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March 5th, 2011, 02:58 AM  #11 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs  Re: Differential equations 
March 5th, 2011, 04:01 AM  #12  
Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0  Quote:
 
March 6th, 2011, 08:01 AM  #13 
Senior Member Joined: Feb 2011 Posts: 150 Thanks: 0  Re: Differential equations
I got to a question which I'm struggling with. The equation for current i: i R+L di/dt = E, Where R,L and E are constants, arises in circuit theory. Find the solution which satisfies i(0)=0. Will separation of variables work with this? 
March 6th, 2011, 12:16 PM  #14 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs  Re: Differential equations
No, this is not a separable equation, but what is known as a linear equation, and requires an integrating factor. First let's rewrite the equation in standard form (using I instead of i to avoid confusion with the imaginary unit): Now, on the left, we have the sum of a function and a constant times its derivative, so it is reasonable to assume the integrating factor will be of the form: and observing the constant in front of the function is we should try Multiplying the equation by we get: Now, we may rewrite the left side as: Now we may integrate: Dividing through by gives: Now, we are given thus: so our solution is: 
March 7th, 2011, 01:38 AM  #15 
Senior Member Joined: Feb 2011 Posts: 150 Thanks: 0  Re: Differential equations
Well, I wouldn't have come up with that solution even after a few attempts, but great work. Thanks a lot. 
March 10th, 2011, 05:00 AM  #16  
Senior Member Joined: Feb 2011 Posts: 150 Thanks: 0  Re: Differential equations Quote:
 
March 10th, 2011, 08:17 AM  #17 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs  Re: Differential equations
You can't solve explicitly for y, but you could for x. However, the implicit relationship is fine as a solution unless directed otherwise.

March 10th, 2011, 08:25 AM  #18 
Senior Member Joined: Feb 2011 Posts: 150 Thanks: 0  Re: Differential equations
Really ? What about a) y^2= 2e^x + c y= sqrt(2e^x +c a) y^2= 12cos(x) +c y = sqrt(12cos(x) +c 
March 10th, 2011, 08:56 AM  #19 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs  Re: Differential equations
When you have: then you get: y is not a function then, as there are two y's for a given x. 

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