Solving Problem on AP Calc Test: Was My Answer Wrong?

[rpc]

One of the problems on my AP Calc test:

The point (1,9) lies on the graph of an equation y=f(x) for which dy/dx = 4x*y^(1/2) where x> or = to 0 and y > or = 0

When x=0 y=?

separation of variables:

dy/y^(1/2) = 4x dx

Integrate :

2 y^(1/2) = 2x^2 +C now, if you do C now:

2 * (9)^(1/2) = 2 (1)^2 + C
6 = 2 + C
C= 4 ,

plug in 0 for X and 4 for C:

2 y^(1/2) = 2 (0)^2 + 4

y^1/2 = 2
y = 4 when x =0 <--- that's what the answer key said/ teacher marked


What I did:

separation of variables:

dy/y^(1/2) = 4x dx

Integrate :

2 y^(1/2) = 2x^2

solve for Y

y^(1/2) = x^2
y = x^4 + C

solve for C

9 = (1)^4 + C
C = 8

y = x^4 + 8
Solve for y(0):
y=8 <--- that's the answer I got, and its a multiple choice question

Since it satisfies the initial condition (1,9) and the separation, then 8 is a correct answer, right?

If you take the deriv of where I got y = x^4 from, y^(1/2) = x^2, you still get dy/dx = 4x*y^(1/2), and I merely simplified the equation to put it in terms of Y like it says in the intro: "The point (1,9) lies on the graph of an equation y=f(x)"

Any thoughts? -Thanks

 

[Muzza]

Since it satisfies the initial condition (1,9) and the derivative, then 8 is a correct answer, right?

Have you actually checked that your solution satisfies dy/dx = 4x*y^(1/2) for all x >= 0? (It doesn't btw).

The problem lies between these two steps:

derivative of variables:

dy/y^(1/2) = 4x dx

Integrate :

2 y^(1/2) = 2x^2

You forgot the constant of integration. You can't "add it in later" like you did now. Try it and you'll see why.

 

[JasonRox]

Don't you have to add C first?

You should know better than that. ;)

Like Muzza said, you can't add it later.

 

[rpc]

But, the deriv is not unique to 1 equation,

and isn't the C somewhat arbitrary

for example:

separation of variables:

dy/y^(1/2) = 4x dx

Integrate :

2 y^(1/2) = 2x^2 +C1

Simplify

y^(1/2) = x^2 + C2 <--- and the deriv of this is still dy/dx = 4x*y^(1/2), then all I did was solve for Y

solve for Y
y = x^4 + C3

What am I doing wrong here?, or do you have to solve for C before you can simplify?

 

[StatusX]

y^(1/2) = x^2 + C2

solve for Y
y = x^4 + C3

come on, this is easy. howd you get this second equation? by squaring both sides? try it again.