Mathematics, 13.10.2019 23:20 omar2334

Ap calculus, me, i dont understand

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Answer from: ashlyo

$\displaystyle y = Ce^\bigg{\frac{x^2}{2}} - 1$

General Formulas and Concepts:

Pre-Algebra

Equality Properties

Algebra I

FunctionsFunction NotationExponential Rule [Multiplying]: $\displaystyle b^m \cdot b^n = b^{m + n}$

Algebra II

Natural Logarithms ln and Euler's number e

Calculus

Derivatives

Derivative Notation

Basic Power Rule:

f(x) = cxⁿ f’(x) = c·nxⁿ⁻¹

Slope Fields

Solving differentials Separation of Variables

Antiderivatives - Integrals

Integration Constant C

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

U-Substitution

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

Step-by-step explanation:

Step 1: Define

$\displaystyle y' = x(1 + y)$

Step 2: Redefine

Separation of Variables. Get differential equation to a form where we can integrate both sides.

[Division Property of Equality] Isolate x: $\displaystyle \frac{1}{1 + y}y' = x$ Rewrite derivative notation: $\displaystyle \frac{1}{1 + y} \ \frac{dy}{dx} = x$ Rewrite: $\displaystyle \frac{1}{1 + y} \ dy = x \ dx$

Step 3: Find General Solution Pt. 1

[Equality Property] Integrate both sides: $\displaystyle \int {\frac{1}{1 + y}} \, dy = \int {x} \, dx$ [Right Integral] Integrate [Integration Rule - Reverse Power Rule]: $\displaystyle \int {\frac{1}{1 + y}} \, dy = \frac{x^2}{2} + C$

Step 4: Find General Solution Pt. 2

Identify variables for u-substitution.

Set: $\displaystyle u = 1 + y$ Differentiate [Basic Power Rule]: $\displaystyle du = dy$

Step 5: Find General Solution Pt. 3

[Integral] U-Substitution: $\displaystyle \int {\frac{1}{u}} \, du = \frac{x^2}{2} + C$ [Integral] Integrate [Logarithmic Integration]: $\displaystyle ln|u| = \frac{x^2}{2} + C$ Back-Substitute: $\displaystyle ln|1 + y| = \frac{x^2}{2} + C$ [Equality Property] e both sides: $\displaystyle e^\bigg{ln|1 + y|} = e^\bigg{\frac{x^2}{2} + C}$ Simplify: $\displaystyle |1 + y| = e^\bigg{\frac{x^2}{2} + C}$ Rewrite: $\displaystyle |1 + y| = Ce^\bigg{\frac{x^2}{2}}$ Rewrite: $\displaystyle 1 + y = \pm Ce^\bigg{\frac{x^2}{2}}$ [Subtraction Property of Equality] Isolate y: $\displaystyle y = \pm Ce^\bigg{\frac{x^2}{2}} - 1$

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

Unit: Slope Fields

Book: College Calculus 10e

Answer from: Quest

step-by-step explanation:

Answer from: Quest

the answer is b.)

< 1=< 2

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