Question

calc 1 problem, solve the differential equation y'=3y(x+1)^2/(y-1) HELP PLEASEEEEEEEEEEEE

Answer 1

It appears we want to separate our variables...
Change y' for dy/dx. Then get all the y's on the left side with dy and the x's with dx on the right. That allows us to integrate each side. Is that part clear?

Answer 2

\( \dfrac{dy}{dx} = \dfrac{3y (x + 1)^2}{y - 1} \\
\dfrac{y - 1}{y} \ dy = 3(x + 1)^2 \ dx \qquad \text{Distribute 1/y to y and -1.}\\
\left(1 - \dfrac{1}{y} \right) \ dy = 3 (x + 1)^2 \ dx \)

Answer 3

And then we can integrate both sides...
\( \displaystyle \int \left( 1 - \dfrac{1}{y} \right) \ dy = \int 3(x+1)^2 \ dx \)

Answer 4

yeah i got to the integrals

Answer 5

You know how to evaluate each integral?

Answer 6

i evaluated the integral on the right side i cannot figure out how to do the left

Answer 7

right now I have y - ln|y| = x^3 +3x^2+3x

Answer 8

I think you're just missing the constant of integration +C
Otherwise that looks correct to me. I don't believe you can solve for y in this case, so it has to be left as an implicit solution.

Answer 9

so the answer is y - ln|y| = x^3 +3x^2+3x + c?

Answer 10

I wolframed this as well and this is the answer y(x) = c_1+x^3+3 x^2+2 x not sure how to get it got that 2x

Answer 11

I'm not sure you entered it correctly there...
http://www.wolframalpha.com/input/?i=y%27+%3D+3y+%28x%2B1%29%5E2+%2F+%28y+-+1%29
Gives y(x) = -W(e^(-x^3 - 3x ^2 - 3x) Where W is the lambert w function, or product log function.

Answer 12

oh alright thanks but y - ln|y| = x^3 +3x^2+3x + c is the answer?

Answer 13

Yup. Unless you know what the W function is, there is no way to simplify that to get y alone. :)

Answer 14

thank you so much i've been stuck on this problem for like 2 hours trying to figure out how to get y alone thank thanks thanks

Answer 15

In general I think if you have a y and some exponential/logarithm of y added together, it is likely impossible to continue solving for y. Since any operation you do just shifts one to y and the other to the exponential of y/logarithm of y again.

Answer 16

Glad to help, though! :)