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Question
Show that the given differential equation is homogeneous and solve them.
`{xcos(y/x) + ysin(y/x)}ydx = {ysin (y/x) - xcos(y/x)}xdy`
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Solution
Given differential equation
`{x cos (y/x) + y sin (y/x)} y dx = {y sin (y/x) - x cos (y/x)} x dy`
or `dy/dx = ({x cos (y/x) + y sin (y/x)} y)/({y sin (y/x) - x cos (y/x)} x)`
and `dy/dx = ((y/x) {cos (y/x) + y/x sin (y/x)})/({y/x sin (y/x) - cos (y/x)} x) = g (y/x)` (say) .... (i)
The right side of the differential equation is in the form of `g (y/x)`. Therefore, this is an even exponential differential equation of zero degree.
∴ Putting y = vx
v + x `(dv)/dx = ((cos v + v sin v) v)/(v sin v - cos v)`
⇒ x `(dv)/dx = (v cos v + v^2 sin v)/(v sin v - cos v) - v`
= v cos v + v2 sin v
⇒ x `(dv)/dx = (- v^2 sin v + v cos v)/(v sin v - cos v)`
⇒ x `(dv)/dx = (2v cos v)/(v sin v - cos v)`
`= (v sin v - cos v)/(v cos v) dv = 2/x dx`
`= (tan v - 1/v) dv = 1/x dx`
On integrating
log sec v - log v = 2 log x + log C
log `((sec v)/v)` = log x2 = log C
log `((sec v)/v)` = log cx2
sec v = v. Cx2
Finally, on putting `y/x` in place of v
`sec (y/x) = (y/x). Cx^2`
`sec (y/x) = Cxy`
`xy cos |y/x| = C`
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|
An equation involving derivatives of the dependent variable with respect to the independent variables is called a differential equation. A differential equation of the form `dy/dx` = F(x, y) is said to be homogeneous if F(x, y) is a homogeneous function of degree zero, whereas a function F(x, y) is a homogeneous function of degree n if F(λx, λy) = λn F(x, y). To solve a homogeneous differential equation of the type `dy/dx` = F(x, y) = `g(y/x)`, we make the substitution y = vx and then separate the variables. |
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