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Mr.MathBest ResponseYou've already chosen the best response.2
This is a first order DE, and can be rewritten as: \[T'+0.85T=29.75\sin t+63.75\] In order to make this ODE separable, we multiply by the integrating factor [ http://en.wikipedia.org/wiki/Integrating_factor ] The integrating factor here is \(\large e^{\int0.85dt}=e^{0.85t}\). Multiply both sides by the integration factors, you get: \[e^{0.85t}T'+0.85e^{0.85t}T=e^{0.85t}(29.75\sin t+63.75).\] This is equivalent to: \[\frac{d}{dt} (e^{0.85t}T)=e^{0.85t}(29.75\sin t+63.75).\] Integrate both sides and you're done!
 one year ago
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