Solve Differential Equation Using Variation of Parameters

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SUMMARY

The discussion centers on solving the differential equation y'' + 25y = 10sec(5t) using the method of variation of parameters. The user initially proposed a particular solution, yp = (2/5)log(cos(5t))cos(5t) + 2tsin(5t), which was flagged as incorrect by the Webwork platform. Upon further investigation, the user discovered that the logarithmic term should include an absolute value, correcting the solution to align with the integral of the tangent function.

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Homework Statement



Solve y''+25y=10sec(5t)

Homework Equations



NA

The Attempt at a Solution



I believe I have the correct answer for yp which is:

2/5log(cos(5t))cos(5t)+2tsin(5t)

When I plug this into the Webwork field, it says it is incorrect. I checked my answer against Wolfram Alpha and they look the same.

I was hoping someone could look at it and identify if it was actually correct or not. If not, I can go through the steps here and try to work it out.

Thanks
 
Last edited:
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I figured it out. The value within the log should be an absolute value as it came from the integral of the tangent.
 

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