Answer:
[tex]\displaystyle y = \sqrt{2}e^{\frac{1}{2} ln|2 + x^2|}[/tex]
General Formulas and Concepts:
Pre-Algebra
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtraction Property of Equality
Algebra I
- |Absolute Value|
- Functions
- Function Notation
- Exponential Rule [Multiplying]: [tex]\displaystyle b^m \cdot b^n = b^{m + n}[/tex]
Algebra II
- Logarithms and Natural Logs
- Euler's number e
Calculus
Derivatives
Derivative Notation
Derivative of a constant is 0
Differential Equations
Antiderivatives - Integrals
Integration Constant C
Integration Property [Multiplied Constant]: [tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]
U-Substitution
Logarithmic Integration
Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle y' = \frac{xy}{2 + x^2}[/tex]
[tex]\displaystyle y(0) = 2[/tex]
Step 2: Rewrite
Separation of Variables
- Rewrite Derivative Notation: [tex]\displaystyle \frac{dy}{dx} = \frac{xy}{2 + x^2}[/tex]
- [Division Property of Equality] Isolate y's: [tex]\displaystyle \frac{1}{y} \frac{dy}{dx} = \frac{x}{2 + x^2}[/tex]
- [Multiplication Property of Equality] Rewrite Derivative Notation: [tex]\displaystyle \frac{1}{y} dy = \frac{x}{2 + x^2} dx[/tex]
Step 3: Find General Solution Pt. 1
Integration
- [Equality Property] Integrate both sides: [tex]\displaystyle \int {\frac{1}{y}} \, dy = \int {\frac{x}{2 + x^2}} \, dx[/tex]
- [1st Integral] Integrate [Logarithmic Integration]: [tex]\displaystyle ln|y| = \int {\frac{x}{2 + x^2}} \, dx[/tex]
Step 4: Identify Variables
Identify variables for u-substitution for 2nd integral.
u = 2 + x²
du = 2xdx
Step 5: Find General Solution Pt. 2
- [2nd Integral] Rewrite [Integration Property - Multiplied Constant]: [tex]\displaystyle ln|y| = \frac{1}{2}\int {\frac{2x}{2 + x^2}} \, dx[/tex]
- [2nd Integral] U-Substitution: [tex]\displaystyle ln|y| = \frac{1}{2}\int {\frac{1}{u}} \, du[/tex]
- [2nd Integral] Integrate [Logarithmic Integration]: [tex]\displaystyle ln|y| = \frac{1}{2} ln|u| + C[/tex]
- [Equality Property] e both sides: [tex]\displaystyle e^{ln|y|} = e^{\frac{1}{2} ln|u| + C}[/tex]
- Simplify: [tex]\displaystyle |y| = e^{\frac{1}{2} ln|u| + C}[/tex]
- Rewrite [Exponential Rule - Multiplying]: [tex]\displaystyle |y| = e^{\frac{1}{2} ln|u|} \cdot e^C[/tex]
- Simplify: [tex]\displaystyle |y| = Ce^{\frac{1}{2} ln|u|}[/tex]
- Back-Substitute: [tex]\displaystyle |y| = Ce^{\frac{1}{2} ln|2 + x^2|}[/tex]
Our general solution is [tex]\displaystyle |y| = Ce^{\frac{1}{2} ln|2 + x^2|}[/tex].
Step 6: Find Particular Solution
- Substitute in point: [tex]\displaystyle |2| = Ce^{\frac{1}{2} ln|2 + 0^2|}[/tex]
- Evaluate |Absolute Value|: [tex]\displaystyle 2 = Ce^{\frac{1}{2} ln|2 + 0^2|}[/tex]
- |Absolute Value| Evaluate exponents: [tex]\displaystyle 2 = Ce^{\frac{1}{2} ln|2 + 0|}[/tex]
- |Absolute Value| Add: [tex]\displaystyle 2 = Ce^{\frac{1}{2} ln|2|}[/tex]
- |Absolute Value| Evaluate: [tex]\displaystyle 2 = Ce^{\frac{1}{2} ln(2)}[/tex]
- [Division Property of Equality] Isolate C: [tex]\displaystyle \sqrt{2} = C[/tex]
- Rewrite: [tex]\displaystyle C = \sqrt{2}[/tex]
- Substitute in C [General Solution]: [tex]\displaystyle y = \sqrt{2}e^{\frac{1}{2} ln|2 + x^2|}[/tex]
∴ Our particular solution is [tex]\displaystyle y = \sqrt{2}e^{\frac{1}{2} ln|2 + x^2|}[/tex].
Topic: AP Calculus AB/BC (Calculus I/II)
Unit: Differential Equations
Book: College Calculus 10e
