Advanced Placement (AP), 31.08.2019 04:00, gamboaserg

Bc calculus
if dy / dx = (2x - 1)y , y(1) = e, find y(2).

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answered: hasshh

$\displaystyle y(2) = e^3$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right

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

Separation of VariablesSolving Differentials

Integrals

Antiderivatives

Integration Constant C

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

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

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

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

Explanation:

*Note:

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

Step 1: Define

$\displaystyle \frac{dy}{dx} = (2x - 1)y$

$\displaystyle y(1) = e$

Step 2: Rewrite

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

[Separation of Variables] Rewrite Leibniz Notation: $\displaystyle dy = (2x - 1)y \ dx$ [Separation of Variables] Isolate y's together: $\displaystyle \frac{1}{y} \ dy = (2x - 1) \ dx$

Step 3: Find General Solution

[Differential] Integrate both sides: $\displaystyle \int {\frac{1}{y}} \, dy = \int {(2x - 1)} \, dx$ [dy Integral] Integrate [Logarithmic Integration]: $\displaystyle ln|y| = \int {(2x - 1)} \, dx$ [dx Integral] Rewrite [Integration Property - Addition/Subtraction]: $\displaystyle ln|y| = \int {2x} \, dx - \int {} \, dx$ [1st dx Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle ln|y| = 2\int {x} \, dx - \int {} \, dx$ [dx Integrals] Integrate [Integration Rule - Reverse Power Rule]: $\displaystyle ln|y| = 2(\frac{x^2}{2}) - x + C$ Simplify: $\displaystyle ln|y| = x^2 - x + C$ [Equality Property] e both sides: $\displaystyle e^\bigg{ln|y|} = e^\bigg{x^2 - x + C}$ Simplify: $\displaystyle |y| = Ce^\bigg{x^2 - x}$ Rewrite: $\displaystyle y = \pm Ce^\bigg{x^2 - x}$

General Solution: $\displaystyle y = \pm Ce^\bigg{x^2 - x}$

Step 4: Find Particular Solution

Substitute in function values [General Solution]: $\displaystyle e = \pm Ce^\bigg{1^2 - 1}$ Simplify: $\displaystyle e = \pm C$ Rewrite: $\displaystyle C = e$ Substitute in C [General Solution]: $\displaystyle y = e \bigg( e^\bigg{x^2 - x} \bigg)$ Simplify [Exponential Rule - Multiplying]: $\displaystyle y = e^\bigg{x^2 - x + 1}$

Particular Solution: $\displaystyle y = e^\bigg{x^2 - x + 1}$

Step 5: Solve

Substitute in x [Particular Solution]: $\displaystyle y(2) = e^\bigg{2^2 - 2 + 1}$ Simplify: $\displaystyle y(2) = e^3$

∴ our final answer is $\displaystyle y(2) = e^3$ .

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

Unit: Differentials and Slope Fields

Book: College Calculus 10e

answered: Guest

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answered: Guest

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