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Princeton Review AP Calculus BC, Chapter 9: Differential Equations

Introduction & Slope Fields

  • In related rates we saw how we can model the change in one thing related to another with derivatives, and differential equations are similar!

    • Oftentimes, variables are not constant, so we have to represent their change using a derivative (Ex. the change in y is dy/dx)

    • A differential equation models the change in one variable with respect to another

  • Slope fields show us what the slopes look like at points on a graph

  • This is for the equation dy/dx = x

    • Remember that this is a derivative so it will show us the slope at these points!

    • All you have to do to construct a slope field is plug in your x/y (or both) values into your differential equation and draw that as your slope

      • Ex. The slope at x = -1 would be -1 (because dy/dx = x)

dy/dx = x solution

  • The AP exam might also require you to sketch a solution curve given a slope field!

  • All you have to do is “flow” with the slopes

    • Make sure you don’t cross abruptly or draw a line that doesn’t follow the slope

    • Because this is by hand, it doesn’t have to be exact, just try and go with the tangent lines!

Differential Equations

  • If you’re given a differential equation where the derivative of a function is equal to some other function, you have to solve for the original! You can do this by taking the integral (antiderivative) of both sides!

  • A good memory trick is that differential equation problems will be SIPPY problems

    • S: separate (dy and dx on separate sides)

    • I: integrate (remove the derivative)

    • P: Plus C (add your c value to your integral)

    • P: Plug in your initial condition

    • Y: Y equals (solve to find what y is)

  • Example: If dy/dx = 4x/y and y(0) = 5 we need to solve for y

    • Start by separating → ydy = 4xdx

    • Then integrate → ∫ydy = ∫4xdx → y^2/2 = 2x^2 + C

      • (Make sure you add C!)

    • Plug in → (5)^2/2 = 2(0)^2 + C

      • C = 25/2

    • Now set y equals → y = 2x^2 + 25/2

Princeton Review AP Calculus BC, Chapter 9: Differential Equations

Introduction & Slope Fields

  • In related rates we saw how we can model the change in one thing related to another with derivatives, and differential equations are similar!

    • Oftentimes, variables are not constant, so we have to represent their change using a derivative (Ex. the change in y is dy/dx)

    • A differential equation models the change in one variable with respect to another

  • Slope fields show us what the slopes look like at points on a graph

  • This is for the equation dy/dx = x

    • Remember that this is a derivative so it will show us the slope at these points!

    • All you have to do to construct a slope field is plug in your x/y (or both) values into your differential equation and draw that as your slope

      • Ex. The slope at x = -1 would be -1 (because dy/dx = x)

dy/dx = x solution

  • The AP exam might also require you to sketch a solution curve given a slope field!

  • All you have to do is “flow” with the slopes

    • Make sure you don’t cross abruptly or draw a line that doesn’t follow the slope

    • Because this is by hand, it doesn’t have to be exact, just try and go with the tangent lines!

Differential Equations

  • If you’re given a differential equation where the derivative of a function is equal to some other function, you have to solve for the original! You can do this by taking the integral (antiderivative) of both sides!

  • A good memory trick is that differential equation problems will be SIPPY problems

    • S: separate (dy and dx on separate sides)

    • I: integrate (remove the derivative)

    • P: Plus C (add your c value to your integral)

    • P: Plug in your initial condition

    • Y: Y equals (solve to find what y is)

  • Example: If dy/dx = 4x/y and y(0) = 5 we need to solve for y

    • Start by separating → ydy = 4xdx

    • Then integrate → ∫ydy = ∫4xdx → y^2/2 = 2x^2 + C

      • (Make sure you add C!)

    • Plug in → (5)^2/2 = 2(0)^2 + C

      • C = 25/2

    • Now set y equals → y = 2x^2 + 25/2

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