AP Calculus AB: Finding y with dy/dx = 2y^2 and given x=1, y=-1

[MercuryRising]

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 that's probably wrong and not going to help...

 

[d_leet]

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 that's probably 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.

 

[dicerandom]

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]

It's seperable (as dicerandom 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 that's probably 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

 

[MercuryRising]

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