# Differential eq (implicit sol)

1. Sep 20, 2008

### Qyzren

1. The problem statement, all variables and given/known data
i have a differential equation.

∂h/∂t + [g sin a h²/v]∂h/∂x = 0. where h = h(x,t).
i need to show by substitution that the (implicit) general solution for h is h = f(x - (g sin a h²t/v)) where f is an arbitrary differential function of a single variable.

3. The attempt at a solution
∂h/∂x = ∂f/∂x
∂h/∂t = ∂f/∂t * (-g sin a h²/v)
subbing in gives
∂f/∂t*(-g sin a h²/v) + (g sin a h²/v)*∂f/∂x = 0
cancelling g sin a h²/v
-∂f/∂t + ∂f/∂x = 0
so now i have to show... ∂f/∂t = ∂f/∂x ???
seems like i'm going in circles...

any help will be appreciated

2. Sep 20, 2008

### xaos

try:

h(x,t)=f(y(x,t))

3. Sep 20, 2008

### bdforbes

I'm not sure where the brackets should be, so let me just confirm that this is the problem:

$$\frac{\partial h}{\partial t} + \frac{gsin(a)h^2}{v}\frac{\partial h}{\partial x} = 0$$

$$h=f(x-\frac{gsin(a)h^{2}t}{v})$$

4. Sep 20, 2008

### bdforbes

Okay I worked through it, that must be the correct form. Your problem is that you're interpreting f as a function of two variables. As xaos said, write h=f(y(x,t)). Then:

$$\frac{\partial h}{\partial t} = f'(y)\frac{\partial}{\partial t}(y)$$

5. Sep 20, 2008

### HallsofIvy

Staff Emeritus
f is a function of a single variable so "$\partial f/\partial x$" and "$\partial f/\partial t$" are meaningless. Having that "h" inside f makes it complicated: More correctly
$$\partial h/\partial x= f'(x- (g sin(ah^2t/v))(1- 2aght/v)(-g cos(ah^2t/v))(2aht/v)\partial h/\partial x$$
and
$$\partial h/\partial t= f'(x- (g sin(ah^2t/v)(-2aght/v)(-gcos(ah^2t/v))((2aht/v)\partial h/\partial t+ ah^2/v)$$