Question

solve the initial value

y'=(x^3)(1-y), y(0)=3

Answer 1

Any ideas on how you would start this one? We first need to solve the DE, and then we use the initial value for finding the constant of variation. :)

Answer 2

I believe im supposed to separate the variables

Answer 3

Yes. There are two distinct functions of x and y that are being multiplied, so we can separate this.

Answer 4

\( \displaystyle \frac{\text{d}y}{\text{d}x} = x^3 \left( 1 - y\right) \)
Put the y's with the dy, the x's with the dx.
\( \displaystyle \frac{1}{1 - y} \; \text{d}y = x^3 \; \text{d}x \)

Answer 5

then integrate correct?

Answer 6

Yes. We use indefinite integration on both sides, which will produce that constant of variation in the end.

Answer 7

okay let me see
so
\[-\ln(1-y)=x^4/4\]

Answer 8

Looks almost good, we still have to account for that constant of variation for integrating:
\( \displaystyle -ln (1 - y) = \frac{1}{4} x^4 \color{green}{+ C} \)
The idea of that constant is that we will be solving for it using our initial value problem.

Answer 9

ah yes sorry...so next step..plug in the initial value for y and solve for x?

Answer 10

or do I exponential the eq

Answer 11

We could solve for y first and then plug it in. I think if we had used absolute values for the natural log after integrating, that would also work. Otherwise, it looks like we get 1 - 3 = -2 on the natural log.

\( \displaystyle -\ln \lvert 1 - y \rvert = \frac{1}{4} x^4 + C \)

Answer 12

If you are not entirely familiar with the ln with absolute values, I looked it up and it is explained fairly well here:
http://clem.mscd.edu/~talmanl/PDFs/APCalculus/OnAnIntegral.pdf

Anyways, if we use the absolute values, we can solve for c immediately.
\( \displaystyle - \ln \lvert 1 - y \rvert = \frac{1}{4} x^4 + C \), \(y(0) = 3 \)
\( \displaystyle - \ln \lvert 1 - 3 \rvert = \frac{1}{4} \times 0 + C \)
\ \( \displaystyle - \ln \lvert -2 \rvert = C \)
\( - \ln 2 = C \)

Like that.

We can then plug this back into the equation while solving for y.

Answer 13

so then i plug in -ln2 for c and solve for y ..ending the problem?

Answer 14

Yes.

Sorry, I was looking at how the absolute values would affect the end solution. Since to be mindful of the absolute values means to account for the positive and negative cases. I figure it would be easier to just avoid it entirely and solve for y before finding the constant of variation.

Answer 15

\( \displaystyle - \ln \lvert1 - y \rvert = \frac{1}{4} x^4 + C_1 \)
\( \displaystyle \ln \lvert 1 - y \rvert = - \frac{1}{4} x^4 + C_2 \)
\( \displaystyle \lvert 1 - y \rvert = e^{-\frac{1}{4} x^4 + C_2} \)
\( \displaystyle 1 - y = \pm e^{-\frac{1}{4} x^4 + C_2} \)
\( \displaystyle y - 1 = \mp e^{-\frac{1}{4} x^4 + C_2} \)
\( \displaystyle y = \color{#aa5555}{\mp e^{C_2}} e^{-\frac{1}{4} x^4} + 1\)
\( \displaystyle y = \color{#aa5555}{A} e^{-\frac{1}{4} x^4} + 1\)

C1 = -ln 2, what we found. C2 = ln 2, opposite of C1,

Answer 16

I suppose it is easiest to ignore the negative half and just use y = +e^(C_2) e^(-1/4 x^4) + 1. Sorry, I feel like I am over-complicating something here. >.>

Answer 17

If we had solved in the last equation for A, it will just turn out to be A=2, which is e^(ln 2).

Answer 18

Basically, we should end up with \( \displaystyle y = 2 e^{- \frac{1}{4} x^4} + 1\) for a solution.

Answer 19

thank you...awesome explanation

Answer 20

You're welcome! I'm glad it makes sense. I haven't actually run into any situations where absolute values mattered in a DE until now, it was a bit weird. :P