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If the solution of the differential equation`(dy)/(dx)=1/(xcosy+sin2y)`is `x=c e^(siny)-k(1+siny),`then thevalue of `k`is_______ |
Answer» Correct Answer - 2 `(dy)/(dx) = 1/(xcosy+2sinycosy)` `therefore (dx)/(dy) = xcosy+2sinycosy` `therefore (dx)/(dy) +(-cosy)x=2sinycosy` `therefore I.F. =e^(-intcosydx)=e^(-siny)` Thus, the solution is `x.e^(-siny)=2inte^(-siny).sinycosydy = -2sinye^(-siny)+2inte^(-siny)cosydy` `=-2sinye^(-siny)+2inte^(-siny)cosydy` `=-2sinye^(-siny)-2e^(-siny)+c` i.e., `x=-2siny-2+ce^(siny)=ce^(siny)-2(1+siny)` `therefore k=2` |
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