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Solve (dy)/(dx)+ycotx=sinx...

Solve `(dy)/(dx)+ycotx=sinx`

Text Solution

Verified by Experts

The correct Answer is:
`ysinx=1/4[2x-sin2x]+c`
Given equation is linear and
`P=cotx, Q=sinx`
`therefore I.F. = e^(int(cotxdx))=e^(logsinx)=sinx`
Hence, the solution is
`ysinx=int(sinxsinxdx+c)`
`=1/2int(1-cos2x)dx+c`
`=1/2[x-1/2sin2x]+c`
`therefore ysinx=1/y[2x-sin2x]+c`
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