Solution to a partial differential equation example
I was wondering on how to deal with the following PDE. I can see it is on the form of a heat equation, but I just want to know how to solve this concrete example by "hand", i.e. without computer programs. The equation is given below.
$$u_x=2u_{yy}, \hskip 0.5cm u(2,y)=y^2.$$
The solution in Maple 2020 is$$u=y^2+4x-8.$$
Any suggestion, books, methods are welcome. Thanks in advance.
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$\begingroup$Hint: You can suppose that $$u(x,y)=X(x)Y(y)$$so, you can see that $$u_{x}=YX' \quad u_{yy}=XY''$$Then, $$u_{x}=2u_{yy} \iff YX'=2XY'' \iff \frac{X'}{2X}=\frac{Y''}{Y}=-\lambda$$ and you can the use method of separation of variables that you know in ODE.
$\endgroup$ 3 $\begingroup$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$\begin{align} &\on{u}\pars{x,y} \equiv \on{a}\pars{x} + \on{b}\pars{y} \\[5mm] &\on{a}'\pars{x} = 2\on{b}''\pars{y} = \alpha = \mbox{constant} \\[5mm] &\ \on{a}\pars{x} = \alpha x + \beta,\qquad \on{b}\pars{y} = {1 \over 4}\,\alpha y^{2} + cy + d \\ &\ y^{2} = \on{u}\pars{2,y} = \pars{2\alpha + \beta} + \pars{{1 \over 4}\,\alpha y^{2} + cy + d} \\ &\ {1 \over 4}\alpha = 1,\quad \bbx{c = 0},\quad 2\alpha + \beta + d = 0 \\ &\ \implies \bbx{\alpha = 4},\quad \bbx{\beta + d = -8} \\ &\ \bbx{\on{u}\pars{x,y} = 4x + y^{2} - 8} \\ & \end{align}
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