Question

The
exact solusion of the IVP

Answer 1

`\displaystyle y = √(2)e^2 + x^2`

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right

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]:
  `\displaystyle b^m \cdot b^n = b^(m + n)`

Algebra II

• Logarithms and Natural Logs
• Euler's number e

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Differential Equations

• Separation of Variables

Antiderivatives - Integrals

Integration Constant C

Integration Property [Multiplied Constant]:
`\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx`

U-Substitution

Logarithmic Integration

Explanation:

Step 1: Define

Identify

`\displaystyle y' = (xy)/(2 + x^2)`

`\displaystyle y(0) = 2`

Step 2: Rewrite

Separation of Variables

1. Rewrite Derivative Notation:
   `\displaystyle (dy)/(dx) = (xy)/(2 + x^2)`
2. [Division Property of Equality] Isolate y's:
   `\displaystyle (1)/(y) (dy)/(dx) = (x)/(2 + x^2)`
3. [Multiplication Property of Equality] Rewrite Derivative Notation:
   `\displaystyle (1)/(y) dy = (x)/(2 + x^2) dx`

Step 3: Find General Solution Pt. 1

Integration

1. [Equality Property] Integrate both sides:
   `\displaystyle \int {(1)/(y)} \, dy = \int {(x)/(2 + x^2)} \, dx`
2. [1st Integral] Integrate [Logarithmic Integration]:
   `\displaystyle ln|y| = \int {(x)/(2 + x^2)} \, dx`

Step 4: Identify Variables

Identify variables for u-substitution for 2nd integral.

u = 2 + x²

du = 2xdx

Step 5: Find General Solution Pt. 2

1. [2nd Integral] Rewrite [Integration Property - Multiplied Constant]:
   `\displaystyle ln|y| = (1)/(2)\int {(2x)/(2 + x^2)} \, dx`
2. [2nd Integral] U-Substitution:
   `\displaystyle ln|y| = (1)/(2)\int {(1)/(u)} \, du`
3. [2nd Integral] Integrate [Logarithmic Integration]:
   `\displaystyle ln|y| = (1)/(2) ln|u| + C`
4. [Equality Property] e both sides:
   `\displaystyle e^(ln|y|) = e^u`
5. Simplify:
   `\displaystyle |y| = e^u`
6. Rewrite [Exponential Rule - Multiplying]:
   `\displaystyle |y| = e^u \cdot e^C`
7. Simplify:
   `\displaystyle |y| = Ce^u`
8. Back-Substitute:
   `\displaystyle |y| = Ce^(1)/(2) ln`

Our general solution is
`\displaystyle |y| = Ce^(1)/(2) ln`.

Step 6: Find Particular Solution

1. Substitute in point:
   `\displaystyle |2| = Ce^(1)/(2) ln`
2. Evaluate |Absolute Value|:
   `\displaystyle 2 = Ce^2 + 0^2`
3. |Absolute Value| Evaluate exponents:
   `\displaystyle 2 = Ce^2 + 0`
4. |Absolute Value| Add:
   `\displaystyle 2 = Ce^(1)/(2) ln`
5. |Absolute Value| Evaluate:
   `\displaystyle 2 = Ce^{(1)/(2) ln(2)}`
6. [Division Property of Equality] Isolate C:
   `\displaystyle √(2) = C`
7. Rewrite:
   `\displaystyle C = √(2)`
8. Substitute in C [General Solution]:
   `\displaystyle y = √(2)e^2 + x^2`

∴ Our particular solution is
`\displaystyle y = √(2)e^(1)/(2) ln`.

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

Unit: Differential Equations

Book: College Calculus 10e