Solve Differential Equation Using Variation of Parameters

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The differential equation y'' + 25y = 10sec(5t) was solved using the method of variation of parameters. The proposed particular solution was 2/5log(cos(5t))cos(5t) + 2tsin(5t), but it was initially marked incorrect in Webwork. After checking with Wolfram Alpha, the user realized the mistake was in the logarithmic term, which should include an absolute value. The solution was corrected to reflect this adjustment, resolving the issue. The discussion highlights the importance of careful attention to detail in solving differential equations.
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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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