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.....
$$\partial_r (r e^{a(r) })=1$$ 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 $: $$r (e^{a(r) }-1)=C(\theta)$$