# 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

Staff Emeritus
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

Staff Emeritus
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!