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If y = Axcos(Inx) + Bxsin(Inx), then y': (A+B)cos(In x)+(B-A)sin(In x) -(A+B) sin(In x)/x+(B-A)cos(In x)/x 0 -COS X . Hence, xy-xy' +2y=[ Question 2
If y = Axcos(Inx) + Bxsin(Inx), then y': (A+B)cos(In x)+(B-A)sin(In x) -(A+B) sin(In x)/x+(B-A)cos(In x)/x 0 -COS X . Hence, xy"-xy' +2y=[ Question 2 If y = Axcos(Inx) + Bxsin(Inx), then y' = [ A Moving to another question will save this response. ' =C (A+B)cos(In x)+(B-A)sin(In x) -(A+B) sin(In x)/x+(B-A)cos(In x)/x 0 -COS X Hence, xy" xy' + 2y = 10 points Save Answer Question 2 of 12 Question 2 If y = Axcos(Inx) + Bxsin(Inx), then y': = Moving to another question will save this response. Hence, xy" -xy' + 2y = 10 points Save Answer (A+B)cos(In x)+(B-A)sin(In x) -(A+B) sin(In x)/x+(B-A)cos(In x)/x 0 -COS X Question 2 of 12 > >>
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1 Given question y Axcoslog x Bxsinlog x Lets find y and y first y A Bcoslog x B Asinlog x A Bsinlog xx B Acoslog xx Now differentiate y to find y y sinlog xA B coslog xB A A Bcoslog xx B Asinlog xx N...
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Differential Equations And Linear Algebra
Authors: C. Edwards, David Penney, David Calvis
4th Edition
013449718X, 978-0134497181
