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Question
Show that the given differential equation is homogeneous and solve them.
(x2 – y2) dx + 2xy dy = 0
Solution
(x2 - y2) dx + 2xy dy = 0
Which can be written as
`dy/dx = (y^2 - x^2)/(2 xy)`
`= ((y/x)^2 - 1)/(2 (y/x))` ....(1)
Since R.H.S is of the form `g(y/x)`, and so it is a homogeneous function of degree zero
Therefore equation (1) is a homogeneous differential equation.
⇒ `dy/dx = v + x (dv)/dx`, then (1) become
`v + x (dv)/dx = (v^2 - 1)/(2v)`
⇒ `x (dv)/dx = (v^2 - 1)/(2v) - v`
⇒ `(2vdv)/(v^2 + 1) = -dx/x` ....(2)
Integrating (2) both sides, we get
log |v2 + 1| = - log |x| + C
⇒ log |(v2 + 1) x | = C
⇒ `log |(y^2 + x^2)/x| = C_1` ...`(∵ v = y/x)`
⇒ `|(y^2 + x^2)/x| = e^(C_(1))`
⇒ `(x^2 + y^2)/x =pm e^(C_(1)) = C` (say)
⇒ `x^2 + y^2 = Cx`
which is the required general solution of the given differential equation.
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