How can I solve this Bernoulli Equation with a given initial condition?

Of course the reason is that the set of solutions is a vector space. This is also why you can get the general solution by adding the fundamental system of solutions (to the homogeneous equation).In summary, the conversation discusses the solution process for the Bernoulli equation x^2y' + 2xy = 5y^3. The first step is to substitute v = y^-2 and y = v^-1/2, which leads to the homogeneous equation v' - 4v/x = 0. The general solution for this equation is v = Cx^4, and a particular solution for the inhomogeneous equation is v = 2/x. By adding these two solutions, the final solution is found to be
  • #1
CaptainJames
18
0

Homework Statement


x^2y' + 2xy = 5y^3

Homework Equations


Bernoulli Equation

The Attempt at a Solution


so... v=y^-2, y=v^(-1/2) y' = -1/2v'v^(-3/2)

moving everything over...
y'/y^3 + 2/(xy^2) = 5/x^2

and plugging everything in...
-1/2v' + 2v/x = 5/x^2
v' - 4v/x = -10/x^2

v' - 4v/x = 0
dv/dx = 4v/x
dv/v = 4dx/x
ln v = 4 ln x + C
v = C(x)x^4
v' = C'(x)x^4 + C(x)4x^3

C'(x)x^4 + C(x)4x^3 - 4C(x)x^4/x = 5/x^2
C'(x)x^4 = 5/x^2
C'(x) = 5/x^6
C(x) = -x^-5 + C

v = (-x^-5 + C)x^4

y = 1/(v)^(1/2)

y^2 = 1/((-x^-5 + C)x^4)
... which is wrong. Could someone point me in the right direction?
 
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  • #2
CaptainJames said:

Homework Statement


x^2y' + 2xy = 5y^3

Homework Equations


Bernoulli Equation

The Attempt at a Solution


so... v=y^-2, y=v^(-1/2) y' = -1/2v'v^(-3/2)

moving everything over...
y'/y^3 + 2/(xy^2) = 5/x^2

and plugging everything in...
-1/2v' + 2v/x = 5/x^2
v' - 4v/x = -10/x^2

v' - 4v/x = 0
dv/dx = 4v/x
dv/v = 4dx/x
ln v = 4 ln x + C
v = C(x)x^4

I think you are doing great up to the last line here. Why the x dependence in C?
I would just say v=C*x^4. This the general solution to the homogeneous equation v'-4v/x=0. Now you just want to add this to a particular solution (any one) of the inhomogeneous equation v' - 4v/x = -10/x^2. Can you find one?
 
  • #3
I think 2/x is a particular solution?

cause... you got, let's see...

-2/x^2 - 4*2/x^2 = -10/x^2

Okay so now what...??

Cx^4 + 2/x?

v=Cx^4 + 2/x

y = v^(-1/2)

y^2 = 1/(Cx^4+2/x)

or...

y^2 = x/(Cx^5 + 2) which is... CORRECT! thanks a bunch!

So all you have to do is find the general solution, then a particular solution? When is this applicable (only in cases where there is one derivative??)?
 
Last edited:
  • #4
Finding a "general" solution (to the corresponding homogeneous equation), then adding a specific solution (to the entire equation), works only for linear equations. The Bernoulli equation is not linear (because of the y3) but the equation you get for v,v' - 4v/x = -10/x2 is linear.
 
  • #5
Nope, any number of derivatives. You only need that the ODE is linear. So you CAN add solutions and still get a solution.
 

1. What is the Bernoulli Equation?

The Bernoulli Equation is a fundamental principle in fluid mechanics that describes the relationship between pressure, velocity, and elevation in a fluid flow. It states that as the velocity of a fluid increases, the pressure decreases, and vice versa.

2. How is the Bernoulli Equation derived?

The Bernoulli Equation is derived from the principle of conservation of energy, where the total energy of a system remains constant. It takes into account the kinetic energy, potential energy, and internal energy of the fluid.

3. What are the applications of the Bernoulli Equation?

The Bernoulli Equation has various applications in engineering and physics, including the study of fluid dynamics, aerodynamics, and hydraulics. It is used in designing pipes, pumps, and aircraft wings, as well as in calculating the velocity of fluids in different systems.

4. What are the limitations of the Bernoulli Equation?

The Bernoulli Equation assumes that the fluid is incompressible, inviscid, and flows without turbulence. It also neglects factors such as friction, viscosity, and external forces, which can affect the accuracy of its predictions.

5. How can I use the Bernoulli Equation to solve problems?

To use the Bernoulli Equation to solve problems, you first need to identify the variables involved, such as pressure, velocity, and elevation. Then, you can use the equation to calculate the unknown variable, assuming the other variables are known. It is important to consider the limitations of the equation and make necessary adjustments for a more accurate solution.

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