Solve the initial value problems  

` (x + ye^(y/x))dx - (xe^(y/x))dy = 0, y(1) = 0`

Solution:

This is a first-order non-exact differential equation. However, we can
still find the general solution using the method of substitution.

A common substitution in this kind of equation is to let `u = xe^(y/x)`.
Then we can express x and y in terms of u:

`x = ue^(-y/u)`

`y = (u/e)e^(-y/u)`

Now we can substitute these expressions into the original differential
equation:

`(ue^(-y/u) + (u/e)e^(-y/u)e^(y/ue^(-y/u))(de^(-y/u)) -
ue^(-y/u)e^(y/ue^(-y/u))dy = 0`

after simplifying we get:

`(ue^(-y/u) + u/e)du - u^2 dy = 0`

Now we can separate the variables and integrate both sides:

`∫(ue^(-y/u) + u/e)du = ∫-u^2 dy`

On the left side we can integrate and get:

`u^2/2 + u = F(y) + C`

On the right side we can integrate and get:

`-u^3/3 = G(x) + C`

Where F(y) and G(x) are arbitrary functions of x and y respectively and C
is a constant of integration.

We can see that both sides are equal and equal to `-u^3/3 + u^2/2 + u +
C`.

Solving for u in terms of y and vice versa we get:

`u = 2Ce^(y/2)`

and

`y = 2ln(u/2C)`

This is the general solution of the differential equation. However, we have
one initial condition y(1) = 0. To use this initial condition we have to
express x in terms of y,

`x = ue^(-y/u)`

`x(1) = 2Ce^(1/2)e^(-y/2Ce^(1/2))`

Now we can substitute the initial condition `y(1) = 0` and get:

`x(1) = 2Ce^(1/2)e^(-0) = 2Ce^(1/2)`

So,

`C = x(1)/2e^(1/2)`

Hence the solution is

`y = 2ln(xe^(-y/x)/x(1))`

Note that the constant of integration C can be replaced with any
constant.

Also See the Solution of this

Find the tangential and normal components of the acceleration of
a point describing the curve `y^2 – 2x = 1 -2y` with the
uniform speed v when the particle is at (0, -1 ± `\sqrt 2 \)`.