Question

Find the particular solution of the differential equation:

`2y  e^(x/y) dx + (y - 2x  e^(x/y)) dy = 0` given that y(0) = 1

Answer 1

Step 1: Finding `dx/dy`:

`2y  e^(x/y) dx + (y - 2x  e^(x/y)) dy = 0`

`2y  e^(x/y) dx = -(y - 2x  e^(x/y)) dy`

`2y  e^(x/y) dx = (2x  e^(x/y) - y) dy`

`dx/dy = (2 x  e^(x/y) - y)/(2y  e^(x/y))`

Step 2: Put F(x, y) = `dx/dy` and find F(λx, λy):

F(x, y) = `(2 x  e^(x/y) - y)/(2y  e^(x/y))`

Find F(λx, λy):

F(λx, λy) = `(2(λx)e^((λx)/(λy) - λy))/(2λy  e^((λx)/(λy)))`

= `(λ(2x  e^(x/y) - y))/(λ * 2y  e^(x/y))`

= `(2 x  e^(x/y) - y)/(2y  e^(x/y))`

= F(x, y)

So, F(λx, λy) = F(x, y)

= λ° F(x, y)

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

Therefore, the given differential equation is a homogeneous differential equation.

Step 3: Solving `dx/dy` by putting x = vy:

`dx/dy = (2 x  e^(x/y) - y)/(2y  e^(x/y))`   ....(1)

put x = vy

Differentiate with respect to y:

`dx/dy = d/dy (vy)`

`dx/dy = y * (dv)/dy + v dy/(dy)`

`dx/dy = y * (dv)/dy + v`

Putting values of `dx/dy` and x in (1):

`dx/dy = (2 x  e^(x/y) - y)/(2y  e^(x/y))`

`v + y (dv)/dy = (2 v  e^v - 1)/(2  e^v)`

`y (dv)/dy = (2v  e^v - 1)/(2  e^v) - v`

`y (dv)/dy = (2 v  e^v - 1 - 2 v  e^v)/(2  e^v)`

`y (dv)/dy = (-1)/(2  e^v)`

`2  e^v dv = (-dy)/y`

Integrating both sides:

`int 2  e^v dv = int (-dy)/y`

2 e^v = −log |y| + C

Putting back v = `x/y`

`2 e^(x/y) = -log |y| + C`

`2 e^(x/y) + log |y| = C`  ....(2)

Given that at x = 0, y = 1

Putting x = 0 and y = 1 in (2)

`2e^(0/1) - log |1| = C`

2 × 1 + 0 = C  ...(As log |1| = 0)

C = 2

Put the value of C in (2):

`2 e^(x/y) + log |y| = C`

`2 e^(x/y) + log |y| = 2`