Calculus help please

Question

thatonegirl_ · Answer

The original problem is dy/dx=x^2(1-y) and I have to find the particular solution. 

I checked my answer and work with the answer key and I had the last step the same:
$$\ln \left| 1-y \right|=-\frac{ x^3 }{ 3 }-c$$
But then the answer key jumped ahead to the answer and I didn't get the same thing.. So could someone help me solve this for y only?

thatonegirl_ · Answer

Ohh nvm i figured out what i did :')

thatonegirl_ · Answer

i made e^(-c)=a instead of -a

sunnnystrong · Answer

Okay so:

$$\frac{ dy }{ dx }=x^2(1-y)$$

Rearrange -->
$$\frac{ dy }{ (1-y) }=x^2dx$$

Integrate

$$\ln(1-y)=\frac{ x^3 }{ 3 }+C$$
Solve for y -->
$$1-y=e^{1/3x^3+C}$$
$$y=-e^{1/3x^3+C}+1$$

thatonegirl_ · Answer

Thank you sooo much, I appreciate it!!! :D

zepdrix · Answer

Hmm why would it be -a?
Exponential can't give you a negative value.

thatonegirl_ · Answer

idk but it worked?

sunnnystrong · Answer

@zepdrix can you help?

thatonegirl_ · Answer

I will show what i did

thatonegirl_ · Answer

$$e ^{\ln \left| 1-y \right|}=e ^{-\frac{ x^3 }{ 3 }-c}$$

thatonegirl_ · Answer

$$1-y=-ae ^{\frac{ -x^3 }{ 3 }}$$

thatonegirl_ · Answer

$$y=1+ae ^{\frac{ -x^3 }{ 3 }}$$

thatonegirl_ · Answer

solve for a using the initial condition (0,2) and a=1

sunnnystrong · Answer

Oh okay XD well you didn't post that part --> than its different

thatonegirl_ · Answer

$$y=1+e ^{\frac{ -x^3 }{ 3 }}$$

thatonegirl_ · Answer

Sorry I didn't even think about that D:

zepdrix · Answer

Ya I was forgetting about the absolute value :))$$\large\rm |1-y|=Ae ^{\frac{ -x^3 }{ 3 }}$$So ya, that definitely gives the potential for a negative A value,$$\large\rm 1-y=\pm Ae ^{\frac{ -x^3 }{ 3 }}$$Oh and you have initial data? cool :)

sunnnystrong · Answer

@zepdrix lol XD

thatonegirl_ · Answer

so does the negative a come from the absolute value or because it was e^-c ?

zepdrix · Answer

$\large\rm e^{-c}>0$ for all values of c.
So it has to come from the absolute value or more accurately from the context of the problem. Was it a physics problem of some sort, where a negative would only make sense?

thatonegirl_ · Answer

Oh okay thx for clearing tht up. And no it was just an old AP question that asked to find the particular solution of dy/dx=x^2(1-y) in the form of y=f(x)  with the initial condition f(0)=2

zepdrix · Answer

Well if the A is still in the equation, then you've only found the `general solution` up to this point. Did the book show the general solution with a negative A?

You would plug in your initial data in order to replace A with some particular value, giving you your particular solution.

thatonegirl_ · Answer

No it didn't even show the steps after it had the ln|1-y| part. That's why I was confused

thatonegirl_ · Answer

But when i made a negative i got the right answer

zepdrix · Answer

Oh weird :3

thatonegirl_ · Answer

Yup lol xD I'll ask my teacher monday about tht part

thatonegirl_ · Answer

Thanks for the help!

zepdrix · Answer

c: