Question

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

Answer 1

Answer:

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

General Formulas and Concepts:

Pre-Algebra

• Order of Operations
• Equality Properties

Algebra I

• Functions
• Function Notation
• Exponential 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:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

Slope Fields

• Separation of Variables

• Solving 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.

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

Step 3: Find General Solution Pt. 1

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

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

Step 5: Find General Solution Pt. 3

1. [dy Integral] U-Substitution:                                                                             $\displaystyle \int {\frac{1}{u}} \, du = \frac{x^3}{3} + C$
2. [dy Integral] Integrate [Logarithmic Integration]:                                            $\displaystyle ln|u| = \frac{x^3}{3} + C$
3. [Equality Property] e both sides:                                                                     $\displaystyle e^\bigg{ln|u|} = e^\bigg{\frac{x^3}{3} + C}$
4. Simplify:                                                                                                             $\displaystyle |u| = Ce^\bigg{\frac{x^3}{3}}$
5. Rewrite:                                                                                                             $\displaystyle u = \pm Ce^\bigg{\frac{x^3}{3}}$
6. Back-Substitute:                                                                                               $\displaystyle y - 1 = \pm Ce^\bigg{\frac{x^3}{3}}$
7. [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

1. Substitute in function values [General Form]:                                                $\displaystyle 3 = \pm Ce^\bigg{\frac{0^3}{3}} + 1$
2. Simplify:                                                                                                             $\displaystyle 3 = \pm C + 1$
3. [Equality Property] Isolate C:                                                                           $\displaystyle 2 = \pm C$
4. Rewrite:                                                                                                             $\displaystyle C = 2$
5. 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