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!

  • $\begingroup$ $\ln f(y)$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.