Mathematics, 14.01.2021 02:50 alexanderickh1

Need help please its Calculus. Ill give the 5 stars as well.

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Answer from: pickelswolf5938

$\displaystyle y = 2e^\bigg{\frac{x^3}{3}} + 1$

General Formulas and Concepts:

Pre-Algebra

Order of OperationsEquality Properties

Algebra I

FunctionsFunction NotationExponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$

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

U-Substitution

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

Step-by-step explanation:

*Note:

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

Step 1: Define

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

$\displaystyle f(0) = 3$

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

Step 3: Find General Solution Pt. 1

[Differential] Integrate both sides: $\displaystyle \int {\frac{1}{y - 1}} \, dy = \int {x^2} \, dx$ [dx Integral] Integrate [Integration Rule - Reverse Power Rule]: $\displaystyle \int {\frac{1}{y - 1}} \, dy = \frac{x^3}{3} + C$

Step 4: Find General Solution Pt. 2

Identify variables for u-substitution for dy.

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

Step 5: Find General Solution Pt. 3

[dy Integral] U-Substitution: $\displaystyle \int {\frac{1}{u}} \, du = \frac{x^3}{3} + C$ [dy Integral] Integrate [Logarithmic Integration]: $\displaystyle ln|u| = \frac{x^3}{3} + C$ [Equality Property] e both sides: $\displaystyle e^\bigg{ln|u|} = e^\bigg{\frac{x^3}{3} + C}$ Simplify: $\displaystyle |u| = Ce^\bigg{\frac{x^3}{3}}$ Rewrite: $\displaystyle u = \pm Ce^\bigg{\frac{x^3}{3}}$ Back-Substitute: $\displaystyle y - 1 = \pm Ce^\bigg{\frac{x^3}{3}}$ [Equality Property] Isolate y: $\displaystyle y = \pm Ce^\bigg{\frac{x^3}{3}} + 1$

General Form: $\displaystyle y = \pm Ce^\bigg{\frac{x^3}{3}} + 1$

Step 6: Find Particular Solution

Substitute in function values [General Form]: $\displaystyle 3 = \pm Ce^\bigg{\frac{0^3}{3}} + 1$ Simplify: $\displaystyle 3 = \pm C + 1$ [Equality Property] Isolate C: $\displaystyle 2 = \pm C$ Rewrite: $\displaystyle C = 2$ Substitute in C [General Form]: $\displaystyle y = 2e^\bigg{\frac{x^3}{3}} + 1$

∴ our particular solution is $\displaystyle y = 2e^\bigg{\frac{x^3}{3}} + 1$ .

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

Unit: Differentials and Slope Fields

Book: College Calculus 10e

Answer from: Quest

since your problem is x^4 = 81, what you have to undeo is the "to the fourth power" part.

the inverse (opposite operation) of "to the fourth power" is "to the fourth root".

(x^4) ^ ¼ = 81 ^ ¼

81 = 3^4 so (3 ^ 4) ^ ¼ = 3 ^ 1 = 3

by hand: (3 ^ 4) ^ ¼ = 3 ^ (4 x ¼) = 3 ^ 1 = 3 (when you take a power to a power, multiply the exponents.)

or, do it by calculator. one way of finding the fourth root is to use ^ (1/4)

step-by-step explanation: