Diffeq initial value problem

[Guest]

dy/dx=3y-6-(y-2)^(1/3) y(1)=2

Does the existence and uniqueness theorem guarantee a unique solution? Explain, and show that there are two distinct solutions.


I understand why there is no unique solution guaranteed, but in showing that there are two distinct solutions I did the following:

Solve the seperable equation dy/(3y-6-(y-2)^(1/3))=dx
which equals: 1/2*ln(3(y-2)^(2/3)-1)=x + C
when I used my calculator to solve for y in that equation (I was too lazy to do it myself), it gave me: y=2 +/- (sqrt(3)(e^2x + 1)^(3/2)) / 9

yet my teacher told me this wasn't the right answer. What did I do wrong?

 

[galactus]

Did you forget to use your initial conditions and just give a general

solution?.

 

[Guest]

You mean solve for C? No, I didn't do that, but that wasn't what the problem asked.
He just wrote "your solution is incorrect". I'm guessing maybe my calculator put it in a format he didn't recognize.

 

[Guest]

Nobody can help me with this? Maybe verify my answer is right?

 

[Matt]

Solve the seperable equation dy/(3y-6-(y-2)^(1/3))=dx
which equals: 1/2*ln(3(y-2)^(2/3)-1)=x + C

You're okay up to here.

when I used my calculator to solve for y in that equation...
it gave me: y=2 +/- (sqrt(3)(e^2x + 1)^(3/2)) / 9

But I don't know how you found these solutions. Neither of them satisfies the initial condition y(1)=2. Double-check your work (without using the calculator).