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Fundamental Solutions of Linear homogeneous equations

  • Thread starter 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 eqution 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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  • #2
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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 eqution 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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  • #3
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Thanks!
 

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