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प्रश्न
Solve the following differential equation.
`(x + y) dy/dx = 1`
उत्तर
`(x + y) dy/dx = 1`
∴ `dy/dx = 1/(x + y)`
∴ `dx/dy = x + y`
∴ `dx/dy − x = y`
∴ `dx/dy + (− 1)x = y` ...(I)
The given equation is of the form `dx/dy + Px = Q`
where, P = − 1 and Q = y
∴ `"I.F." = e int ^"pdy" = e int ^("(−1)dy") = e^-"y"`
∴ Solution of the given equation is
`"x (I.F.)" = int "Q (I.F.) dy" + C`
∴ `"xe"^(−"y") = ubrace(int y.e^(−y))_(("I")) dy + C` ...(II)
Let I = `int y. e^(−y) dy`
Using Integration by parts,
I = `y int e^(−y) dy - int [ d/dy y int e^(−y) dy] dy`
I = `y (e^(−y))/(-1) − int 1. (e^(−y))/(-1) dy`
I = `− y. e^(−y) + int e^(−y) dy`
I = `− y. e^(−y) + e^(−y)/(- 1) dy`
I = `− y. e^(−y) − e^(−y)`
Putting value of I in (2),
∴ `"xe"^(−"y") = int y.e^(−y) dy + C`
∴ `"xe"^(−"y") = − y. e^(−y) − e^(−y) + C`
Dividing by e−y,
∴ x = − y − 1 + Cey
∴ x + y + 1 = Cey
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