Quadratic Solution to Homogeneous Second-Order ODEs

In summary, the conversation discusses finding a solution for the differential equation ax''+bx'+cx=0, with the condition that b^2-4ac=0 and using the substitution k=-b/(2a). The solution x=e^(kt) is suggested and the derivatives are plugged into the equation, resulting in e^((-b/2a)t)(-((b^2)/2)+c)=0. After factoring and using the given conditions, the final answer is determined to be 0.
  • #1
epsilonzero
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Homework Statement


Consider ax''+bx'+cx=0 for b^2-4ac=0 with k= -b/(2a).

Show x=e^(kt) is a solution.

The Attempt at a Solution



x=e^(kt)
x'=(-b/2a)e^((-b/2a)t)
x''=(-b^2/4a^2)e^((-b/2a)t)

Then i plugged these into ax''+bx'+cx=0 and simplified to get e^((-b/2a)t)(-((b^2)/2)+c)=0

I'm stuck at what to do next.

Thank you for any help.
 
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  • #2
To make things easier, don't use the k=-b/2a substitution until you're ready to compute the final answer. This helps to keep it a lot more tidier. Just subsitute the derivatives of x into the DE, factor out e^kt and make use of k = -b/2a as well as b^2 - 4ac = 0. It'll all cancel out to give 0.
 
  • #3
Ok, just got it. Thanks.
 
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FAQ: Quadratic Solution to Homogeneous Second-Order ODEs

What is a quadratic solution to an ODE?

A quadratic solution to an ODE (ordinary differential equation) is a solution that can be written in the form of a quadratic equation, where the dependent variable and its derivatives are raised to the first and second powers, respectively.

How do you solve a quadratic ODE?

To solve a quadratic ODE, you can use various methods such as separation of variables, substitution, or the quadratic formula. The method used will depend on the specific equation and its initial conditions.

What are the advantages of using a quadratic solution to an ODE?

Quadratic solutions to ODEs have several advantages. They are often easier to manipulate and work with compared to other types of solutions. Additionally, they can provide a more accurate representation of the behavior of a system or process being modeled.

Can a quadratic solution to an ODE have complex roots?

Yes, a quadratic solution to an ODE can have complex roots. This means that the solution will involve complex numbers, which are numbers with both real and imaginary components. In some cases, complex roots can provide more information about the behavior of a system.

Are there any limitations to using a quadratic solution to an ODE?

While quadratic solutions to ODEs have many benefits, they may not always be the most appropriate or accurate solution for a given problem. In some cases, a higher order solution may be needed to accurately model the behavior of a system. Additionally, quadratic solutions may not be suitable for non-linear ODEs.

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