Suppose we have a differential: $$\mathrm{d}u=y\mathrm{d}x + (x+2y)\mathrm{d}y\tag{1}$$ and a general differential: $$\mathrm{d}u=\underbrace{P(x,y)}_{\color{#F80}{\dfrac{\partial u}{\partial x}}}\mathrm{d}x +\underbrace{Q(x,y)}_{\color{#A0F}{\dfrac{\partial u}{\partial y}}}\mathrm{d}y\tag{2}$$ where $u$, $P$ and $Q$ are unknown functions of $x$ and $y$; the partial derivatives below the underbraces show that $(2)$ is also the total derivative. Comparing $(1)$ with $(2)$ we have that $P=y$ and $Q=x+2y$.

$(1)$ is an exact differential since $$\frac{\partial P}{\partial y}=1=\frac{\partial Q}{\partial x}=1$$

In my notes it says in order to find $u$ we integrate and $\color{red}{\fbox{match}}$ such that $$P=\color{#F80}{\frac{\partial u}{\partial x}}=y$$ $$\begin{align}\implies \color{#180}{u = xy +\color{black}{\overbrace{f(y)}^{\Large ?}}\quad\quad\quad\quad\tag{A}}\end{align}$$ where $f(y)$ is an unknown function of $y$.

Similarly, $$Q=\color{#A0F}{\frac{\partial u}{\partial y}}=x+2y$$ $$\begin{align} \implies \color{#180}{u=xy + y^2 + \color{black}{\overbrace{g(x)}^{\Large?}}\quad\quad\tag{B}}\end{align}$$ where $g(x)$ is an unknown function of $x$.

Comparing both requirements we see that $\color{blue}{u=xy + y^2 + c}$ where $c$ is a constant.

The first part that is confusing me about these notes is the word $\color{red}{\fbox{match}}$. What are we matching?

My interpretation of the word $\color{red}{\fbox{match}}$ in this context means that if $u = xy +f(y)$ AND $u=xy + y^2 + g(x)$ then $u=xy+y^2 + f(y) +f(x)$ but this is obviously not the same as $\color{blue}{u=xy + y^2 + c}$. Why is my interpretation wrong?

My second query is marked with question marks above the overbraces of equations $\color{#180}{(\mathrm{A})}$ and $\color{#180}{(\mathrm{B})}$.

After we integrate $\color{#180}{(\mathrm{A})}$ and $\color{#180}{(\mathrm{B})}$ with respect to $x$ and $y$ respectively; I don't understand why we need an unknown function of $x$; $f(x)$, and an unknown function of $y$; $f(y)$. Why must these functions be present in $\color{#180}{(\mathrm{A})}$ and $\color{#180}{(\mathrm{B})}$?

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    $\begingroup$ I know you want to understand this method, but I think a much better way of going about it is to differentiate (A) with respect to $y$ and then compare with $Q$. Doing this, you determine $f'(y)$ and hence $f(y)$ up to a constant. $\endgroup$ – Michael Albanese Jan 13 '16 at 2:35
  • $\begingroup$ @Michael Okay so working with your method for $\color{#180}{(\mathrm{A})}:$ ${\cfrac{\partial u}{\partial y}}=x+f^{\prime}(y)$ and now I compare with $Q$ you're saying. Can you show me this method in an answer please? It seems like a decent idea. $\endgroup$ – BLAZE Jan 13 '16 at 2:48
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    $\begingroup$ There is a different method which involves computing a single integral that gives you the answer directly. I describe it briefly here. It’s worth knowing because it generalizes to $k$-forms. $\endgroup$ – amd Jan 13 '16 at 3:54
  • $\begingroup$ Thanks @amd It was your answer with the method right? and not H.R's well I upvoted you anyway :) $\endgroup$ – BLAZE Jan 13 '16 at 4:13

We have two equations

\begin{align*} u &= xy + f(y) && \color{#180}{(\mathrm{A})}\\ u &= xy + y^2 + g(x) && \color{#180}{(\mathrm{B})} \end{align*}

Both of the expressions on the right hand side are equal to $u$ and hence each other (we are 'matching' the two expressions for $u$), so we see that

\begin{align*} xy + f(y) &= xy + y^2 + g(x)\\ f(y) &= y^2 + g(x). \end{align*}

As the left hand side depends only on $y$, but not on $x$, the same must be true of the right hand side. It follows that $g(x)$ must be a constant function, $g(x) = c$, so $f(y) = y^2 + c$. Substituting into either $\color{#180}{(\mathrm{A})}$ or $\color{#180}{(\mathrm{B})}$, we obtain the solution $u = xy + y^2 + c$.

As for your second query, given the equation $\frac{\partial u}{\partial x} = y$, we would like to determine $u$. That is, we want a function $u$ such that its partial derivative with respect to $x$ is $y$. Of course $xy$ is one such function, but so is $xy + f(y)$ for any function $f$ because $\frac{\partial}{\partial x}f(y) = 0$. This is precisely the same reason why we add a constant of integration when antidifferentiating a function of one variable.

Here's the other approach I mentioned in my comment.

We have equation $\color{#180}{(\mathrm{A})}$ and we know $\frac{\partial u}{\partial y} = x + 2y$. Differentiating $\color{#180}{(\mathrm{A})}$ with respect to $y$, we now have the two equations

\begin{align*} \frac{\partial u}{\partial y} &= x + f'(y)\\ \frac{\partial u}{\partial y} &= x + 2y. \end{align*}


\begin{align*} x + f'(y) &= x + 2y\\ f'(y) &= 2y\\ f(y) &= y^2 + c. \end{align*}

Substituting this back into $\color{#180}{(\mathrm{A})}$, we see that $u = xy + y^2 + c$.

  • $\begingroup$ Thank you for an excellent answer; I have understood every step you wrote apart from the $\fbox{$\color{red}{uy + f(y)}$}$ which I marked in your answer (hope that was okay). Could you show me why $\fbox{$\color{red}{uy + f(y)}$}$ is not a valid function for $u$ as I did not understand the proceeding sentence to it: "This is precisely the same reason why we add a constant of integration when antidifferentiating a function of one variable". If you could elaborate a bit more on this in your answer that would be great, thank you. $\endgroup$ – BLAZE Jan 13 '16 at 3:35
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    $\begingroup$ Sorry, that was a typo, it should have been $xy + f(y)$ not $uy + f(y)$. Does that clear up the issue? $\endgroup$ – Michael Albanese Jan 13 '16 at 3:39
  • $\begingroup$ Certainly does, thank you very much for your time. $\endgroup$ – BLAZE Jan 13 '16 at 3:44

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