How would you solve this ODE

  • Thread starter pat666
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In summary, the problem is asking to solve the ODE x^3y'+4x^2y=1/x using an integrating factor or any other appropriate method. So far, the attempt has been made to use the product rule and separate variables, but neither has been successful in finding a solution. The equation can also be written as x^4y'+4x^3y=1, which may be useful in finding the solution.
  • #1
pat666
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Homework Statement



[tex]x^3y'+4x^2y=1/x [/tex]


Homework Equations


NA


The Attempt at a Solution


I've tried separation of variable but I can't get the ys on 1 side and the xs on the other.

Please help the exam is soon and I don't know what method to use?
 
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  • #2
It's a first-order differential equation, so you can always use an integration factor.
 
  • #3
How do I do that for the non homogenous ODE?
 
  • #4
As always, find solution to homogeneous equation. Then find particular solution. Add the two solutions together.
 
  • #5
Ok, ill try to figure it out. I can't find the y_p value for r(x) though. Do you know what it is for 1/x?
 
  • #6
hi pat666! :wink:

hint: the LHS is almost an exact integral, isn't it?

ok, multiply it by something to make it an exact integral (that's vela's :smile: integrating factor)
 
  • #7
Question. what is an exact integral?
 
  • #8
oops! :rolleyes:

i meant an exact derivative :redface:
 
  • #9
Suppose you had a general differential equation:
[tex]
p(x)\frac{dy}{dx}+r(x)y=r(x)
[/tex]
The first thing to do is divide through by p(x) to obtain:
[tex]
\frac{dy}{dx}+\frac{q(x)}{p(x)}y=\frac{r(x)}{p(x)}
[/tex]
Then the trick is to multiply through by:
[tex]
e^{\int\frac{q(x)}{p(x)}dx}
[/tex]
and note that:
[tex]
\frac{d}{dx}e^{\int\frac{q(x)}{p(x)}dx}=e^{\int \frac{q(x)}{p(x)}dx}\frac{q(x)}{p(x)}
[/tex]
and so the LHS of the equation can be written as:
[tex]
\frac{d}{dx}\left( e^{\int\frac{q(x)}{p(x)}dx}y\right) =\frac{r(x)}{p(x)}e^{\int\frac{q(x)}{p(x)}dx}
[/tex]
From this I think you can solve your equation.
 
  • #10
hunt_mat gave the general formula for the integrating factor. I perfer to think like this:
We are looking for a function u(x) so that multiplying by u(x) makes the left side of the equation,
[tex]x^3u(x)y'+ 4x^2u(x)y[/tex]
an "exact derivative". That is, it is of the form
[tex]\frac{d(x^3u(x)y)}{dx}= x^3u(x)y'+ 4x^2u[/tex]
by the product rule, that is the same as
[tex]x^3u(x)y'+ 3x^2uy+ x^3u'y= x^3u(x)y'+ 4x^2u[/tex]
so we want
[tex]3x^2u+ x^3u'= 4x^2u[/tex]

Then
[tex]x^3u'= x^2u[/tex]
which is a separable equation

[tex]\frac{du}{u}= \frac{dx}{x}[/tex]
Which is easily integrable.
 
  • #11
I still can't do this, when I try hund_mat's method I get x^4*y'+4x^3*y=1 I don't know how to solve this either?
 
  • #12
I need help urgently/ ODE

Homework Statement


Solve this ODE

[tex] x^3y'+4x^2y=1/x [/tex]

Separation of variables won't work and I can't find an integrating factor


Homework Equations





The Attempt at a Solution

 
  • #13
hi pat666! :smile:

(try using the X2 icon just above the Reply box :wink:)

hint: what is d/dx (x4y) ? :wink:
 
  • #14
That is 4x^3*y but where did that come from and what do I do with it?

Thanks
 
  • #15
ahhh!

nooo :redface: … use the product rule :wink:
 
  • #16
I'm not sure what you mean? how can the answer change by using a different rule?
 
  • #17
you didn't differentiate the "y" :wink:
 
  • #18


This is a linear first order differential equation. Substitute y by the product f˙g.

ehild
 
  • #19


What is f and g? Does that mean by the product of d/dx(x^3)*?

Thanks
 
  • #20


f(x) and g(x) are two arbitrary functions.
You have a linear differential equation y'+a(x)y=b(x) (The comma ' means d/dx)
Replace y=f(x)g(x), y'=f'g+fg':
f'g+fg'+a(x)fg=b(x), and choose f in such way that f'g+a(x)fg=0. Eliminate g. Solve for f (you need a particular solution only) and plug it into the rest of the original equation:
fg'=b(x) and find the general solution for g(x).
Try. :)

ehild
 
  • #21


Is it an exact differential equation? I looks like, if you multiple by x, you get x4y'+4x3y=1, let f(x, y)=x4y, and you can see df/dx=x4y'+4x3y=1.
 
  • #22
what he's saying is:
[tex]
\frac{d}{dx}(x^{4}y(x))=4x^{3}y(x)+x^{4}\frac{dy}{dx}
[/tex]
 
  • #23


This question was asked before, multiply through by x and notice that you can write the LHS as a total derivative.
 
  • #24


pat666 said:

Homework Statement


Solve this ODE

[tex] x^3y'+4x^2y=1/x [/tex]

Separation of variables won't work and I can't find an integrating factor


Homework Equations





The Attempt at a Solution


Two threads merged. Please do not multiple post the same question.
 

1. What is an ODE?

An ODE, or ordinary differential equation, is a mathematical equation that describes how a function changes over time. It involves an independent variable, such as time, and one or more dependent variables that are related to the independent variable by a derivative.

2. How do you solve an ODE?

There are several techniques for solving ODEs, including separation of variables, integrating factors, and series solutions. The specific method used depends on the type of ODE and its characteristics, such as linearity and homogeneity.

3. What are initial conditions in an ODE?

Initial conditions are the values of the dependent variables at a specific initial point in time. They are necessary for solving an ODE because they provide a starting point for the function and its derivatives.

4. How do you know if you have solved an ODE correctly?

To check if an ODE has been solved correctly, you can substitute the solution into the original equation and see if it satisfies the equation. Additionally, you can plot the solution and compare it to the given initial conditions to ensure it follows the expected behavior.

5. Can ODEs be solved analytically or numerically?

ODEs can be solved using both analytical and numerical methods. Analytical solutions involve finding an exact algebraic expression for the solution, while numerical solutions involve approximating the solution using numerical techniques such as Euler's method or Runge-Kutta methods.

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