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If y = 3 cos (logex) + 4 sin (logex) show that x2y2 + xy1 + y = 0. |
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Answer» y = 3 cos (logex) + 4sin (logex) Diff wrt x y1 = 3(-sin(logx).1/x) + 4 cos (log x). 1/2 xy1 = -3 sin (logx) + 4 cos(logx) diff wrt x xy2 y1. 1 = -3cos(logx).1/x + 4(-sin(logx)).1/x xy2 + y1 = 1/2-(3cos(logx) + 4sin(logx)) x2y2 + xy1 = 1/2-(3cos(logx) + 4sin(logx)) x2y2 + xy1 = -y ∴ x2y2 + xyj1 + y = 0 |
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