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Unit 7.7 – Finding Particular Solutions Using Initial Conditions and Separation of Variables

AP® Calculus AB & BC | From General Solutions to Specific Solutions

Why This Matters: Now we bring everything together! You've learned to solve DEs using separation of variables to get general solutions with arbitrary constants. But real-world problems need SPECIFIC solutions. That's where initial conditions come in! By applying an initial condition (a point the solution must pass through), we find the exact value of the constant, giving us a particular solution. This is the culmination of Unit 7 and appears on every AP® exam—master it!

🎯 Key Definitions

INITIAL VALUE PROBLEM (IVP)

An Initial Value Problem consists of:

  1. A differential equation: \(\frac{dy}{dx} = f(x, y)\)
  2. An initial condition: \(y(x_0) = y_0\)
Standard Form:
\[ \frac{dy}{dx} = f(x, y), \quad y(x_0) = y_0 \]

PARTICULAR SOLUTION

A particular solution is a specific solution to a differential equation that satisfies an initial condition. It contains NO arbitrary constants.

Example:
  • General solution: \(y = Ce^{2x}\) (has constant \(C\))
  • Particular solution: \(y = 3e^{2x}\) (specific, no \(C\))

📋 The Complete Process

Finding Particular Solutions

The 7-Step Method:

  1. Verify separable: Check if DE can be written as \(\frac{dy}{dx} = f(x) \cdot g(y)\)
  2. Separate variables: Get all \(y\)'s with \(dy\), all \(x\)'s with \(dx\)
  3. Integrate both sides: Don't forget +C on ONE side
  4. Find general solution: Solve for \(y\) if possible
  5. Apply initial condition: Substitute \((x_0, y_0)\) into general solution
  6. Solve for constant: Find the value of \(C\) (or \(A\), etc.)
  7. Write particular solution: Substitute constant back into general solution

📖 Comprehensive Worked Examples

Example 1: Complete IVP Solution

Problem: Solve the IVP: \(\frac{dy}{dx} = 3y\), \(y(0) = 5\)

Complete Solution:

Step 1: Check if separable

Yes! Can write as \(3 \cdot y\)

Step 2: Separate variables

\[ \frac{1}{y} \, dy = 3 \, dx \]

Step 3: Integrate both sides

\[ \int \frac{1}{y} \, dy = \int 3 \, dx \]
\[ \ln|y| = 3x + C \]

Step 4: Find general solution

\[ |y| = e^{3x + C} = e^C \cdot e^{3x} \]

Let \(A = \pm e^C\):

\[ y = Ae^{3x} \]

General Solution: \(y = Ae^{3x}\)

Step 5: Apply initial condition \(y(0) = 5\)

Substitute \(x = 0\), \(y = 5\):

\[ 5 = Ae^{3(0)} = Ae^0 = A(1) = A \]

Therefore: \(A = 5\)

Step 7: Write particular solution

Substitute \(A = 5\) back:

\[ y = 5e^{3x} \]

FINAL ANSWER: \(y = 5e^{3x}\)

Example 2: Product Form

Problem: Solve \(\frac{dy}{dx} = xy\), \(y(0) = 2\)

Solution:

Steps 1-3: Separate and integrate

\[ \frac{1}{y} \, dy = x \, dx \]
\[ \ln|y| = \frac{x^2}{2} + C \]

Step 4: General solution

\[ y = Ae^{x^2/2} \]

Steps 5-6: Apply \(y(0) = 2\)

\[ 2 = Ae^{0} = A \]

So \(A = 2\)

PARTICULAR SOLUTION: \(y = 2e^{x^2/2}\)

Example 3: Fraction Form

Problem: Solve \(\frac{dy}{dx} = \frac{x^2}{y}\), \(y(0) = 3\)

Solution:

Steps 1-3: Separate and integrate

\[ y \, dy = x^2 \, dx \]
\[ \frac{y^2}{2} = \frac{x^3}{3} + C \]

Step 4: General solution

Multiply by 2:

\[ y^2 = \frac{2x^3}{3} + 2C \]

Let \(K = 2C\):

\[ y^2 = \frac{2x^3}{3} + K \]

Steps 5-6: Apply \(y(0) = 3\)

\[ 3^2 = \frac{2(0)^3}{3} + K \]
\[ 9 = K \]

Step 7: Particular solution

\[ y^2 = \frac{2x^3}{3} + 9 \]

Or, solving for \(y\) (taking positive root since \(y(0) = 3 > 0\)):

\[ y = \sqrt{\frac{2x^3}{3} + 9} \]

PARTICULAR SOLUTION: \(y = \sqrt{\frac{2x^3}{3} + 9}\)

Example 4: Non-Zero Initial x-Value

Problem: Solve \(\frac{dy}{dx} = \frac{2y}{x}\), \(y(1) = 4\)

Solution:

Steps 1-4: Find general solution

\[ \frac{1}{y} \, dy = \frac{2}{x} \, dx \]
\[ \ln|y| = 2\ln|x| + C = \ln x^2 + C \]
\[ y = Ax^2 \]

Steps 5-6: Apply \(y(1) = 4\)

\[ 4 = A(1)^2 = A \]

PARTICULAR SOLUTION: \(y = 4x^2\)

Example 5: Exponential with Different Base

Problem: Solve \(\frac{dy}{dx} = y(1-y)\), \(y(0) = 0.5\)

Solution:

Steps 1-3: Separate and integrate

\[ \frac{1}{y(1-y)} \, dy = dx \]

Use partial fractions: \(\frac{1}{y(1-y)} = \frac{1}{y} + \frac{1}{1-y}\)

\[ \ln|y| - \ln|1-y| = x + C \]
\[ \ln\left|\frac{y}{1-y}\right| = x + C \]

Step 4: General solution

\[ \frac{y}{1-y} = Ae^x \]

Solving for \(y\):

\[ y = \frac{Ae^x}{1 + Ae^x} \]

Steps 5-6: Apply \(y(0) = 0.5\)

\[ 0.5 = \frac{A}{1 + A} \]
\[ 0.5(1 + A) = A \]
\[ 0.5 + 0.5A = A \]
\[ 0.5 = 0.5A \quad \Rightarrow \quad A = 1 \]

PARTICULAR SOLUTION: \(y = \frac{e^x}{1 + e^x}\)

🔍 Special Cases and Notes

Important Considerations:

  • Implicit vs Explicit: Sometimes particular solution is easier to leave in implicit form
  • Positive vs Negative: When taking square roots, use initial condition to determine sign
  • Domain restrictions: Initial condition helps determine valid domain
  • Multiple steps: Some problems require algebraic manipulation after finding \(C\)

📝 Pro Tip: When the initial condition has \(x_0 \neq 0\), be extra careful with substitution! Make sure to substitute the actual values given, not just plug in zero.

✓ Verifying Your Answer

Two-Part Verification:

  1. Check the DE: Differentiate your particular solution and verify it satisfies the differential equation
  2. Check the IC: Substitute the initial condition point and verify \(y(x_0) = y_0\)

Both must be true!

Example Verification: For \(y = 5e^{3x}\) solving \(\frac{dy}{dx} = 3y\), \(y(0) = 5\)

Check 1: DE

\(y' = 15e^{3x}\) and \(3y = 3(5e^{3x}) = 15e^{3x}\) ✓

Check 2: IC

\(y(0) = 5e^0 = 5\) ✓

Verified!

❌ Common Mistakes to Avoid

  • Mistake 1: Finding general solution but forgetting to apply initial condition
  • Mistake 2: Substituting initial condition into wrong form (before solving for \(y\))
  • Mistake 3: Algebra errors when solving for constant
  • Mistake 4: Not stating the final particular solution clearly
  • Mistake 5: Forgetting about implicit solutions (sometimes explicit form is hard)
  • Mistake 6: Wrong sign when taking square roots
  • Mistake 7: Substituting \(x = 0\) when initial condition is at different \(x\)
  • Mistake 8: Not checking if answer satisfies both DE and IC
  • Mistake 9: Losing track of constants during algebraic manipulation
  • Mistake 10: Not showing work (loses partial credit on AP® exam)

📊 Common Problem Types

Problem Variations

What You Might Encounter
Type Example Key Point
Standard IVP \(\frac{dy}{dx} = ky, y(0) = y_0\) Most common type
Non-zero \(x_0\) \(\frac{dy}{dx} = f(x,y), y(1) = 2\) Substitute actual point
Implicit form \(y^2 = x^2 + C\) May leave as implicit
Find constant only Given general, find \(C\) Direct substitution
Context problem Population, temperature, etc. Interpret initial condition

📝 Practice Problems

Solve these Initial Value Problems:

  1. \(\frac{dy}{dx} = 2xy, \quad y(0) = 3\)
  2. \(\frac{dy}{dx} = \frac{y}{x}, \quad y(1) = 5\)
  3. \(\frac{dy}{dx} = e^{x-y}, \quad y(0) = 0\)
  4. \(\frac{dy}{dx} = y^2, \quad y(0) = 1\)

Answers:

  1. \(y = 3e^{x^2}\)
  2. \(y = 5x\)
  3. \(e^y = e^x + 1\) or \(y = \ln(e^x + 1)\)
  4. \(y = \frac{1}{1-x}\)

✏️ AP® Exam Success Tips

What AP® Graders Look For:

  • Complete separation: Show variables separated clearly
  • Integration with +C: Must show integration step
  • General solution stated: Before applying IC
  • Substitution shown: Explicitly substitute initial condition
  • Constant found: Show work solving for \(C\) or \(A\)
  • Particular solution clearly stated: Final answer should be obvious
  • Verification (bonus): If time, verify both DE and IC

💯 Exam Strategy:

  1. Read problem completely—identify DE and initial condition
  2. Separate variables and integrate (show all work)
  3. Find and clearly state general solution
  4. Write "Using initial condition \(y(x_0) = y_0\):"
  5. Substitute and solve for constant
  6. Write final particular solution
  7. Box or circle your final answer
  8. If time, verify

⚡ Quick Reference Guide

COMPLETE IVP SOLUTION PROCESS

The 7 Steps:

  1. ✓ Check separable: \(\frac{dy}{dx} = f(x) \cdot g(y)\)
  2. ✓ Separate: \(\frac{1}{g(y)}\,dy = f(x)\,dx\)
  3. ✓ Integrate: Both sides (add +C to one side)
  4. ✓ General solution: Solve for \(y\) (with constant)
  5. ✓ Apply IC: Substitute \((x_0, y_0)\)
  6. ✓ Find constant: Solve for \(C\) or \(A\)
  7. ✓ Particular solution: Write final answer (no constants)

Remember:

  • General: Has arbitrary constant (family of solutions)
  • Particular: No arbitrary constant (one specific solution)
  • Always verify: Check both DE and IC if possible

Master Particular Solutions! An Initial Value Problem (IVP) combines a differential equation with an initial condition \(y(x_0) = y_0\). To solve: (1) Use separation of variables to find the general solution (with constant \(C\) or \(A\)), (2) Substitute the initial condition point into the general solution, (3) Solve for the constant, (4) Write the particular solution by substituting the constant value back. The process: separate → integrate → general solution → apply IC → find constant → particular solution. Key difference: general solutions have arbitrary constants (represent family of curves), particular solutions have no constants (one specific curve). Common mistakes: forgetting to apply IC, substituting into wrong form, algebra errors, not stating final answer clearly. Always show: separation, integration with +C, general solution, substitution of IC, solving for constant, final particular solution. Verify by checking both the DE (differentiate) and IC (substitute point). This is THE culmination of Unit 7 and appears on every AP® exam—practice until you can do it in your sleep! 🎯✨