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Answer:

\(\displaystyle y = Ce^\bigg{\frac{x^2}{2}} - 1\)

General Formulas and Concepts:

Pre-Algebra

  • Equality Properties

Algebra I

  • Functions
  • Function Notation
  • Exponential Rule [Multiplying]:                                                                        \(\displaystyle b^m \cdot b^n = b^{m + n}\)

Algebra II

  • Natural Logarithms ln and Euler's number e

Calculus

Derivatives  

Derivative Notation  

Basic Power Rule:  

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

Slope Fields  

  • Solving differentials
  • Separation of Variables

Antiderivatives - Integrals  

Integration Constant C

Integration Rule [Reverse Power Rule]:                                                                   \(\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C\)

U-Substitution

Logarithmic Integration:                                                                                            \(\displaystyle \int {\frac{1}{x}} \, dx = ln|x| + C\)

Step-by-step explanation:

Step 1: Define

\(\displaystyle y' = x(1 + y)\)

Step 2: Redefine

Separation of Variables. Get differential equation to a form where we can integrate both sides.

  1. [Division Property of Equality] Isolate x:                                                         \(\displaystyle \frac{1}{1 + y}y' = x\)
  2. Rewrite derivative notation:                                                                             \(\displaystyle \frac{1}{1 + y} \ \frac{dy}{dx} = x\)
  3. Rewrite:                                                                                                             \(\displaystyle \frac{1}{1 + y} \ dy = x \ dx\)

Step 3: Find General Solution Pt. 1

  1. [Equality Property] Integrate both sides:                                                        \(\displaystyle \int {\frac{1}{1 + y}} \, dy = \int {x} \, dx\)
  2. [Right Integral] Integrate [Integration Rule - Reverse Power Rule]:              \(\displaystyle \int {\frac{1}{1 + y}} \, dy = \frac{x^2}{2} + C\)

Step 4: Find General Solution Pt. 2

Identify variables for u-substitution.

  1. Set:                                                                                                                    \(\displaystyle u = 1 + y\)
  2. Differentiate [Basic Power Rule]:                                                                     \(\displaystyle du = dy\)

Step 5: Find General Solution Pt. 3

  1. [Integral] U-Substitution:                                                                                  \(\displaystyle \int {\frac{1}{u}} \, du = \frac{x^2}{2} + C\)
  2. [Integral] Integrate [Logarithmic Integration]:                                                 \(\displaystyle ln|u| = \frac{x^2}{2} + C\)
  3. Back-Substitute:                                                                                               \(\displaystyle ln|1 + y| = \frac{x^2}{2} + C\)
  4. [Equality Property] e both sides:                                                                     \(\displaystyle e^\bigg{ln|1 + y|} = e^\bigg{\frac{x^2}{2} + C}\)
  5. Simplify:                                                                                                             \(\displaystyle |1 + y| = e^\bigg{\frac{x^2}{2} + C}\)
  6. Rewrite:                                                                                                             \(\displaystyle |1 + y| = Ce^\bigg{\frac{x^2}{2}}\)
  7. Rewrite:                                                                                                             \(\displaystyle 1 + y = \pm Ce^\bigg{\frac{x^2}{2}}\)
  8. [Subtraction Property of Equality] Isolate y:                                                   \(\displaystyle y = \pm Ce^\bigg{\frac{x^2}{2}} - 1\)

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Slope Fields

Book: College Calculus 10e

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