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AP Calculus AB probelm

  • #1
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...
 

Answers and Replies

  • #2
1,074
1
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.
 
  • #3
307
1
Here's a general rule for tackling these sorts of problems:

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

You can treat the differentials [itex]dy[/itex] and [itex]dx[/itex] as if they were any other variable, so you can bring [itex]f(y)[/itex] and [itex]dy[/itex] 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):
[tex]\frac{dy}{f(y)} = g(x) dx[/tex]

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

And you'll get an expression which you should be able to solve for [itex]y(x)[/itex] in terms of [itex]x[/itex] and an arbitrary constant [itex]C[/itex]. You can then use your boundary conditions to determine what [itex]C[/itex] should be.
 
  • #4
benorin
Homework Helper
Insights Author
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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...
[tex]\frac{dy}{dx}=2y^2\Rightarrow\int\frac{dy}{y^2}=\int 2dx\Rightarrow -\frac{1}{y}=2x+C[/tex]
 
Last edited:
  • #5
ahh...i didnt know you can work both sides like that, thank you all very much :biggrin:
 

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