Differential eq (implicit sol)

In summary: 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})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:\
  • #1
Qyzren
44
0

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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  • #2
try:

h(x,t)=f(y(x,t))
 
  • #3
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]
 
  • #4
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]
 
  • #5
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
 

Related to Differential eq (implicit sol)

1. What is a differential equation?

A differential equation is a mathematical equation that describes the relationship between a function and its derivatives. It is commonly used to model physical phenomena in fields such as physics, engineering, and biology.

2. What is an implicit solution to a differential equation?

An implicit solution to a differential equation is a general solution that is expressed in the form of an equation rather than a specific function. It may contain one or more arbitrary constants that need to be determined using initial conditions.

3. How do you solve a differential equation implicitly?

To solve a differential equation implicitly, you can use techniques such as separation of variables or substitution. These methods involve manipulating the equation to isolate the dependent and independent variables, and then integrating both sides to find the general solution.

4. What is the difference between implicit and explicit solutions to a differential equation?

The main difference between implicit and explicit solutions to a differential equation is the form in which they are expressed. An explicit solution is in the form of a specific function, while an implicit solution is in the form of an equation. Implicit solutions are generally more general and complex than explicit solutions.

5. Can you give an example of an implicit solution to a differential equation?

One example of an implicit solution to a differential equation is the general solution to the first-order linear differential equation: dy/dx = y + x + 1. The implicit solution is y = Ce^x - x - 2, where C is an arbitrary constant. This solution satisfies the given differential equation for any value of C.

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