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The solution of differential equation `cos x (dy)/(dx)+y sin x =1`A. `tanx+tany=k`B. `tanx-tany=k`C. `(tanx)/(tany)=k`D. `tanx.tany=k` |
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Answer» Given that, `" "tanysec^(2)xdx+tanxsec^(2)ydy=0` `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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