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I need to solve this eqn $\partial_x g=1 $. If instead of partial derivatives I had total derivative, I would do separation of variables and get the answer.

Is doing something like $\int \partial g =\int \partial x$ correct... Can we do this separation, when the derivatives are partial rather than total.

Is there a thing as partial differential. Can we integrate with respect to a partial differential.....

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  • $\begingroup$ When you introduce a function, please mention its domain. $\endgroup$
    – Rodrigo de Azevedo
    Nov 27, 2020 at 19:34

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$$\partial_r (r e^{a(r) })=1$$ Simply gives after integration: $$r e^{a(r)}=r+C$$ Now it depends on what the second variable is. Then $C$ is just a function of that second variable. $$r e^{a(r)}=r+C(\theta)$$ For the other equation: $$\partial_x g(x,y)=1$$ $$\implies g(x,y)=x+C(y)$$ Where $C(y)$ is any finction of $y$.

Consider the equation as an ordinary differential equation: $$\partial_x g=1$$ $$\dfrac {dg}{dx}=1$$ Integrate: $$\int dg=\int dx$$ $$g=x+C$$ Now consider the constant as a function of the second variable: $$g(x,y)=x+C(y)$$

$$\partial_r (r e^{a(r) })=1$$ Is like an ordinary DE: $$\dfrac {d (r e^{a(r) })}{dr}=1$$ $${d (r e^{a(r) })}=dr$$ Integrate: $$\int {d (r e^{a(r) })}=\int dr$$ $$r e^{a(r) }=r+C$$ Consider the constant $C$ as a function of a second variable $\theta $: $$r (e^{a(r) }-1)=C(\theta)$$

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  • $\begingroup$ Thank you. I have doubt at the second step, where you have written " Simply gives.. ". Could you please show that step explicitly, in that could you please write the integrand and with what you are integrating. I have modified the question a bit to be precise in showing where I am confused. Please have a look. Let me know the precise the integrand and whether you are integrating with respect to partial r..Everything else is clear. I will accept the answer after that. $\endgroup$
    – Shashaank
    Nov 27, 2020 at 19:00
  • $\begingroup$ In fact consider this like an ordinary differential equation. Integrate. Then consider the constant of integration as a function of the second variable. @Shashaank $\endgroup$
    – user577215664
    Nov 27, 2020 at 19:01
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    $\begingroup$ yah but do I write it like $\int \partial g = \int \partial x$ or $\int dg = \int dx$. And similarly for the other one. Could you please write the integrand and with what you are integrating explicitly... That is the step between $\partial_x g = 1 $ and g = x + c(y)... $\endgroup$
    – Shashaank
    Nov 27, 2020 at 19:05
  • $\begingroup$ I added some lines Hope its more clear @Shashaank $\endgroup$
    – user577215664
    Nov 27, 2020 at 19:05
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    $\begingroup$ Thanks yes it is.. I have accepted the answer. Could you please add a similar thing for original example as well..( the step after $\partial_ r ( r e^(a(r))=1$. It will be very clear then. Thank you $\endgroup$
    – Shashaank
    Nov 27, 2020 at 19:08

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