Fundamental Solutions of Linear homogeneous equations

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SUMMARY

The discussion centers on the differential equation y'' + p(t)y' + q(t)y = 0 and the proposed solution y = sin(t^2). It is established that when evaluating the equation at t = 0, the left-hand side simplifies to 2, indicating that y = sin(t^2) cannot satisfy the equation on any interval containing t = 0. The conclusion is definitive: y = sin(t^2) is not a valid solution due to the non-zero result of the equation at that point.

PREREQUISITES
  • Understanding of second-order linear differential equations
  • Familiarity with continuous functions and their derivatives
  • Knowledge of the sine function and its properties
  • Basic calculus, specifically differentiation techniques
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  • Study the theory of linear homogeneous differential equations with variable coefficients
  • Learn about the Wronskian and its role in determining the linear independence of solutions
  • Explore the method of undetermined coefficients for solving differential equations
  • Investigate the implications of initial conditions on the existence of solutions
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Students and professionals in mathematics, particularly those focusing on differential equations, as well as educators seeking to clarify concepts related to linear homogeneous equations.

hsong9
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Homework Statement


Can y = sin(t^2) be a solution on an interval containing t = 0 of an equation y'' + p(t)y' + q(t)y = 0 with continuous coefficients?


Homework Equations





The Attempt at a Solution


y = sin(t^2)
y' = 2tcos(t^2)
y'' = 2cos(t^2) - 4t^2sin(t^2)

2cos(t^2) - 4t^2sin(t^2) + p(t)(2tcos(t^2)) + q(t)sin(t^2) = 0

when t=0, above equation is 2. That is, there does not exist the solution. so y can not be a solution on I containing t=0.
 
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hsong9 said:

Homework Statement


Can y = sin(t^2) be a solution on an interval containing t = 0 of an equation y'' + p(t)y' + q(t)y = 0 with continuous coefficients?


Homework Equations





The Attempt at a Solution


y = sin(t^2)
y' = 2tcos(t^2)
y'' = 2cos(t^2) - 4t^2sin(t^2)

2cos(t^2) - 4t^2sin(t^2) + p(t)(2tcos(t^2)) + q(t)sin(t^2) = 0

when t=0, above equation is 2. That is, there does not exist the solution. so y can not be a solution on I containing t=0.

Your reasoning is good, but let me help you with your conclusion. An equation is not a number, so an equation can't be 2.

When t = 0, the value of the expression on the left side of the equation is 2, so y = sin(t^2) cannot be a solution of the differential equation on any interval I containing t = 0.
 
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Thanks!
 

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