dx)= 2(dy/du)[/itex] and [itex]d^2y/dx^2= (d/dx)(dy/dx)= d/dx(2dy/du)= 4 d^2y/du^2[/itex]. So the equation becomes [itex]4\alpha d^2y/du^2- 2udy/d+ 2y= 0[/itex].

Now look for a power series solution. Let [itex]y= \sum_{n=0}^\infty a_nu^n[/itex].
Content abridged by a mentor.
 
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Related to Solving a 2nd Order Linear ODE: Any Clever Tricks?

1. What is a 2nd order linear ODE?

A 2nd order linear ODE (ordinary differential equation) is a type of mathematical equation that describes the relationship between an unknown function and its derivatives. It is considered "linear" because the unknown function and its derivatives appear in a linear fashion, meaning they are not multiplied or divided by each other.

2. How do you solve a 2nd order linear ODE?

To solve a 2nd order linear ODE, you can use various methods such as the method of undetermined coefficients, variation of parameters, or the Laplace transform. These methods involve manipulating the equation to isolate the unknown function and then using initial conditions or boundary conditions to find a specific solution.

3. What are initial conditions and boundary conditions in a 2nd order linear ODE?

Initial conditions are given values of the unknown function and its derivatives at a specific point. Boundary conditions, on the other hand, are given values of the unknown function and its derivatives at the endpoints of a given interval. These conditions are necessary to find a specific solution to the 2nd order linear ODE.

4. Can a 2nd order linear ODE have multiple solutions?

Yes, a 2nd order linear ODE can have multiple solutions. This is because the general solution to the equation includes two arbitrary constants, which can take on different values depending on the initial or boundary conditions given. However, the existence and uniqueness theorem states that under certain conditions, a 2nd order linear ODE will have a unique solution.

5. What are some real-world applications of 2nd order linear ODEs?

2nd order linear ODEs have many applications in fields such as physics, engineering, and economics. They can be used to describe the motion of a spring-mass system, the behavior of electrical circuits, or the growth and decay of populations. They are also commonly used in modeling and predicting natural phenomena such as weather patterns or chemical reactions.

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