Question

Which of the following is the solution to the differential equation dy/dx=2xy/x^2+1 whose graph contains the points (0,1)?
A. y=ex^2
B. y=x^2+1
C. y=ln(x^2+1)
D. y=1+ln(x^2+1)
E. y=sqrt(1+2ln(x^2+1))

Answer 1

Answer:

B. y = x² + 1

General Formulas and Concepts:

Symbols

• e (Euler's number) ≈ 2.7182

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

• Functions

Algebra II

• Logarithms - ln and e

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

• f(x) = cxⁿ
• f’(x) = c·nxⁿ⁻¹

Slope Fields

Solving Differentials - Integrals

Integration Constant C

U-Substitution

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

Step-by-step explanation:

*Note:

When solving differential equations in slope fields, disregard the integration constant C for variable y.

Step 1: Define

$\displaystyle \frac{dy}{dx} = \frac{2xy}{x^2 + 1}$

Point (0, 1)

Step 2: Rewrite Differential

Rewrite Leibniz Notation using Separation of Variables.

1. [Separation of Variables] Isolate x's together:                                             $\displaystyle dy = \frac{2xy}{x^2 + 1}dx$
2. [Separation of Variables] Isolate y's together:                                             $\displaystyle \frac{1}{y}dy = \frac{2x}{x^2 + 1}dx$

Step 3: Integrate Pt. 1

Solving general form of differential using integration.

1. [Differential] Integrate both sides:                                                                 $\displaystyle \int {\frac{1}{y}} \, dy = \int {\frac{2x}{x^2 + 1}} \, dx$
2. [Left Integral] ln Integration:                                                                           $\displaystyle ln|y| = \int {\frac{2x}{x^2 + 1}} \, dx$

Step 4: Identify Variables

Set up u-substitution for right integral.

u = x² + 1

du = 2xdx

Step 5: Integrate Pt. 2

1. [Right Integral] U-Substitution:                                                                       $\displaystyle ln|y| = \int {\frac{1}{u}} \, du$
2. [Right Integral] ln Integration:                                                                        $\displaystyle ln|y| = ln|u| + C$
3. Back-Substitute:                                                                                             $\displaystyle ln|y| = ln|x^2 + 1| + C$
4. [Equality Property] Raise e on both sides:                                                    $\displaystyle e^{ln|y|} = e^{ln|x^2 + 1| + C}$
5. Simplify:                                                                                                           $\displaystyle |y| = C|x^2 + 1|$

General Form: $\displaystyle |y| = C|x^2 + 1|$

Step 6: Solve Particular Solution

Since both sides have absolute value, assume that the particular solution will be positive.

1. Substitute in point [General Form]:                                                                $\displaystyle |1| = C|(0)^2 + 1|$
2. [Particular] |Absolute Value| Evaluate exponents:                                        $\displaystyle |1| = C|1|$
3. [Particular] Evaluate absolute values:                                                           $\displaystyle 1 = C$
4. [Particular] Rewrite:                                                                                        $\displaystyle C = 1$

Substituting integration constant C into the general form:

Particular Solution: $\displaystyle y = x^2 + 1$

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

Unit: Differentials and Slope Fields

Book: College Calculus 10e