PDE $yu_{yy} + u_{y} = 0$ solution with boundary conditions

Question

I have to solve the following problem:

$yu_{yy} + u_{y} = 0$

$u(x, 1) = x^{2}$

$u(x, e) = 1$

I have tried first taking one $\partial u/\partial y$ and integrating but I obtain a function that depends on $y$ and then I have to do another integration on $y$ but now I have to integrate $u$, so it doesn't work.

Answer 1

Let us substitute $v = u_y$ in the differential equation, so that $y v_y + v = 0$ or maybe $v_y/v = -1/y$. This is a (separable) Cauchy-Euler differential equation w.r.t. $y$, which solution reads $v(x,y) = c_1(x)/y$. Integration w.r.t. $y$ then yields $$ u(x,y) = c_1(x)\ln y + c_2(x) \, . $$ Using the boundary conditions, we have \begin{aligned} u(x,1) &= c_2(x) = x^2 \\ u(x,e) &= c_1(x) + c_2(x) = 1 \end{aligned} from which we can conclude.

Answer 2

$$yu_{yy} + u_{y} = 0$$ $$yu''+u'=0$$ Rewrite it as: $$(yu')'=0$$ Integrate twice. $$yu'=c_1(x)$$ $$u'=\dfrac {c_1(x)}y$$ integrate again: $$u(x,y)= {c_1(x)} \ln |y|+c_2(x)$$