Separation of variables IVP, book wrong?

[rygza]

dy/dx = 3[(y-1)^(1/3)]
with initial values: y(0)=1

ultimately I end up with y=sqrt((2x)^3) + 1 or ((2x)^3/2) + 1
book answer: y= 1+((3x)^3/2)

steps: separation of variables...
((y-1)^-1/3)dy = 3dx

after u-subs, where u = y-1...
3/2((y-1)^2/3) = 3x + C
3/2((y-1)^2/3)-3x = C

after plugging in intial values, C=0
(y-1)^2/3 = 2x then, y-1 = (2x)^3/2
finally... y = ((2x)^3/2) + 1

Teacher has found wrong answers in book before, but unfortunately we aren't told all of the answers that are wrong. Plz let me know if I've done something wrong, as I've been at this like crazy

 

[LCKurtz]

Your answer looks ok to me. But, to be sure, remember you can always plug your answer back into the original equation and initial value to see if it works.

 

[rygza]

Your answer looks ok to me. But, to be sure, remember you can always plug your answer back into the original equation and initial value to see if it works.

Into the dy/dx = ... part?
or the y = ... ?

 

[Mark44]

1. Evaluate your solution to see if y(0) = 1.
2. Substitute your solution into the differential equation dy/dx = 3[(y-1)^(1/3)]

 

[LCKurtz]

Into the dy/dx = ... part?
or the y = ... ?

You started with a differential equation. You have a proposed solution to it. Plug y and its derivative back into the differential equation and see if it works. Also check the initial condition works.

 

[rygza]

oh ok. I plugged into the y = ...
and got 1=1
also tried taking derivative of y (before plugging in initial conditions), substituting into the original equation, and after plugging in initial values got 0=0. checks out

Thanks guys