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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?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.

Also how do I thank you as I can't find the button

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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.

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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 doesnt have it and id like to have it before i finish the question

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

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

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).

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

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.

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