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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: 521 Math Focus: Calculus/ODEs Re: Differential equations $\frac{dx}{dt}=t(x-2)$ $\int\frac{1}{x-2}\,dx=\int t\,dt$ $\ln|x-2|=\frac{1}{2}t^2+C$ $x=Ce^{\frac{t^2}{2}}+2$ $5=C+2\:\therefore\:C=3$ $x=3e^{\frac{t^2}{2}}+2$
March 5th, 2011, 04:01 AM   #12
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 Originally Posted by jakeward Finding the general solution of the equation dx/dt=t(x-2) hence find the particular equation when x(0)=5
I hope this is just a warm up for IMO participants!

 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: 521 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): $\frac{dI}{dt}+\frac{R}{L}I=\frac{E}{L}$ 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: $\mu(t)=e^{kt}$ and observing the constant in front of the function is $\frac{R}{L}$ we should try $k=\frac{R}{L}$ Multiplying the equation by $e^{\frac{R}{L}t}$ we get: $e^{\frac{R}{L}t}\frac{dI}{dt}+Ie^{\frac{R}{L}t}\fr ac{R}{L}=\frac{E}{L}e^{\frac{R}{L}t}$ Now, we may rewrite the left side as: $\frac{d}{dt}$$e^{\frac{R}{L}t}I$$=\frac{E}{L}e^{\f rac{R}{L}t}$ Now we may integrate: $\int\,d$$e^{\frac{R}{L}t}I$$=\frac{E}{L}\int e^{\frac{R}{L}t}\,dt$ $e^{\frac{R}{L}t}I=\frac{E}{R}e^{\frac{R}{L}t}+C$ Dividing through by $e^{\frac{R}{L}t}$ gives: $I=\frac{E}{R}+Ce^{-\frac{R}{L}t}$ Now, we are given $I(0)=0$ thus:  so our solution is: $I=\frac{E}{R}-\frac{E}{R}e^{-\frac{R}{L}t}=\frac{E}{R}$$1-e^{-\frac{R}{L}t}$$$
 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
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Re: Differential equations

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 Originally Posted by MarkFL The problems are a bit ambiguous as typed. If they are: 1) $\frac{dy}{dx}=\frac{e^{-x}}{y}$ 2) $\frac{dy}{dx}=\frac{6\sin(x)}{y}$ Then these are what's known as first order separable differential equation. Separable means the variables may be separated to either side of the equation, and may be solved directly by integration. 1) may be separated as follows: $y\,dy=e^{-x}\,dx$ $\int y\,dy=\int e^{-x}\,dx$ $\frac{1}{2}y^2=-e^{-x}+C$ 2) may be separated as follows: $y\,dy=6\sin(x)\,dx$ $\int y\,dy=6\int\sin(x)\,dx$ $\frac{1}{2}y^2=-6\cos(x)+C$ You may check these solutions through implicit differentiation.
I forgot to ask, would I need to solve for y alone on these two solutions or would what Is there suffice? As the question only asks to solve the equation? Jake

 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: 521 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: 521 Math Focus: Calculus/ODEs Re: Differential equations When you have: $y^2=f(x)$ then you get: $y=\pm\sqrt{f(x)}$ y is not a function then, as there are two y's for a given x.

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