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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Answer from: 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

Answer from: Quest

its c because i have that test too

Answer from: Quest

Answer: the interval on which the curve is concave up is (-∞, -1/2)====== $\displaystyle y = \int_{0}^x \frac{1}{1 + t + t^2} \; dt$ we find the interval where this curve is concave up by finding the interval in which the 2nd derivative is positive.differentiate the curve equation with respect to x $\begin{aligned} \frac{dy}{dx} & = \frac{d}{dx} \int_{0}^x \frac{1}{1 + t + t^2} \; dt \end{aligned}$ let the function f denote be the anti-derivative of the inside expression, i.e. a function such that $(d/dx)f(x) = 1/(1 + t + t^2)$ .then use the fundamental theorem of calculus: $\begin{aligned} \frac{dy}{dx} & = \frac{d}{dx} \int_{0}^x \frac{1}{1 + t + t^2} \; dt \\& = \frac{d}{dx} \big[ f(x) - f(0) \big] & & \text{\footnotesize apply ftc} \\ & = \frac{d}{dx}f(x) - 0 \\ & = \frac{1}{x + x + x^2} & & \text{\footnotesize since $\frac{d}{dx}f(x) = \frac{1}{1 + x + x^2}$ } \end{aligned}$ differentiate again to get second derivative. we can use chain rule with 1/x as the outside function and 1 + x + x² as the inside function $\begin{aligned} \frac{d}{dx} \left( \frac{dy}{dx} \right) & = \frac{d}{dx}\left(\frac{1}{1 + x + x^2} \right) \\ \frac{d^2y}{dx^2} & = \frac{-1}{\left(1 + x + x^2\right)^2} \cdot \tfrac{d}{dx}(1 + x + x^2) \\ & = \frac{-1}{\left(1 + x + x^2\right)^2} \cdot (1 + 2x) \\ & = \frac{-(1 + 2x)}{\left(1 + x + x^2\right)^2} \end{aligned}$ find critical numbers of the second derivative. these numbers are where the 2nd deriv. is undefined or zero.where the 2nd derivative is zero: set (d²y/dx²) to equal 0 and solve for x $\begin{aligned} 0 & = \frac{-(1 + 2x)}{\left(1 + x + x^2\right)^2} \\ 0 & = -(1+2x) \\ -1 & = 2x \\ x & = -1/2 \end{aligned}$ where the 2nd derivative is undefined: if the denominator is equal to zero, the derivative will be undefeined. attempting to solve the equation 1 + x + x² = 0results in a discriminant of $b^2 - 4ac = 1^2 - 4(1)(1) = -3$ meaning that the equal 1 + x + x² = 0 has no real solutions and therefore will never equal zero.======our only critical number is x = -1/2.this forms two intervals to test the sign for: (-∞, -1/2) and (-1/2, ∞)test values into the 2nd derivative for (-∞, -1/2)if we try x = -1, we get $\begin{aligned} \left.\frac{d^2y}{dx^2}\right|_{x = -1} & = \frac{-(1+2(-1))}{\left(1+(-1)+(-1)^2\right)^2} \\ & = \frac{-(1-2)}{1} \\ & = 1 \end{aligned}$ so the since the 2nd derivative of the curve is positive on the interval (-∞, -1/2), the curve is concave up on (-∞, -1/2)test values into the 2nd derivative for (-1/2, ∞)if we try x = 0, we get $\begin{aligned} \left.\frac{d^2y}{dx^2}\right|_{x = 0} & = \frac{-(1+2(0))}{\left(1+(0)+(0)^2\right)^2} \\ & = \frac{-(1)}{1} \\ & = -1 \end{aligned}$ so the since the 2nd derivative of the curve is positive on the interval (-∞, -1/2), the curve is concave down on (-∞, -1/2).the interval on which the curve is concave up is (-∞, -1/2)