Solutions to this PDE?

1. Feb 22, 2013

AntSC

I have seen a couple of solutions to this PDE -

$\frac{\partial x}{\partial u}=\frac{x}{\sqrt{1+y^{2}}}$

One is -

$u=\ln \left | y+\sqrt{1+y^{2}} \right |+f\left ( x \right )$

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 $y=\sinh \theta$ giving -

$u=x\ln \left | \cosh \theta \right | + f\left ( x \right )$

which is where i'm a bit stuck as substituting in $\theta =\sinh^{-1} y$ gives a $\cosh \sinh^{-1} y$ term that i don't know how to simpify.

Any help with both of these would be appreciated

2. Feb 22, 2013

HallsofIvy

I am confused by your notation. You have x differentiated with respect to u but on the right both x and y. If you really mean what you have, it is not a "partial" differential equation at all. We are thinking of x as a function of u, with y a fixed parameter. Further that is a separable equation: $dx/du= x\sqrt{1+ y^2}$ gives
$$\frac{dx}{x}= \frac{du}{\sqrt{1+ y^2}}$$
$$ln|x|= \frac{u}{\sqrt{1+ y^2}}+ c$$
$$x= Ce^{\frac{u}{\sqrt{1+ y^2}}}$$
where $C= e^c$.

Your original equation treats x as a function of u but your answer gives u as a function of x. Which is intended? if the latter, then
$$\frac{du}{dx}= \frac{\sqrt{1+ y^2}}{x}$$
$$\frac{du}{\sqrt{1+ y^2}}= \frac{dx}{x}$$
$$\frac{u}{\sqrt{1+ y^2}}= ln|x|+ C$$
$$u= (ln|x|+ C)\sqrt{1+y^2}$$
essentially just solving the previous equation for u.

3. Feb 23, 2013

AntSC

Sorry, i've written the dependent and independent variables the wrong way round and got the independent variable wrong. It's been a long week!

It should be -

$\frac{\partial u}{\partial y}=\frac{x}{\sqrt{1+y^{2}}}$

for a function $u\left ( x,y,z \right )$

There will be a constant term in the form $f\left ( x,z \right )$ in addition to whatever the result of the integration yields (which i can't work out).

Last edited: Feb 23, 2013
4. Feb 26, 2013

HallsofIvy

Since there is no differentation with respect to x, you can treat x as a parameter:
$$\frac{du}{dy}= \frac{x}{\sqrt{1+ y^2}}$$
is separable
$$du= x\frac{dy}{\sqrt{1+ y^2}}$$

5. Feb 27, 2013

AntSC

I've managed to get it sussed. Thanks everyone for the help. Even if it was for me to see that i've written the question down wrong!