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Question
For the differential equation, find the general solution:
`(x + y) dy/dx = 1`
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Solution
Differential equations,
`(x + y) dy/dx = 1`
`therefore dx/dy = x + y`
or `dx/dy - x = y`
Comparing with the differential equation, `dx/dy + Px = Q`,
P = -1, Q = y
`I.F. = e^(int P dx) = e^(int (- 1)dy) = e^(- y)`
The solution of the differential equation is:
`x × I.F. = int Q xx I.F. dy + C`
`=> x xx e^(- y) = int y * e^(- y) dy + C`
On integrating piecewise,
`xe^(- y) = y (e^(- y)/(-1)) - int 1((e^(- y))/(-1)) dy + C`
`= - ye^(- y) + e^(-y)/(- 1) dy + C`
`= - ye^-y - e^(- y) + C`
or x = - y - 1 + Cey
∴ x + y + 1 = Cey
This is the desired solution.
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