# AP Calculus AB probelm

If dy/dx = 2y^2 and if y=-1 when x=1
then when x=2, y=?

I have no idea how to even start this probelm, i never seen this on any previous practice tests
i thought about using the equation y+1 = dy/dx (x-1), but thats probaly wrong and not going to help...

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MercuryRising said:
If dy/dx = 2y^2 and if y=-1 when x=1
then when x=2, y=?
I have no idea how to even start this probelm, i never seen this on any previous practice tests
i thought about using the equation y+1 = dy/dx (x-1), but thats probaly wrong and not going to help...
Rewrite the equation as dy over y^2 equals 2dx integrate both sides then solve for the constant of integration by plugging in the initial condition and then you can find y when x = 2.

Here's a general rule for tackling these sorts of problems:

Suppose you have an equation of the form
$$\frac{dy}{dx} = f(y) g(x)$$

You can treat the differentials $dy$ and $dx$ as if they were any other variable, so you can bring $f(y)$ and $dy$ on the same side of the equation and move everything else over to the other side: (I forget the exact proof of this, it involves the definition of the differential and chain rule I believe):
$$\frac{dy}{f(y)} = g(x) dx$$

Then you can integrate both sides:
$$\int \frac{dy}{f(y)} = \int g(x) dx$$

And you'll get an expression which you should be able to solve for $y(x)$ in terms of $x$ and an arbitrary constant $C$. You can then use your boundary conditions to determine what $C$ should be.

benorin
Homework Helper
Gold Member
It's seperable (as dicerandom said)

MercuryRising said:
If dy/dx = 2y^2 and if y=-1 when x=1
then when x=2, y=?
I have no idea how to even start this probelm, i never seen this on any previous practice tests
i thought about using the equation y+1 = dy/dx (x-1), but thats probaly wrong and not going to help...
$$\frac{dy}{dx}=2y^2\Rightarrow\int\frac{dy}{y^2}=\int 2dx\Rightarrow -\frac{1}{y}=2x+C$$

Last edited:
ahh...i didnt know you can work both sides like that, thank you all very much 