Two Differential Equation Problems

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SUMMARY

The discussion focuses on solving a second-order linear differential equation represented as \(y^{''}+a_1y^{'}+a_2y=0\) or in operator form as \((D^2+a_1D+a_2)y=0\). The method of exponential shifting is employed, specifically using the transformation \(e^{a_1x/2}\) to simplify the equation. The final form of the equation is expressed as \(f(D-a_1/2)=(D-a_1/2)^2+a_1(D-a_1/2)+a_2\), indicating the completion of the solution process.

PREREQUISITES
  • Understanding of second-order linear differential equations
  • Familiarity with differential operators and notation (e.g., \(D\) for differentiation)
  • Knowledge of exponential functions and their properties in differential equations
  • Basic skills in manipulating algebraic expressions involving operators
NEXT STEPS
  • Study the method of exponential shifting in differential equations
  • Learn about the characteristic equation for second-order linear differential equations
  • Explore the application of the Laplace transform in solving differential equations
  • Investigate the stability of solutions to differential equations
USEFUL FOR

Students and professionals in mathematics, particularly those specializing in differential equations, as well as engineers and physicists who apply these concepts in their work.

JM00404
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Please see the PDF attatchment to view the problems. Thank you for your time.
 

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What diff.eq does [tex]\psi(x)e^{-a_{1}\frac{x}{n}}[/tex] fulfill?
 
JM00404 said:
Please see the PDF attatchment to view the problems. Thank you for your time.

The first one (too late I know but anyway):

Let's do a simple one first:

[tex]y^{''}+a_1y^{'}+a_2y=0[/tex]

or:

[tex](D^2+a_1D+a_2)y=0[/tex]

or:

[tex]f(D)y=0[/tex]

Now, use the exponential shift:

[tex]e^{cx}f(D)y=f(D-c)[e^{cx}y][/tex]

so up there, multiply by:

[tex]e^{a_1x/2}[/tex]

so:

[tex]e^{a_1x/2}f(D)y=f(D-a_1/2)[e^{a_1x/2}\phi]=0[/tex]

can you finish it?

[tex]f(D-a_1/2)=(D-a_1/2)^2+a_1(D-a_1/2)+a_2[/tex]
 

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