Differential equation where rate is $\propto x^2$ I was wondering if someone could tell me where I'm going wrong with my word problem.
The question is:
The rate at which a salt dissolves is approximately equal to the square of its mass. The starting mass is 35 grams. $ y(7)=13.2$. How much of the salt is left after 11 minutes?
My effort is: 
I've set up the equation as   $y'=(35-x)k$ and get the right solution but I don't know how to proceed when its asking the rate is the square of the mass.
Thanks.
 A: I believe we are missing something in the statement of the problem. 
I think (as you have stated the question) we need to solve something like this 
$$ y'(x)=y^2, \\y(0)=35, \\ y(7)=13.2$$
So then we can solve the diff eq and compute the constants.
$$ y(x)=\frac{1}{c_1-x}$$
Does this mean that $c_1=1/35$? but now how will we include this second point? 
I'm not quite sure yet. 
What if we interpreted the line as the rate salt dissolves is approximately proportional to the square of its mass. Then maybe we could look at it like this? 
$$ y'(x)=cy^2, \\y(0)=35, \\ y(7)=13.2$$
Then we should be able to solve this for 
$$y(x)=\frac{1}{k_1-cx}$$
Then we could solve for $k_1=1/35$ and $c$. 
A: When you are told the rate of change is equal to the square of its mass, there has to be a constant because the units don't match.  It would be better to say the rate of change of the mass is proportional to the square of the mass.  
Using $y$ for the mass of the block we have 
$$y'=-ky^2\\\frac 1y=C+kt$$
Now we can plug the data in to evaluate $C,k$
$$y(0)=35\\C=\frac1{35}\\y(7)=13.2\\\frac 1{13.2}=\frac1{35}+7k\\k=\frac 17\left(\frac 1{13.2}-\frac 1{35}\right)\approx 0.00674\\
y(11)\approx\frac 1{0.00674\cdot 11+\frac 1{35}}\approx 9.735$$
