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Question
Show that the given differential equation is homogeneous and solve them.
`(1+e^(x/y))dx + e^(x/y) (1 - x/y)dy = 0`
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Solution
`(1 + e^(x/y))dx + e^(x/y)(1 - x/y) dy = 0`
`=> dx/dy = (e^(x/y)(x/y - 1))/(1 + e^(x/y)) = g(x/y)` say ...(i)
∵ The right side of the equation is in the form of `g (x/y)`, so it is a homogeneous differential equation of zero degrees.
∴Putting x = vy
`dx/dy = v + y (dv)/dy` ...(from equation (i))
`v + y (dv)/dy = (e^v(v - 1))/(1 + e^v)`
or `y (dv)/dy = (e^v(v - 1))/(1 + e^v) - v`
`=> (ve^v - e^v - v - ve^v)/(1 + e^v)`
`=> ((1 + e^v)/(v + e^v))dv = - 1/y dy`
`= int(1 + e^v)/(v + e^v) dv = - int 1/y dy`
⇒ log |ev + v| = - log |y| + C1
⇒ log |(ev + v)y| = C1
⇒ |(ev + v) y| = eC1
⇒ (ev + v)y = ± eC1 = C (say)
⇒ `(e^(x/y) + x/y) y = C`
⇒ `y e^(x/y) + x = C`
which is the required general solution of the given differential equation.
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