Answer:
[tex]\displaystyle y = 2e^\bigg{\frac{x^3}{3}} + 1[/tex]
General Formulas and Concepts:
Pre-Algebra
- Order of Operations
- Equality Properties
Algebra I
- Functions
- Function Notation
- Exponential Rule [Rewrite]: [tex]\displaystyle b^{-m} = \frac{1}{b^m}[/tex]
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
Integrals
Integration Constant C
Integration Rule [Reverse Power Rule]: [tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]
Integration Property [Addition/Subtraction]: [tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]
U-Substitution
Logarithmic Integration: [tex]\displaystyle \int {\frac{1}{u}} \, dx = ln|u| + C[/tex]
Step-by-step explanation:
*Note:
When solving differential equations in slope fields, disregard the integration constant C for variable y.
Step 1: Define
[tex]\displaystyle \frac{dy}{dx} = x^2(y - 1)[/tex]
[tex]\displaystyle f(0) = 3[/tex]
Step 2: Rewrite
Separation of Variables. Get differential equation to a form where we can integrate both sides and rewrite Leibniz Notation.
- [Separation of Variables] Rewrite Leibniz Notation: [tex]\displaystyle dy = x^2(y - 1) \ dx[/tex]
- [Separation of Variables] Isolate y's together: [tex]\displaystyle \frac{1}{y - 1} \ dy = x^2 \ dx[/tex]
Step 3: Find General Solution Pt. 1
- [Differential] Integrate both sides: [tex]\displaystyle \int {\frac{1}{y - 1}} \, dy = \int {x^2} \, dx[/tex]
- [dx Integral] Integrate [Integration Rule - Reverse Power Rule]: [tex]\displaystyle \int {\frac{1}{y - 1}} \, dy = \frac{x^3}{3} + C[/tex]
Step 4: Find General Solution Pt. 2
Identify variables for u-substitution for dy.
- Set: [tex]\displaystyle u = y - 1[/tex]
- Differentiate [Basic Power Rule]: [tex]\displaystyle du = dy[/tex]
Step 5: Find General Solution Pt. 3
- [dy Integral] U-Substitution: [tex]\displaystyle \int {\frac{1}{u}} \, du = \frac{x^3}{3} + C[/tex]
- [dy Integral] Integrate [Logarithmic Integration]: [tex]\displaystyle ln|u| = \frac{x^3}{3} + C[/tex]
- [Equality Property] e both sides: [tex]\displaystyle e^\bigg{ln|u|} = e^\bigg{\frac{x^3}{3} + C}[/tex]
- Simplify: [tex]\displaystyle |u| = Ce^\bigg{\frac{x^3}{3}}[/tex]
- Rewrite: [tex]\displaystyle u = \pm Ce^\bigg{\frac{x^3}{3}}[/tex]
- Back-Substitute: [tex]\displaystyle y - 1 = \pm Ce^\bigg{\frac{x^3}{3}}[/tex]
- [Equality Property] Isolate y: [tex]\displaystyle y = \pm Ce^\bigg{\frac{x^3}{3}} + 1[/tex]
General Form: [tex]\displaystyle y = \pm Ce^\bigg{\frac{x^3}{3}} + 1[/tex]
Step 6: Find Particular Solution
- Substitute in function values [General Form]: [tex]\displaystyle 3 = \pm Ce^\bigg{\frac{0^3}{3}} + 1[/tex]
- Simplify: [tex]\displaystyle 3 = \pm C + 1[/tex]
- [Equality Property] Isolate C: [tex]\displaystyle 2 = \pm C[/tex]
- Rewrite: [tex]\displaystyle C = 2[/tex]
- Substitute in C [General Form]: [tex]\displaystyle y = 2e^\bigg{\frac{x^3}{3}} + 1[/tex]
∴ our particular solution is [tex]\displaystyle y = 2e^\bigg{\frac{x^3}{3}} + 1[/tex].
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Differentials and Slope Fields
Book: College Calculus 10e
