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March 22nd, 2011, 01:02 PM   #1
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2nd Order Differential Equation

Hey, I have this problem that I have come to a holt.

Find the general solution of the second order ordinary differential equation

d^2 y/dx^2 - 2 dy/dx + y = sinh x

My attempt so far:

Find complimentary solution:

d^2 y/dx^2 -2 dy/dx + y = 0

Aucillary equation = m^2 -2m + 1 = 0 ---> (m-1)(m-1) = 0 m= 1 twice

Complimentary solution: Yc = (A+Bx)e^x

Particular solution: Yp: pcos ? + psin ? I just thought of the definition of sinhx in terms of the exponential fucntion, would this come into play in terms of what goes in the way of the (?)'s ...

thanks
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March 22nd, 2011, 01:23 PM   #2
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Re: 2nd Order Differential Equation



The method of undetermined coefficients then suggests something of the form


Notice that the first exponential is multiplied by a power of x. That's because the exponent 1x has the coefficient 1, which is a root of the ancillary equation. And because that root has multiplicity 2, you must multiply by the 2nd power, x.
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March 22nd, 2011, 01:44 PM   #3
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Re: 2nd Order Differential Equation

Thanks! I will run that through and post my answers for confirmation
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March 22nd, 2011, 02:36 PM   #4
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Re: 2nd Order Differential Equation

Ok so I didnt get as far as the solution outcome ...im a bit new with this.

Yp= p1 x^2 e^x + p2 e^-x

Y'p= e^x (x^2 -2x+2)p1 - e^-x p2

Y''p= e^x (x^2 - 4x + 6)p1 + e^-x p2

Then do I substitute into the original equation ...?

Jake
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March 22nd, 2011, 02:59 PM   #5
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Re: 2nd Order Differential Equation

Check your differentiations:



And yes.
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March 22nd, 2011, 03:08 PM   #6
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Re: 2nd Order Differential Equation

Oh, I integrated ! Thanks guys
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March 22nd, 2011, 04:03 PM   #7
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Re: 2nd Order Differential Equation

so subbing into original equation (bare with me) got to get my heas around this.

C1 e^x (x^2 +4x+2) + C2 e^-x -2[C1 e^x (x^2 +2x) - C2 e^-x ] + C1 x^2 e^x +C2 e^-x = 1/2 e^x - 1/2 e^-x

Compare co efficients of e^x and e^-x

e^x : 0c1 =1/2 --> c1 = 1/2

e^-x : 4c2 =-1/2 --> c2 = -1/8 My best attempt !
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March 22nd, 2011, 04:14 PM   #8
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Re: 2nd Order Differential Equation

You should get, after simplification:



Thus, equating coefficients gives:



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March 22nd, 2011, 05:07 PM   #9
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Re: 2nd Order Differential Equation

I hate simplification!

So, the terminolgy i have been taught is obviously different to what you use but anyhow,

yc= (A+Bx)e^x

yp= 1/4 e^x - 1/8 e^-x

General solution:

y= yc + yp

y= (A+Bx)e^x + 1/4 e^x - 1/8 e^-x
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March 22nd, 2011, 05:11 PM   #10
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Re: 2nd Order Differential Equation

You've almost got it...it should be:

y(x) = (A + Bx)e^x + (1/4)(x^2)e^x - (1/e^-x
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