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AP Calculus AB: 11.1.3 Finding a Particular Solution

Mathematics12 CardsCreated 3 months ago

This content focuses on using initial conditions to determine the particular solution from the general solution of a differential equation. It demonstrates how to apply given points to solve for the constant of integration and provides examples with various types of separable and non-separable differential equations.

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Finding a Particular Solution

Given a point, you can use the general solution to a differential equation to find the particular solution that satisfies that point.

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Key Terms

Term
Definition

Finding a Particular Solution

Given a point, you can use the general solution to a differential equation to find the particular solution that satisfies that point.

note

  • This differential equation can be solved by separating the variables and integrating each side.

  • Now that you have the genera...

If dy/dx=4x^3+2x/3y^2+1 and y(0)=−1,which of the following is the particular solution?

y ^3 + y = x ^4 + x^ 2 − 2

If dy/dx=e^2x−y and y(0)=3,which of the following is the particular solution?

e^y=1/2e^2x+e^3−1/2

If x^2dy/dx=y+3 and y(−2)=3,which of the following is the particular solution?

ln|y+3|=−1/x+ln6−1/2

If dy/dx=sinx and y(0)=1, which of the following is the particular solution?

y = 2 − cos x

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AP Calculus AB: 11.1.3 Finding a Particular Solution
TermDefinition

Finding a Particular Solution

Given a point, you can use the general solution to a differential equation to find the particular solution that satisfies that point.

note

  • This differential equation can be solved by separating the variables and integrating each side.

  • Now that you have the general solution, suppose you were also given an initial condition that the solution must satisfy. The notation y(1) = 2 means that y = 2 when x = 1.

  • Use these values in the general solution.

  • Solving for C and substituting in the general solution produces this particular solution.

  • You might not think that this equation is separable since the x- and y-variables seem interlocked.

  • However, you can use rules of exponents to separate the expression.

  • Since the y-expressions can be collected with dy and the x-expressions can be collected with dx, the original equation is separable. Integrate both sides to arrive at the general solution.

  • Now use the initial condition that y = 1 when x = 0. Initial conditions get their name from the fact that they originally represented the beginning value of a variable representing time.

  • Use the values of x and y to solve for C. Then you can write the particular solution corresponding to the given initial condition.

If dy/dx=4x^3+2x/3y^2+1 and y(0)=−1,which of the following is the particular solution?

y ^3 + y = x ^4 + x^ 2 − 2

If dy/dx=e^2x−y and y(0)=3,which of the following is the particular solution?

e^y=1/2e^2x+e^3−1/2

If x^2dy/dx=y+3 and y(−2)=3,which of the following is the particular solution?

ln|y+3|=−1/x+ln6−1/2

If dy/dx=sinx and y(0)=1, which of the following is the particular solution?

y = 2 − cos x

If dy/dx=x^2+2/y^2+4 and y(1)=2,which of the following is the particular solution?

y^3/3+4y=x^3/3+2x+25/3

If y2+yx2+2x+2dydx=x/y and y(0)=2,which of the following is the particular solution?

x^4/4+2x^3/3+x^2=y^4/4+y^3/3−20/3

If dy/dx=x3+x and y(2)=−2,which of the following is the particular solution?

y=x^44/+x^2/2−8

If dy/dx=√ycosx and y(π/2)=2,which of the following is the particular solution?

2√y=sinx+2√2−1

If dy/dx=1√1−x^2 and y(1/2)=0,which of the following is the particular solution?

y=arcsinx−π/6

If dy/dx=y and y(−1)=2, which of the following is the particular solution?

y = 2e^ x + 1