Solving Separable Equations: Tips and Tricks

  • MHB
  • Thread starter Kris1
  • Start date
  • Tags
    Separable
In summary, the conversation was about separating a differential equation into a separable equation. The order in which the variables are separated does not matter as long as they are on different sides. The expert provided guidance on how to separate the variables and showed how to get the final solution through integration. The final solution is y^2 = (2(4x+1))/(C(4x+1)+27) with the possibility of a trivial solution being y = 0.
  • #1
Kris1
29
0
( 4*x+1 )^2 dy/dx = 27*y^3

I'm trying to separate this into a separable equation. Does it matter which way I do it? I.e taking all xs to the left or all ys to the left or does it not matter as long as x and y are on different sides?
 
Physics news on Phys.org
  • #2
As long as you have:

\(\displaystyle f(y)\,dy=g(x)\,dx\) or \(\displaystyle f(x)\,dx=g(y)\,dy\) it does not matter. I tend to like this first form, with $y$ on the left and $x$ on the right, but that's just the way I was taught.
 
  • #3
MarkFL said:
As long as you have:

\(\displaystyle f(y)\,dy=g(x)\,dx\) or \(\displaystyle f(x)\,dx=g(y)\,dy\) it does not matter. I tend to like this first form, with $y$ on the left and $x$ on the right, but that's just the way I was taught.

Hi Mark, I have tried separating the variables yet I can't figure out how to move the bracketed term across. Could you advise how to do this and show what the final solution would be?

Also how do I thank you as I can't find the button
 
  • #4
I will help you separate the variables, and then guide you to get the solution...you will get more from the problem that way. (Sun)

We are given:

\(\displaystyle (4x+1)^2\frac{dy}{dx}=27y^3\)

See what you get when you divide through by \(\displaystyle (4x+1)^2y^3\) (bearing in mind that in doing so we are eliminating the trivial solution $y\equiv0$).

edit: You should see a Thanks link at the lower right of each post, except your own.
 
  • #5
Thanks I can get the answer from here just wasn't sure if you could divide through by the whole bracket :)
 
  • #6
dy/(27*y^3)=dx/((4*x+1)^2)

This is my final separation can you please tell me if this is the correct result. Can this be simplified down even further?

At this point it is okay to integrate right?edit: Do you know what the answer will be for the integration because my work sheet doesn't have it and id like to have it before i finish the question
 
  • #7
Kris said:
dy/(27*y^3)=dx/((4*x+1)^2)

This is my final separation can you please tell me if this is the correct result. Can this be simplified down even further?

At this point it is okay to integrate right?edit: Do you know what the answer will be for the integration because my work sheet doesn't have it and I'd like to have it before I finish the question.

Yes, that's fine, although I would probably choose to write:

\(\displaystyle y^{-3}\,dy=27(4x+1)^{-2}\,dx\)

I would use a $u$-substitution on the right side:

\(\displaystyle u=4x+1\,\therefore\,du=4\,dx\) and we have (after multiplying through by 4):

\(\displaystyle 4\int y^{-3}\,dy=27\int u^{-2}\,du\)

Now, we are just a couple of steps from the solution (and don't forget to include the trivial solution we eliminated when separating variables when you state the final solution, if this trivial solution is not included in the general solution for a suitable choice of the constant of integration).
 
  • #8
Ive got a solution at y = -54 + c but i don't think this is right

I went with -1/2*y^2 * y = -1/u * u? as my integrations and then rearranged

Is this the correct answer or have I integrated something wrong which leads to my poor solution?
 
  • #9
Using the power rule for integration, and using the form in my post above, you should get:

\(\displaystyle 4\left(\frac{y^{-2}}{-2} \right)=27\left(\frac{u^{-1}}{-1} \right)+C\)

Multiply through by -1 and simplify a bit:

\(\displaystyle 2y^{-2}=27u^{-1}+C\)

Note: the sign of the parameter $C$ does not change as it can be any real number, negative or positive.

At this point, I would rewrite using positive exponents, and combine terms on the right:

\(\displaystyle \frac{2}{y^2}=\frac{Cu+27}{u}\)

Invert both sides:

\(\displaystyle \frac{y^2}{2}=\frac{u}{Cu+27}\)

\(\displaystyle y^2=\frac{2u}{Cu+27}\)

Back-substitute for $u$:

\(\displaystyle y^2=\frac{2(4x+1)}{C(4x+1)+27}\)

This is the general solution, and the only way we can get the trivial solution is for:

\(\displaystyle 4x+1=0\)

but we eliminated that possibility during the separation of variables as well.
 
  • #10
Thanks so much :) I see where i went wrong because i tried to integrate and invert before I multiplied out. It makes much more sense to multiply and flip then turn into a positive rather than trying to do it all at once :)
 

FAQ: Solving Separable Equations: Tips and Tricks

1.

What is a separable equation?

A separable equation is a type of differential equation where the independent variable and dependent variable can be separated and written on opposite sides of the equation. This makes it easier to solve the equation and find the solution.

2.

How do you know if an equation is separable?

A separable equation can be identified by the variables being separated on opposite sides of the equation and there being no cross-terms with both variables present in the same term.

3.

What is the process for solving a separable equation?

The process for solving a separable equation involves separating the variables, integrating both sides, and then solving for the constant of integration. Finally, the solution is checked for validity by plugging it back into the original equation.

4.

What are some common applications of separable equations in science?

Separable equations are commonly used in physics, engineering, and other scientific fields to model natural phenomena such as radioactive decay, population growth, and chemical reactions.

5.

Are there any limitations to using separable equations?

While separable equations can be useful in many situations, they may not always provide an accurate representation of a system. For example, they do not take into account external factors or interactions between variables, so they may not be appropriate for more complex systems.

Similar threads

Back
Top