Question

solve the differential equation
​

Answer 1

Question:

$(1 + {e}^{ \frac{x}{y} } )dx + {e}^{ \frac{x}{y} } (1 - \frac{x}{y} )dy = 0$

Solution:

The given equation is:

$(1 + {e}^{ \frac{x}{y} } )dx + {e}^{ \frac{x}{y} } (1 - \frac{x}{y} )dy = 0 \\ \\ = > \frac{dx}{dy} = - \frac{ {e}^{ \frac{x}{y} (1 - \frac{x}{y} )} }{1 + {e}^{ \frac{x}{y} } } \\ \\ = > \frac{dx}{dy} = \frac{ (\frac{x}{y} - 1) {e }^{ \frac{x}{y} } }{1 + {e}^{ \frac{x}{y} } } ...............(1).$

Here

$f(x,y) = \frac{( \frac{x}{y } - 1) {e}^{ \frac{x}{y} } }{1 + {e}^{ \frac{x}{y} } } \\ \\ f(λx,λy) = \frac{( \frac{λx}{λy} - 1) {e}^{ \frac{λx}{λy} } }{1 + {e}^{ \frac{λx}{λy} } } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ (\frac{x}{y} - 1) {e}^{ \frac{x}{y} } }{1 + {e}^{ \frac{x}{y} } } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = λ ^{0} f(x,y)$

Thus f(x,y) is homogeneous function of degree zero.

To solve:

Put x= vy

so that

$\frac{dx}{dy} = v + y \frac{dv}{dy}$

(1) becomes.

$v + y \frac{dv}{dy} = \frac{ (\frac{vy}{y} - 1) {e}^{ \frac{vy}{y} } }{1 + {e}^{ \frac{x}{y} } } \\ \\ = > v + y \frac{dv}{dy} = \frac{(v - 1) {e}^{v} }{1 + {e}^{v} } \\ \\ = > y \frac{dv}{dy} = \frac{(v - 1) {e}^{v} }{1 + {e}^{v} } - v \\ \\ = > y \frac{dv}{dy} = \frac{v {e}^{v} - {e}^{v} - v - v {e}^{v} }{1 + {e}^{v} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = - \frac{ {e}^{v} + v}{1 + {e}^{v} } \\ \\ = > \frac{1 + {e}^{v} }{v + {e}^{v} } dv = - \frac{dy}{y}$

Integrating

$∫ \frac{1 + {e}^{ \frac{x}{y} } }{v + {e}^{v} } dv = - ∫ \frac{1}{y} dy + log |c|$

Put

$v + {e}^{v} = t$

so that

$( 1 + {e}^{v} )dv = dt$

$∫ \frac{dt}{t} = - ∫ \frac{1}{y} dy + log |c| \\ \\ = > log |t| = - log |y| + log |c| \\ \\ = > log |t| + log |y| = log |c| \\ \\ = > log |ty| = log |c| \\ \\ = > ty = c = > (v + {e}^{ \frac{x}{y} }) y = c \\ \\ = > ( \frac{x}{y} + {e}^{ \frac{x}{y} } )y = c \\ \\ = > x + y {e}^{ \frac{x}{y} } = c$

Which is required solution.