Frustrating Bernoulli Equation

In summary: This is easy (but slightly messy) to do, but it's even easier to guess and verify a particular solution as in the earlier method.In summary, we have the equation \frac{du}{dx} - 3u = -3x which can be solved either by finding a general solution of \frac{du}{dx} - 3u = 0 plus a particular solution of \frac{du}{dx} - 3u = -3x, or by finding an integrating factor and using integration by parts to solve the resulting equation. Both methods lead to the same solution u(x) = Ce^{3x} + x + 1/3.
  • #1
1d20
12
0
I've been unable to fully solve this: [itex]\frac{dy}{dx} + y = xy^4[/itex]

The U: [itex]u = y^{-3}[/itex], so [itex]y = u^\frac{-1}{3}[/itex], and [itex]\frac{dy}{dx} = \frac{-1}{3}u^\frac{-4}{3}\frac{du}{dx} [/itex]

The substitution: [itex]\frac{-1}{3}u^\frac{-4}{3}\frac{du}{dx} + u^\frac{-1}{3} = xu^\frac{-4}{3}[/itex]

Simplified: [itex]\frac{du}{dx} - 3u = -3x[/itex]

AKA: [itex]\frac{du}{dx} + 3x = 3u[/itex]

I've tried solving the resulting equation as a linear:

[itex]p(x) = 3x[/itex], so the integrating factor is [itex]e^{\frac{3}{2}x^2}[/itex].

Which creates this unworkable equation: [itex]e^{\frac{3}{2}x^2}\frac{du}{dx} + 3xe^{\frac{3}{2}x^2} = 3ue^{\frac{3}{2}x^2}[/itex]

And I've tried solving it as a homogenous ([itex]3udx - 3xdx - du = 0[/itex]):

[itex]u = vx[/itex], so [itex]du = vdx + xdv[/itex]

Subing those in creates this unworkable mess: [itex]3vxdx - 3xdx - vdx - xdv = 0[/itex]

Which leaves me stuck, because I've only learned to solve separable, exact, homogenous, linear and bernoullis.
 
Physics news on Phys.org
  • #2
The equation is most easily solved at this stage: [itex]\frac{du}{dx} - 3u = -3x[/itex]

The integrating factor is just [itex]e^{-3x}[/itex]. Solve from there.
 
  • #3
1d20 said:
Simplified: [itex]\frac{du}{dx} - 3u = -3x[/itex]

AKA: [itex]\frac{du}{dx} + 3x = 3u[/itex]

I've tried solving the resulting equation as a linear:

[itex]p(x) = 3x[/itex], so the integrating factor is [itex]e^{\frac{3}{2}x^2}[/itex].

You don't need an integrating factor here. This is a linear equation with constant coefficients. Find the general solution of
[itex]\frac{du}{dx} - 3u = 0[/itex]
plus a particular integral of
[itex]\frac{du}{dx} - 3u = -3x[/itex]
which is going to be of the form u(x) = a + bx, for some constants of a and b.
 
  • #4
Both methods solve the equation in about 2 lines, but yeah you can do complimentary function and particular integral if you prefer.
 
  • #5
Also the particular integral is u(x) = a - 3x. You don't get two arbitrary constants from a first order ODE
 
  • #6
Sorry it's a + x
 
  • #7
Ben M said:
Also the particular integral is u(x) = a - 3x. You don't get two arbitrary constants from a first order ODE

You don't get any arbitary constants in a particular integral. My a and b were values to be found, not arbitrary constants.

Ben M said:
Sorry it's a + x

No, see above. N.B. I was trying not to give the OP the complete answer to something that might be a homework question.
 
  • #8
Thanks for the help!

A final question: I thought all integrating factors must have the p(x) part integrated, but this seems to be an exception. How come?
 
  • #9
1d20 said:
A final question: I thought all integrating factors must have the p(x) part integrated, but this seems to be an exception. How come?

The way I did it is a standard method, but maybe you haven't learned it yet. This logic behind it is:
For a linear equation like this one, any solution of ##du/dx -3u = -3x## is the sum of the solutions of two equations, the "given" equation
##du/dx -3u = -3x##
and the simpler equation
##du/dx - 3u = 0##
You can easily find the general solution of the equation with 0 on the right hand side, and that solution ##Ce^{3x}## contans an arbitrary constant.

You know there is only one arbitrary constant in the complete solution, so you don't need another one. Instead of finding the general solution of ##du/dx -3u = -3x##, now we only need one particular solution. You can usually guess the form of a particular solution from the function on the right hand side, and then equate the coefficients of different terms in the equation to find the unknown values.

In this case, we can guess that ##u = ax + b## is a solution for some values of a and b. Then ##du/dx = a##, and the substituting in the differential equation we have
##a -3(ax + b) = -3x##
To satisfy that equation for all values of ##x##, the constant terms and the coefficients of ##x## must both cancel out. So we have
##a - 3b = 0##
and
##-3ax = -3x## for all ##x##
which gives ##a = 1## and ##b = 1/3##
and the general solution is ##u = Ce^{3x} + x + 1/3##.
That method might seem longwinded (or even sneaky) but the real benefit comes because most of the "interesting" differential equations in physics and engineering are second order not first order, and finding integrating factors for second order equations is hard. (And most of the longwindedness was explaining it, not actually doing it).

Alternatively you can solve this equation using an integrating factor without any "trickery".
##du/dx -3u = -3x##
The integrating factor is ##e^{-3x}##
##e^{-3x}du/dx - 3u e^{-3x} = -3xe^{-3x}##
##d/dx (ue^{-3x}) = -3xe^{-3x}##
##ue^-3x = -3\int x e^{-3x}\,dx##
And you need to integrate by parts to do the integral on the right hand side.
 

1. What is the Bernoulli Equation and why is it frustrating?

The Bernoulli Equation is a fundamental equation in fluid mechanics that relates the pressure, velocity, and elevation of a fluid. It can be frustrating because it is often difficult to solve analytically and requires simplifying assumptions that may not always hold true in real-world applications.

2. Can the Bernoulli Equation be applied to all types of fluids?

No, the Bernoulli Equation is derived for ideal fluids, which are assumed to be incompressible, inviscid, and have no turbulent flow. In reality, most fluids do not meet these criteria, so the Bernoulli Equation may not accurately describe their behavior.

3. How can the Bernoulli Equation be used in practical applications?

The Bernoulli Equation can be used to analyze and predict the behavior of fluids in various systems, such as in aerodynamics, hydraulics, and ventilation. It can also be used to design and optimize systems for efficiency.

4. Are there any limitations to the Bernoulli Equation?

Yes, the Bernoulli Equation has several limitations, including its assumptions of ideal fluid behavior and steady, laminar flow. It also does not take into account factors such as viscosity, friction, and turbulent flow, which can significantly affect the behavior of real fluids.

5. How can one overcome the frustrations of the Bernoulli Equation?

One way to overcome the frustrations of the Bernoulli Equation is to use numerical methods and computer simulations to solve it. These methods can take into account more complex factors and provide more accurate results. It is also important to carefully consider the assumptions and limitations of the equation and apply it appropriately in practical applications.

Similar threads

Replies
2
Views
986
Replies
2
Views
1K
  • Differential Equations
Replies
27
Views
2K
  • Calculus and Beyond Homework Help
Replies
19
Views
672
  • Differential Equations
Replies
14
Views
2K
Replies
7
Views
2K
  • Differential Equations
Replies
9
Views
2K
  • Differential Equations
Replies
1
Views
5K
  • Differential Equations
Replies
7
Views
2K
  • Differential Equations
Replies
3
Views
1K
Back
Top