Quick Question on Variation of Parameters Differential Equations

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SUMMARY

The discussion addresses the issue of handling a zero root in the characteristic equation of a differential equation when applying the method of variation of parameters. The specific differential equation in question is y" - y' = 4t, leading to the characteristic equation r² - r = 0, which yields roots r = -1 and r = 0. The challenge arises as one solution, y2, becomes zero, resulting in a Wronskian that is also zero. The alternative solution proposed is y2 = e^(0*t) = 1, which resolves the issue of the Wronskian being undefined.

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Homework Statement



What do you do if one of the roots to the characteristic equation of a differential equation is zero when using variation of parameters?

Homework Equations


The Attempt at a Solution



The problem I encountered this in is

y" - y' = 4t

Characteristic equation

r2 - r = 0

so

r = -1 , 0

Therefore

y1 = e-t

but

y2 = 0

which would make the Wronskian zero.

Any thoughts?
 
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Your other solution would be y2=e^(0*t) = 1
 

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