Question

find the particular solution satisfying the initial condition indicated.

y = x exp(y-x') when x=0, y=0

I tried using ln but it seems not to be working.

Answer 1

i dont understand the "x prime" , does that mean dx/dt ?
isn't the goal to find y(x) ?

Answer 2

Maybe \(x\) is the dependent variable and it's a function of \(y\)? @moongazer

Answer 3

it has a typo it should be

y' = x e^(y-x^2)

Answer 4

The equation is separable:
\[\begin{align*}
y'&=xe^{y-x^2}\\
y'&=x\frac{e^y}{e^{x^2}}\\
e^{-y}~dy&=xe^{-x^2}~dx
\end{align*}\]