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The
exact solusion of the IVP

The Exact Solusion Of The IVP class=

Sagot :

Space

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

  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]:                                                                        [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

  • Separation of Variables

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

  1. Rewrite Derivative Notation:                                                                            [tex]\displaystyle \frac{dy}{dx} = \frac{xy}{2 + x^2}[/tex]
  2. [Division Property of Equality] Isolate y's:                                                       [tex]\displaystyle \frac{1}{y} \frac{dy}{dx} = \frac{x}{2 + x^2}[/tex]
  3. [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

  1. [Equality Property] Integrate both sides:                                                        [tex]\displaystyle \int {\frac{1}{y}} \, dy = \int {\frac{x}{2 + x^2}} \, dx[/tex]
  2. [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

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

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

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