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 December 5th, 2010, 02:22 PM #1 Member   Joined: Oct 2010 Posts: 30 Thanks: 0 first order differential equations Could someone help me with these questions, please? 1)Solve the following separable equation and verify your solution dy/dt+(e^y)=(e^y)sint
 December 5th, 2010, 02:28 PM #2 Member   Joined: Jul 2009 From: New Jersey Posts: 65 Thanks: 0 Re: first order differential equations How would you separate this equation?
 December 5th, 2010, 02:33 PM #3 Member   Joined: Oct 2010 Posts: 30 Thanks: 0 Re: first order differential equations I have been trying to do it but can't figure out how; would you be able to work through it for me?
 December 5th, 2010, 03:24 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: first order differential equations You have: $\frac{dy}{dt}+e^y=e^y\sin t$ Subtract through by $e^y$ $\frac{dy}{dt}=e^y\sin t-e^y$ Factor right side: $\frac{dy}{dt}=e^y\left(\sin t-1\right)$ Multiply through by $e^{-y}dt$ $e^{-y}dy=\left(\sin t-1)dt$ Integrate: $\int e^{-y}\,dy=\int \sin t-1\,dt$ $-e^{-y}=-\cos t-t+C$ $e^{-y}=\cos t+t+C$ $y=\ln\left(\frac{1}{\cos t+t+C}\right)$ To verify the answer, we can differentiate y with respect to t: $\frac{dy}{dt}=\frac{\sin t-1}{\left(\cos t+t+C\right)}=e^{y}(\sin t-1)$ $\frac{dy}{dt}+e^y=e^y\sin t$
 December 6th, 2010, 08:20 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,618 Thanks: 2072 y = -ln(cos(t) + t + C), where t exceeds the root of cos(t) + t + C = 0, so that the logarithm is real.

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