`(y+1)cosx dx - dy = 0` Solve the first-order differential equation

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Given,

`(y+1)cosx dx-dy=0`

=>` (y+1)cosx dx=dy`

=>`(y+1)cosx =dy/dx`

=> `(dy/dx )*(1/(y+1))=cosx`

integrating on both side we get

`int (dy/dx )*(1/(y+1))=int cosx`

=> ` int (1/(y+1)) dy= int cosx dx`

`ln(y+1)=sinx +c `

=> `y+1= e^(sinx+c)`

=> `y=e^(sinx+c) -1`

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