Question

The solution of the differential equation `(1 + e^(x/y)) dx + e^(x/y) (1 + x/y) dy` = 0 is

पर्याय

• ye^x + x = C

• xe^y + y = C

• `ye^(y/x) + y` = C

• `ye^(x/y) + y` = C

Answer 1

`ye^(x/y) + y` = C

Explanation:

The given differential equation is `(1 + e^(x/y)) dx + e^(x/y) (1 - x/y) dy` = 0

`(dx)/(dy) = (e^(x/y) (x/y - 1))/((e^(x/y) + 1))`  ......(i)

= `g(x/y)`  ∵ `(dx)/(dy) = g(x/y)`

∴ Equation (i) is the homogeneous differential equation so, put `x/y` = v i.e., x = vy

⇒ `(dx)/(dy) = v + y (dv)/(dy)`

Then, equation (i) becomes

`v + y (dv)/(dy) = (e^v(v - 1))/(e^v + 1)`

⇒ `y (dv)/(dy) = (e^v (v - 1))/(e^v + 1) - v`

⇒ `y (dv)/(dy) = (ve^v - e^v - ve^v - v)/(e^v + 1) - v`

⇒ `(e^v + 1)/(e^v + v) dv = - 1/y  dy`

On integrating both sides, we get

`int (e^v + 1)/(e^v + v) dv = -int 1/y dy`

Put e^v + v = t

⇒ `e^v + 1 = (dt)/(dv)`

⇒ dv = `(dt)/(e^v + 1)`

∴ `int (e^v + 1)/t (dt)/(e^v + 1) - log |y| + log C`

⇒ `log |t| + log |y| = log C`

⇒ `log |e^v + v| + log |y| = log C`  .....(∵ t = e^v + v)

⇒ `log|(e^v + v)y|` = C

⇒ `|(e^v + v)y|` = C

⇒ `(e^v + v)y` = C.

So, put v = `x/y`, we get

`(e^(x/y) + x/y)y` = C

⇒ `ye^(x/y) + x` = C

This is the required solution of the given differential equation.