Question

xdy/dx–y=e^(dy/dx)

Answer 1

$x\frac{dy}{dx}-y=e^{\frac{dy}{dx}}$

Let's rewrite the equation in form

$y=xy'-e^{y'}$

It's a Clairault's equation.

Differentiation with respect to x will yield

$y'=y'+xy''+e^{y'}y''$

Therefore

$y''\left(x-e^{y'}\right)=0$

Hence, either

$y''=0\iff y=c_1x+c_2$ - which is the general solution of Clairault's equation.

or

$x-e^{y'}=0$

$\ln x=y'$

$y=x\ln x-x+c$ - which is the singular solution of Clairault's equation.