anonymous
  • anonymous
y'-2y=t^2e^2t
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
primeralph
  • primeralph
What methods do you know?
primeralph
  • primeralph
To explain I need to know what you already know. I will leave this question in 2 minutes if you don't reply.
dumbcow
  • dumbcow
multiply equation by integrating factor....e^-2t \[e^{-2t}y' -2e^{-2t}y = t^{2}\] rewrite right side as derivative of product \[(e^{-2t} y)' = t^{2}\] integrate \[\int\limits (e^{-2t} y)' = \int\limits t^{2}\] \[e^{-2t} y = \frac{1}{3} t^{3} +C\] \[y = \frac{e^{2t}}{3}(t^{3} +C)\]

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anonymous
  • anonymous
thanks dumbcow, sorry for the delay primeralph
anonymous
  • anonymous
@dumbcow how did you go from step 1 to 2 in your method where you rewrite it as a product?
dumbcow
  • dumbcow
by recognizing that it was in the form of: \[fg' + f'g\] which is the product rule for derivatives...you can say \[(fg)'\] in general if the diff equ is in the form \[y' + p(t)y = q(t)\] then multiplying by \[\large e^{\int\limits p(t) dt}\] will always allow you to write right side as derivative of a product more in depth analysis can be found here: http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx
dumbcow
  • dumbcow
@FutureMathProfessor
anonymous
  • anonymous
Thanks Dumbcow! I don't know why you picked that name because you're sure not dumb!

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