Integral of exponential over polynomial

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SUMMARY

The discussion focuses on solving the ordinary differential equation (ODE) y'' + x*y' - y = 0 using the method of reduction of order. The user assumes a solution of the form y = c1*y1 + c2*y2, where y1 = x. The integral v = ∫(x^-2 * exp(-0.5*x^2) dx) is derived for the second solution y2. Participants confirm that integration by parts leads to an expression involving the error function, which is essential for completing the solution.

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  • Understanding of ordinary differential equations (ODEs)
  • Familiarity with the method of reduction of order
  • Knowledge of integration techniques, specifically integration by parts
  • Basic understanding of the error function and its properties
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  • Study the method of reduction of order in ODEs
  • Learn about integration by parts and its applications
  • Explore the properties and applications of the error function
  • Practice solving variable coefficient ODEs with different methods
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Students and professionals in mathematics, particularly those studying differential equations, as well as educators seeking to enhance their understanding of integration techniques and error functions.

AAO
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Homework Statement


Solve the ODE: y''+x*y'-y=0

Homework Equations

The Attempt at a Solution


Since this is a variable coefficient ODE, I have used the method of reduction of order, and assumed the solution in the form: y=c1*y1+c2*y2

In this case: y1=x, and I have the reached the integral below for the second solution (y2), Can anyone tell me how to approach this integral:

v=integral[ (x^-2) * exp (-0.5*x^2) dx]
 
Last edited by a moderator:
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AAO said:

Homework Statement


Solve the ODE: y''+x*y'-y=0

Homework Equations

The Attempt at a Solution


Since this is a variable coefficient ODE, I have used the method of reduction of order, and assumed the solution in the form: y=c1*y1+c2*y2

In this case: y1=x, and I have the reached the integral below for the second solution (y2), Can anyone tell me how to approach this integral:

v=integral[ (x^-2) * exp (-0.5*x^2) dx]
Integration by parts leaves you with an integral that is (up to some constants) the error function.
 
Last edited by a moderator:

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