Question

Solve the differential equation the product of 1 over 8 times x and dy dx equals y times the square root of the quantity 4 times x squared minus 1.

Answer 1

https://i.gyazo.com/6f4938df76b6199c384f7c5f9f42436e.png

Answer 2

@jim_thompson5910 @VeritasVosLiberabit

Answer 3

Separated the variables.

Answer 4

Then took the integral.

Answer 5

Answer should be C no?

Answer 6

That does seem to be the only answer that makes sense

Answer 7

@jim_thompson5910 can you confirm?

Answer 8

one moment

Answer 9

No problem.

Answer 10

\[\Large \frac{1}{8x} \frac{dy}{dx} = y\sqrt{4x^2-1}\]

\[\Large \frac{dy}{y} = 8x\sqrt{4x^2-1}dx\]

\[\Large \int \frac{dy}{y} = \int 8x\sqrt{4x^2-1}dx\]

\[\Large \ln(|y|) = \frac{2}{3}\sqrt{(4x^2-1)^3}+C\]

\[\Huge |y| = e^{{}^{\frac{2}{3}\sqrt{(4x^2-1)^3}+C}}\]

\[\Huge |y| = e^{{}^{\frac{2}{3}\sqrt{(4x^2-1)^3}}}*e^{C}\]

\[\Huge |y| = e^{C}*e^{{}^{\frac{2}{3}\sqrt{(4x^2-1)^3}}}\]

\[\Huge |y| = Ce^{{}^{\frac{2}{3}\sqrt{(4x^2-1)^3}}}\]

If y >= 0, then we can drop the absolute values

\[\Huge y = Ce^{{}^{\frac{2}{3}\sqrt{(4x^2-1)^3}}}\]

so the answer is actually A

Answer 11

Oooohhhh you bring the C along, god bless.

Answer 12

Well then, okay, thanks @jim_thompson5910

Answer 13

yes, `C` is a constant, so `e^C` is a constant too. Which is why I was able to replace `e^C` with just `C` to make things simpler.