Variation of Parameters - 142 Part II: Surveying Second and...

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you know that: ln ln ycxcx x xxxx 2 12 22 =+ - - - ^^ hh In fact, you can absorb the final – x term into c 1 x , giving you the general solution: ln ln x xxx 2 - - What a pair! The variation of parameters method meets the Wronskian As noted in the previous sections, the method of variation of parameters allows you to tackle linear differential equations, such as this second order differential equation: y" + p ( x ) y' + q ( x ) y = g ( x ) The method of variation of parameters relies on the solution to the homoge- neous equation: y" + p ( x ) y' + q ( x ) y = 0 The solution to the homogeneous equation is: y h = c 1 y 1 ( x ) + c 2 y 2 ( x ) The method of variation of parameters says that you then try to find a partic- ular solution of the following form: y p = u 1 ( x ) y 1 ( x ) + u 2 ( x ) y 2 ( x ) Substituting y p into the differential equation gives you these two equations: u' 1 ( x ) y 1 ( x ) + u' 2 ( x ) y 2 ( x ) = 0 u' 1 ( x ) y 1 ' ( x ) + u' 2 ( x ) y 2 ' ( x ) = g ( x ) You can formally solve these equations for u' 1 ( x ) and u' 2 ( x ) as follows: ux yxy x y xyx yxgx 1 1 2 2 = - - l ll ^ ^ ^ h h h and: 2 1 2 1 = - l ^ ^ ^ h h h 142
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Variation of Parameters - 142 Part II: Surveying Second and...

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