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Question
Solve:
`1 + (dy)/(dx) = cosec (x + y)`; put x + y = u.
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Solution
`1 + dy/dx = cosec (x + y)` ...(i)
Put x + y = u
⇒ `1 + dy/dx = du/dx`
∴ 1 + `"dy"/("d"x)="du"/("d"x)`
Given differential equation (i) becomes
`(du)/dx` = cosec u
∴ `(du)/("cosec" u) = dx`
`therefore int 1/("cosec" u) du = int dx`
∴ ∫sin u du = ∫dx
∴ −cos u = x + c
∴ x + cos u + c = 0
∴ x + cos (x + y) + c = 0 ...(∵ x + y = u)
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