# PDEMOC3

$\displaystyle y^{-1} u_x + u_y = 0, u(x,1)=x^2\,$

Let $\displaystyle x = x(y)\,$ so $\displaystyle u = u(x(y),y)\,$ and let $\displaystyle x(1) = x_1\,$ .

Now $\displaystyle u_y = u_x x'(y) + u_y\,$ .

Let $\displaystyle u_y = u_x x'(y) + u_y = y^{-1} u_x + u_y = 0\,$ so $\displaystyle u\,$ is constant with respect to $\displaystyle y\,$ and $\displaystyle u(x(y),y) = u(x(1),1) = x_1^2\,$ .

$\displaystyle x'(y) = y^{-1}\,$

$\displaystyle x(y) = \ln y + x_1\,$

The solution is:

$\displaystyle u(x,y) = (x-\ln y)^2\,$