- #1
AntSC
- 65
- 3
I have seen a couple of solutions to this PDE -
[itex]\frac{\partial x}{\partial u}=\frac{x}{\sqrt{1+y^{2}}}[/itex]
One is -
[itex]u=\ln \left | y+\sqrt{1+y^{2}} \right |+f\left ( x \right )[/itex]
I have no idea how this is arrived at or if it's correct. This is what i want to know.
The solution I've checked out makes the substitution of [itex]y=\sinh \theta [/itex] giving -
[itex]u=x\ln \left | \cosh \theta \right | + f\left ( x \right )[/itex]
which is where I'm a bit stuck as substituting in [itex]\theta =\sinh^{-1} y[/itex] gives a [itex]\cosh \sinh^{-1} y[/itex] term that i don't know how to simpify.
Any help with both of these would be appreciated
[itex]\frac{\partial x}{\partial u}=\frac{x}{\sqrt{1+y^{2}}}[/itex]
One is -
[itex]u=\ln \left | y+\sqrt{1+y^{2}} \right |+f\left ( x \right )[/itex]
I have no idea how this is arrived at or if it's correct. This is what i want to know.
The solution I've checked out makes the substitution of [itex]y=\sinh \theta [/itex] giving -
[itex]u=x\ln \left | \cosh \theta \right | + f\left ( x \right )[/itex]
which is where I'm a bit stuck as substituting in [itex]\theta =\sinh^{-1} y[/itex] gives a [itex]\cosh \sinh^{-1} y[/itex] term that i don't know how to simpify.
Any help with both of these would be appreciated