Understanding a solution of $\sqrt{y^2+1}\,dx = xy\,dy$: where does the constant come from after integrating?

Question

They got $$\frac{dx}{x} = \frac{y \, dy}{\sqrt{y^2+1}},$$ and in the next line $$ \int \frac{dx}{x} = \int \frac{y \, dy}{\sqrt{y^2+1}} + C.$$

Where is the $C$ from?

Answer 1

Summarizing the above comments:

As an indefinite integral has a constant (you should have learnt indefinite integrals before learning differential equations), consequently there must be a constant to cover all solutions for the differential equation.

Answer 2

Strictly speacking, the integration constant at that place is superfluous. The indefinite integrals already implicitly contain an indeterminate constant. So a better way is to have the constant only appear after actually selecting one of the anti-derivative functions, $$ \int \frac{dx}x=\int\frac{y\,dy}{\sqrt{1+y^2}}\\~\\ \implies \ln|x|=\ln\sqrt{1+y^2}+C. $$

Answer 3

Equaling two expressions in $dx$ and $dy$, actually make sense (implicitly means) that we are considering $x$ and $y$ as functions of a same variable, say $t$.
Where the $t$ can then be the $x$ or the $y$ itself.

That premised, we intend that $$ dx = d\left( {x + c_{\,x} } \right)\;:dx(t) = d\left( {x(t) + c_{\,x} } \right) $$ and same for $y$.

Therefore the expression $$ {{d\xi } \over {\left( {\xi - c_{\,x} } \right)}} = {{\left( {\eta - c_{\,y} } \right)d\eta } \over {\sqrt {\left( {\eta - c_{\,y} } \right)^{\,2} + 1} }} $$ shall be understood with $\xi , \, \eta$ as functions of $t$, and it shall be integrated over the same interval for the parameter, i.e.: $$ \int_{t = t_{\,0} }^{\,t} {{{d\xi } \over {\left( {\xi - c_{\,x} } \right)}}} = \int_{t = t_{\,0} }^{\,t} {{{\left( {\eta - c_{\,y} } \right)d\eta } \over {\sqrt {\left( {\eta - c_{\,y} } \right)^{\,2} + 1} }}} $$

Since the constants are arbitrary we can write $$ \xi (t_{\,0} ) - c_{\,x} = a\quad \xi (t) - c_{\,x} = b(t)\quad \eta (t_{\,0} ) - c_{\,y} = c\quad \eta (t) - c_{\,y} = d(t) $$ and therefore $$ \int_{\xi = a}^{\,b(t)} {{{d\xi } \over \xi }} = \int_{\eta = c}^{\,d(t)} {{{\eta d\eta } \over {\sqrt {\eta ^{\,2} + 1} }}} $$ which means $$ \int {{{d\xi } \over \xi }} + C_{\,\xi } = \int {{{\eta d\eta } \over {\sqrt {\eta ^{\,2} + 1} }}} + C_{\,\eta } $$ or $$ \int {{{d\xi } \over \xi }} = \int {{{\eta d\eta } \over {\sqrt {\eta ^{\,2} + 1} }}} + C $$

It is clear that, in the general case, the range of $C$ will be subject to limitations.