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 March 22nd, 2011, 12:02 PM #1 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 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
 March 22nd, 2011, 12:23 PM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: 2nd Order Differential Equation $\sinh x= \frac12e^x-\frac12e^{-x}$ The method of undetermined coefficients then suggests something of the form $c_1x^2e^x+c_2e^{-x}$ 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, x².
 March 22nd, 2011, 12:44 PM #3 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: 2nd Order Differential Equation Thanks! I will run that through and post my answers for confirmation
 March 22nd, 2011, 01:36 PM #4 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 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
 March 22nd, 2011, 01:59 PM #5 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: 2nd Order Differential Equation Check your differentiations: $y=c_1x^2e^x+c_2e^{-x}\\ \quad\\ y'=c_1(2xe^x+x^2e^x)+c_2(-1)e^{-x}\\ =c_1e^x(x^2+2x)-c_2e^{-x}\\ \quad\\ y''=c_1[(e^x+xe^x)(2+x)+(xe^x)(1)]-c_2(-1)e^{-x}\\ =c_1e^x(x^2+4x+2)+c_2e^{-x}$ And yes.
 March 22nd, 2011, 02:08 PM #6 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: 2nd Order Differential Equation Oh, I integrated ! Thanks guys
 March 22nd, 2011, 03:03 PM #7 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 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 !
 March 22nd, 2011, 03:14 PM #8 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: 2nd Order Differential Equation You should get, after simplification: $2c_1e^x+4c_2e^{-x}=\frac{1}{2}e^x-\frac{1}{2}e^{-x}$ Thus, equating coefficients gives: $2c_1=\frac{1}{2}\:\therefore\:c_1=\frac{1}{4}$ $4c_2=-\frac{1}{2}\:\therefore\:c_2=-\frac{1}{8}$
 March 22nd, 2011, 04:07 PM #9 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 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
 March 22nd, 2011, 04:11 PM #10 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: 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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