1.

The solution of differential equation cos x (dy)/(dx)+y sin x =1

Answer»

`tanx+tany=k`
`tanx-tany=k`
`(tanx)/(tany)=k`
`tanx.tany=k`

Solution :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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