Initial-Value Problems

Learn initial conditions, checking a particular solution.

Lesson 3 of 32 parts Practice Flashcards

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Initial Conditions

Objective: Explain initial conditions, apply the stated method, and check its conditions.

A derivative equation determines an antiderivative family. An initial value y(x₀)=y₀ selects the constant for a particular solution. Integrate first, substitute the initial condition, solve for the constant, and then evaluate the requested input. The initial condition determines the integration constant.

Formulas

$y′=k ⇒ y=kx+C$

Worked examples

Try itSolve y′=6, y(0)=7. Find y(7).Show answer

Step 1: Integrate: y=6x+C. Step 2: The initial condition gives C=7. Step 3: y(7)=49.

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Initial Conditions

Objective: Explain initial conditions, apply the stated method, and check its conditions.

Formulas

$y′=k ⇒ y=kx+C$

Worked examples

Try itSolve y′=6, y(0)=7. Find y(7).Show answer

Step 1: Integrate: y=6x+C. Step 2: The initial condition gives C=7. Step 3: y(7)=49.

Checking a Particular Solution

Objective: Explain checking a particular solution, apply the stated method, and check its conditions.

For a polynomial derivative equation, integrate term by term and use the initial value to determine C. Check the final solution by differentiating it and verifying the original initial point. Both checks are needed to confirm the intended particular solution. A correct derivative alone does not guarantee the correct initial value.

Formulas

$y′=2ax ⇒ y=ax²+C$

Worked examples

Try itSolve y′=12x with y(0)=7. Find y(7).Show answer

Step 1: Integrate: y=6x²+C. Step 2: Use y(0)=7 to get C=7. Step 3: y(7)=301.

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