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Question
Show that the given differential equation is homogeneous and solve them.
`y dx + x log(y/x)dy - 2x dy = 0`
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Solution
`y dx + x log (y/x) dy - 2x dy = 0`
⇒ `y/x + log (y/x) dy/dx - 2 dy/dx = 0`
⇒ `dy/dx = (y/x)/(2 - log (y/x))` .....(1)
Since R.H.S. is of the form `g(y/x)`, it is a homogeneous function of degree zero.
Therefore equation (1) is a homogeneous differential equation.
To solve this, put y = vx
⇒ `dy/dx = v + x (dv)/dx,` then (1) becomes
`v + x (dv)/dx = v/ (2 - log v)`
⇒ `x (dv)/dx = v/ (2 - log v) - v`
⇒ `((2 - log v)dv)/(v log v - v) = dx/x`
⇒ `(1 - (log v - 1))/(v (log v - 1)) dv = dx/x` .....(2)
Integrating (2) both sides, we get
`int (1/ (v (logv - 1)) - 1/v) dv`
`= int dx/x + C_1`
⇒ `int (1/v)/(log v - 1) dv - int 1/v dv = int dx /x + C_1`
⇒ `log |log v - 1| - log |v|`
= log |x| + C1
⇒ `log |(log v - 1)/(v x)| = C_1`
⇒ `|(log v - 1)/ (v x)| = e^(C_(1))`
⇒ `(log v - 1)/(v x) = pm e^(C_(1)) = C` (say)
⇒ `log (y/x) - 1 = Cy` ...`(∵ v = y/x)`
which is the required general solution.
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- Solve the above equation to find its general solution. (2)
