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| Author | Message / Information |
| herbert Quote | Reply | | differential equation posted on: 11/9/2006 12:25:45 AM It is required to solve y´´ + k*y = y*cos(x) ; y=y(x) most efficiently. The two accents denoting twofold differentiation in x. Perhaps somebody of the community might be of valuable assistance in this special case. regards herbert |
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Grogler
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differential equation
replied on: 11/9/2006 12:47:56 AM Maybe, y''= y*cos(x) - k*y y''/y = cos(x) - k (dy/dx)*(y'/y)= cos(x) - k (y'/y)*dy = (cos(x) - k)*dx ln|y| = sin(x) - k*x + C1 |y| = e^(sin(x) - k*x + C1) y = C*e^(sin(x)-k*x) Although, I'm not sure about it. |
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herbert
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differential equation
replied on: 11/9/2006 3:29:38 AM The fault lies within the assumption that putting y´´ => (y´  ^2/y which is not legitimate in the case at hand.regards |
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Euler
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differential equation
replied on: 11/11/2006 2:27:35 AM Well, you could start off by looking at the characteristic equation of the homogenous solution: r^2+k=0, which has roots, r=(+/-i)*sqrt(k) so the homogenous solution would be: yc=c1*cos(kx)+c2sin(kx), as for the particular solution, try to think of a "guess" equation which could match the secong side. |
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herbert
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differential equation
replied on: 11/11/2006 1:13:11 PM Thank you very much for your suggestion. Thus I have to look again more deeply into special books to make my guessing more effective. |
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Euler
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differential equation
replied on: 11/12/2006 12:18:32 AM It is possible that you can use: yp=Aycos(x)+Bysin(x), but I haven't tried this, so I am not sure it would work. |
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HallsofIvy
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differential equation
replied on: 4/6/2007 1:31:56 PM Euler, that would work IF this were y"+ ky= cos(x). Unfortunately, here we have ycos(x) which makes the problem much harder! |
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