Differential equations

Can you guys just check my answers? Thanks.

Suppose the rate at which the volume in a tank decreases is proportional to the square root of the volume present. The tank initially contains 25 gallons, but has 20.25 gallons after 3 minutes. Answer the following.

1) Write a differential equation that models this situation. Let V represent the volume (in gallons) in the tank and t represent the time (in minutes).
My answer: dV/dt = k√V

2) Solve for the general solution (do not solve for V).
My answer: 2√V = kt + C

3) Use the initial condition to find the constant of integration, then write the particular solution (do not solve for V).

Attempt at solution: 2√(25)=k(0) + C
My answer: 2√(V)=kt+10


4) Use the second condition to find the constant of proportion.

Attempt at solution: 2√(20.25)=k(3) + 10 --> 4=3k+10 --> -6=3k
My answer: k = -2


5) Find the volume at t = 5 minutes. Round your answer to two decimal places.

My attempt at the solution: 2√(V)=(-2)(5) + 10 --> 2√(V)=0
My answer: V(5)=0

^^^
For number 5... I'm not sure my answer is right. Using common sense, I don't think it's possible for the tank to be at zero gallons at 5 minutes. I think I might have done something wrong at number 4.

You went from 2√(20.25)=k(3) + 10 to 4 = 3k + 10 but thats only true if √(20.25) = 2 but its not.

√(20.25) = √(81/4) = 9/2

2√(20.25)=k(3) + 10
2(9/2) = 3k + 10
9 = 3k + 10
-1 = 3k
k = -1/3

Then do number 5.

2√(V)=(-1/3)t + 10
2√(V)=(-1/3)(5) + 10
2√(V)=(-5/3) + (30/3)
2√(V)=(25/3)
√(V) = 25/6
V = (25/6)^2 = 625/36 = 17.36

1, 2 and 3: I get the same things you do.

4) V(3) = 20.25 so:

2√(20.25)=k(3) + 10

Since the square root of 20.25 is 4.5, we have

2*4.5 = 3k + 10
9 = 3k + 10
-1 = 3k

k = -1/3

From here on out, we'll obviously get different answers.