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The solution of differential equation cos x (dy)/(dx)+y sin x =1 |
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Answer» `tanx+tany=k` `rArr""tansec^(2)xdx=-tanxsec^(2)ydy` `rArr""(sec^(2)x)/(tanx)DX=(-sec^(2)y)/(tany)dy""`...(i) On integrating both sides, we have `""INT(sec^(2)x)/(tanx)dx=-int(sec^(2)y)/(tany)dy` Put `tanx=t` in LHS integral, we get `""sec^(2)xdx=dtrArrsec^(2)xdx=dt` and `""tany=u` in RHS integral, we get `""sec^(2)ydy=du` On SUBSTITUTING these values in Eq. (i), we get `""int(dt)/(t)=-int(du)/(u)` `""logt=-logu+LOGK` `rArr""log(t*u)=logk` `rArr""log(tanxtany)=logk ` `rArr""tanxtany=k` |
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