Equation

[vast sail]

Does someone know how to solve it? Or tell me the metod? Or tell me how
Top left is the equation, down left is the initial conditions and right is the result

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[cold marlin]

Express the second derivative, then integrate twice.

[gritty oracle]

check that this Vx satisfies the definition

there is a typo tho, what they mean in the usual sense is

$$ \frac{d^2 V_x}{dy^2} $$

[mossy summit]

aL

[cold marlin]

Ah, yeah.

[cold marlin]

Or that, yeah.

Though, it's still pretty easy to solve.

[gritty oracle]

@vast sail

[vast sail]

I Still don't how to do it

[vast sail]

Yes that's what I meant

Sorry if the image was wrong

[cold marlin]

Well, we have:
d^2 v/dy^2 = -ΔP/(μL)
First, integrate this once.

[vast sail]

By variable separation?

[cold marlin]

You don't need to do that, you can just integrate.

[vast sail]

How

I don't know how to do it without variable separation

[cold marlin]

Well, just integrate...
If you have d^2 y/dx^2 = A, then you get dy/dx = Ax + C.

[gritty oracle]

$$ -\mu\frac{dV_x}{dy} = \frac{\Delta P}{L}y + C_1 $$

[mossy summit]

aL

[vast sail]

$$\frac-{ΔP}{µL}$$ = $$ \frac{dV_x}{dy} $$

[mossy summit]

Gluwxy
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[vast sail]

Hmm

Anyway

You mean this?

• C1

[gritty oracle]

that just means ... = V_x + C

which is not true

integral and derivative cancel each other

[vast sail]

So?

[cold marlin]

Hm... Well, alright. Then do this:
d(dv/dx) = (-ΔP/(μL))dx
Then integrate.

[vast sail]

[cold marlin]

A little odd, though. Surely you know that ∫((dy/dx)dx) = y + C.

[vast sail]

Oh I see just eliminate integral symbol

[vast sail]

Whith the first image I sent or the second one

Of the result in my notebook

[gritty oracle]

$$ -\mu \frac{d^2 V_x}{dy^2} = \frac{\Delta P}{L} \Rightarrow -\mu\frac{dV_x}{dy} = \frac{\Delta P}{L}y + C_1 $$

[mossy summit]

aL

[gritty oracle]

that's what you should get after integrating once

[vast sail]

Why you kept µ in the left side

[gritty oracle]

la definicion

[cold marlin]

Doesn't really matter.

[cold marlin]

It's a constant, we can keep it on either side.

[gritty oracle]

integrate again to get

$$ -\mu V_x = \frac{\Delta Py^2}{2L} + C_1y + C_2 $$

[mossy summit]

aL

[gritty oracle]

and apply

[cold marlin]

It's probably better to apply the initial condition first, then integrate.

[gritty oracle]

the initial conditions

[cold marlin]

Then we won't get an equation with two constants.

[vast sail]

Sure

[gritty oracle]

how would you apply IC after integrating once?

just because you know Vx = 0 says nothing about how the derivative behaves at that point

[cold marlin]

Ah, sorry! I didn't notice. We have boundary conditions, not really initial ones.

[vast sail]

Are you sure that it's not a diferential equation?

By order reduction

[cold marlin]

Well, of course this is a differential equation. An elementary one, though.

Solved by just integrating.

[gritty oracle]

when was this ever in doubt?

[vast sail]

Are you sure that the result of doing that metod is the same as the result in the image I sent?

[gritty oracle]

which image

[vast sail]

Well is because someone told me that is by order reduction

[vast sail]

This one

Where the result is the one in the top right

[gritty oracle]

which is another way of saying "integrating"

you start with 2nd order equation, you integrate, order reduces to 1

integrate again, order reduces to 0

[gritty oracle]

I showed you two ways of "demonstrating the solution" as is required by the exercise

[vast sail]

But with 2 constants how i'll apply the CI