Differential Equations Ordinary and Partial Differential Equations Math Forum

 January 27th, 2010, 05:48 AM #1 Newbie   Joined: Jan 2010 Posts: 3 Thanks: 0 please help with solving some differential equations please explain how to solve these 3 differential equations [color=#8040BF]thanx in advance [/color]
 January 27th, 2010, 06:13 AM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: please help with solving some differential equations The first one is an ordinary second-order non-homogeneous linear differential equation, and the procedure is fairly straightforward, but long and hard to explain. The second one needs a complicated change in variables. For the third one just add x to both sides, take the square root of both sides, then integrate twice. Where did these problems come from and what do you already know about differential equations?
 January 27th, 2010, 08:45 PM #3 Newbie   Joined: Jan 2010 Posts: 3 Thanks: 0 Re: please help with solving some differential equations Thanks for quick reply! I just joined a degree on math and they are teaching differential equations now. The professor gave these problems and told us to try them. I tried but getting a different answer. I will tell briefly how I solved it. Try to tell where I went wrong.... I assumed dy/dx=p calculated the integrating factor = (e^x) then used the formula , y e^x = int[Q*IF+c] to get p(e^x)=int[-1(e^x)dx+c1] after integration we get, e^x(p+1)=c1 using p=dy/dx and simplifying we get, dy=[(c1/e^x)-1]dx integrating both sides and simplifying we get, y=-c1(e^-x)-x +c2 but the answer given is y= ln sin(x-c1)+c2 please try to find out where I went wrong. I did not get ln and sin in the answer......
 January 28th, 2010, 05:21 AM #4 Newbie   Joined: Jan 2010 Posts: 3 Thanks: 0 Re: please help with solving some differential equations For the 2nd problem, I followed similar steps. I divided the whole equation by (1-x^2) and assumed dy/dx=p calculated the IF to be (1-x^2)^0.5 then I integrated twice to get y=(asin x)^2 +c1asin x +c2 but the answer given is (x^2+y^4-1)e^{x^2}=c thanks
January 28th, 2010, 05:29 AM   #5
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Quote:
 y=-c1(e^(-x))-x +c2
Your answer is right and their answer is wrong, as you can easily tell by differentiating it twice and substituting the results into the original equation.

"y=ln sin(x-c1)+c2" is a solution of y ' ' + (y ' )² + 1 = 0. Did you make a mistake in copying it down?

January 28th, 2010, 12:54 PM   #6
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Quote:
 Originally Posted by shakgoku For the 2nd problem, . . .
You were correct for |x| < 1. For |x| > 1, there are similar solutions involving $^{\cosh^{-1}(|x|).}$ There are "combined" solutions that are continuous for all real values of x, but they are not differentiable at x = ±1.

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