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I have trouble about this differential equation. I don't know if the following holds:

The function $u$ depends from $x$ and $y$ i.e $ u=u(x,y)$

$$ \begin{aligned} \frac{du}{u}&=\dfrac{dx}{x} \\ \int\frac{du}{u}&=\int\dfrac{dx}{x} \\ \ln(u)&=\ln(x)+\ln\left(f(y)\right) \end{aligned}$$

I'm not sure because the function $u$ has the variables $x$ and $y$ can I take the function $f(y)$ after integrating as $\ln\left(f(y)\right)$, because I need it in this form in the other part of the solution.

Thank you very much!

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  • $\begingroup$ $\ln f(y)$ restricts the class of functions you can use to those that have a domain $(0, +\infty)$, even though any function could've been used. Proper way is to write it as $\ln u(x, y) = \ln x + f(y)$, and take exponent. $\endgroup$ – Kaster May 31 '17 at 22:35
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HINT: one can easily check your computations by differentiating the answer.

Indeed, if we want to check your answer we rewrite equation $\;\dfrac{du}{u}=\dfrac{dx}{x}\,$ as $\,\dfrac{du}{dx}=\dfrac{u}{x}$ which is much easier to check. Starting from where you left we have

\begin{alignedat}{3} \frac{du}{u}&=\dfrac{dx}{x} &\implies \int\frac{du}{u}&=\int\dfrac{dx}{x} &\implies \ln(u)&=\ln(x)+\ln\left(f(y)\right) &\iff u(x,y) &= x\cdot f\left(y\right) \end{alignedat}

Then, assuming $y$ does not depend on $x$, we have

$$\require{enclose} \dfrac{\partial}{\partial x} u\left(x,y\right) = \dfrac{\partial}{\partial x} \big[\,x\cdot f\left(y\right)\big] = f(y) = \dfrac{x\cdot f\left(y\right)}{x} = \dfrac{u\left(x,y\right)}{x} \qquad %\color{green}{\LARGE \checkmark} \enclose{circle}[mathcolor="LimeGreen"]{\LARGE \,\color{LimeGreen}{\checkmark}\,} $$

So your computations seem to be correct.

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