Mathematics, 18.03.2021 03:00 ashtonbillups

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))

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Answer from: homework1911

B. y = x² + 1

General Formulas and Concepts:

Symbols

e (Euler's number) ≈ 2.7182

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition 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.

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

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

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

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

Answer from: Quest

depends on how many days you are talking about

step-by-step explanation:

Answer from: Quest

dalila is incorrect. the whisker on the left is very long, going all the way out to 5. the whisker goes from 5 to 30, so it has a spread of 25. the rest of the box plot values are closer together, so 5 is an outlier

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