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January 27th, 2010, 05:48 AM   #1
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please help with solving some differential equations

please explain how to solve these 3 differential equations




[color=#8040BF]thanx in advance [/color]
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January 27th, 2010, 06:13 AM   #2
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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?
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January 27th, 2010, 08:45 PM   #3
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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......
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January 28th, 2010, 05:21 AM   #4
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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
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January 28th, 2010, 05:29 AM   #5
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Re: please help with solving some differential equations

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?
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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 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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