How should I solve the following bvp?

Question

Q: How should I solve the following exercise:

Give the particular solution to the following boundary value problem:

$$\frac{dy}{dx} = ({\ln{e^x}})^{1/y}, \, \hspace{3mm} y(1) = -1$$

I know that $\ln(e^x) = x$ so that the problem comes down to solving

$$y = x^{1/y}dx, \hspace{5mm} y(1) = -1$$

But I can't get much further..

Answer 1

$$dy/dx = ln(e^x)^{1/y}, y(1) = -1$$ $$dy/dx = ln(e^x)^{1/y}$$

$$(e^x)^{1/y}=e^{x/y}$$ $$dy/dx = ln(e^{x/y})=\frac{x}{y}$$ $$\frac{dy}{dx}=\frac{x}{y}$$ $$\frac{ y^2}{ 2}=\frac{x^2}{2}+C$$

Given $y(1)=-1$

$$\frac{ (-1)^2}{ 2}=\frac{1^2}{2}+C$$ $$C=0$$

$$\frac{ y^2}{ 2}=\frac{x^2}{2}$$ $$y^2-x^2=0$$

Answer 2

Do you mean this ? $$y'=\frac 1y \ln e^x $$ then

$$ \frac 12 (y^2)'=x$$ Integrate simply