Solving Separable Equations: Tips and Tricks

[Kris1]

( 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?

 

[MarkFL]

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.

 

[Kris1]

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

 

[MarkFL]

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.

 

[Kris1]

Thanks I can get the answer from here just wasn't sure if you could divide through by the whole bracket :)

 

[Kris1]

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

 

[MarkFL]

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).

 

[Kris1]

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?

 

[MarkFL]

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.

 

[Kris1]

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 :)