# xy' = y − xe^(y/x)

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You should use the following substitution such that:

`y = x*f(x)`

Differentiating both sides with respect to x yields:

`(dy)/(dx) = (d(x*f(x)))/(dx)`

You need to use product rule to the right side such that:

`y' = (dy)/(dx) = f(x) + (x*d(f(x)))/(dx)`

You need to substitute `f(x) + (x*d(f(x)))/(dx)` for `y` ' and `x*f(x)` for `y` , in the given equation, such that:

`x(f(x) + (x*d(f(x)))/(dx)) =x*f(x)- xe^((x*f(x))/x)`

You need to factor out x to the right side such that:

`x(f(x) + (x*d(f(x)))/(dx)) = x*(f(x)- e^((x*f(x))/x))`

Equating like parts both sides such that:

`(x*d(f(x)))/(dx)) = -e^((f(x))/x)`

Dividing both sides by x yields:

`(d(f(x)))/(dx)) = (-e^((f(x))/x))/x`

You should divide both sides by `e^((f(x)))` such that:

`((d(f(x)))/(dx))/(e^((f(x)))) = -1/x`

Integrating both sides yields:

`int ((d(f(x)))/(dx))/(e^((f(x)))) dx= - int 1/x dx`

`- e^((-f(x)) = - ln|x| + c => e^(-f(x)) = ln|x| - c`

Taking logarithms both sides yields:

`ln(e^((-f(x)))) = ln(ln|x| - c)`

Using logarithmic identities yields:

`-f(x)ln e = ln(ln|x| - c) => -f(x) = ln(ln|x| - c)`

You may evaluate the solution to the given equation such that:

`y = x*f(x) => y = -x*(ln(ln|x| - c))`

**Hence, evaluating the solution to the given first order nonlinear differential equation yields `y = -x*(ln(ln|x| - c)).` **

The y/x within the exponential looks problematic, so we can try substituting u = y/x

u' = y'/x - y/x^2

=> u'x^2 = xy' - y

Hence substituting into the original equation we get

u'x^2 = -xe^u => u'x = -e^u

We can arrange this as e^-u du = -1/x dx

so integrating each side gives

e^-u = log(x) + C

Substituing u = y/x back into the equation gives

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

-y/x = log(log(x) + C)

y = -x*log (log(x) + C)