Chapter 6W2(x)=(−1)3−2±±±±±xx−11−x−2±±±±±=2x−1,W3(x−1)3−3±±±±±515x4±±±±±=4x5.Now we divide both sides of the given equation byx3to obtain an equation in standard form,that is,y0−2x−1y0−5x−2y0+5x−3y=x−5.Hence, the right-hand side,g(x), in formula (1) on page 339 of the text equals tox−5. Applyingformula (11), page 340 of the text, yieldsv1(x)=Zx−5(−6x3)48x2dx=−18Zx−4dx=124x−3,v2(xZx−5(2x−1)48x2dx=124Zx−8dx=−1168x−7,v3(xZx−5(4x5)48x2dx=112Zx−2dx=−112x−
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.