Differential eq (implicit sol)

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Homework Help Overview

The discussion revolves around a differential equation given by ∂h/∂t + [g sin a h²/v]∂h/∂x = 0, where h is a function of both x and t. The original poster seeks to demonstrate that the implicit general solution can be expressed as h = f(x - (g sin a h²t/v), with f being an arbitrary function of a single variable.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss substituting expressions for ∂h/∂x and ∂h/∂t in terms of the function f. There is a focus on the implications of treating f as a function of a single variable versus multiple variables. Some participants express uncertainty about the correct interpretation of the function f and its derivatives.

Discussion Status

Several participants are actively engaging with the problem, attempting substitutions and clarifying the form of the function f. There is recognition of potential confusion regarding the variables involved and the derivatives of f, but no explicit consensus has been reached on the next steps.

Contextual Notes

Participants note the complexity introduced by having h within the function f, which complicates the differentiation process. There is also mention of the need to clarify the placement of brackets in the equation, indicating potential ambiguity in the problem setup.

Qyzren
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Homework Statement


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.


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
 
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try:

h(x,t)=f(y(x,t))
 
I'm not sure where the brackets should be, so let me just confirm that this is the problem:

[tex]\frac{\partial h}{\partial t} + \frac{gsin(a)h^2}{v}\frac{\partial h}{\partial x} = 0[/tex]

[tex]h=f(x-\frac{gsin(a)h^{2}t}{v})[/tex]
 
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:

[tex]\frac{\partial h}{\partial t} = f'(y)\frac{\partial}{\partial t}(y)[/tex]
 
Qyzren said:

Homework Statement


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.


The Attempt at a Solution


∂h/∂x = ∂f/∂x
∂h/∂t = ∂f/∂t * (-g sin a h²/v)
f is a function of a single variable so "[itex]\partial f/\partial x[/itex]" and "[itex]\partial f/\partial t[/itex]" are meaningless. Having that "h" inside f makes it complicated: More correctly
[tex]\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[/tex]
and
[tex]\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)[/tex]

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
 

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