Advanced Placement (AP), 15.02.2021 23:50 darius12318

Which of the following is the solution to the differential equation dy/dx=e^(y+x) with initial condition y(0) = -ln4 A) y= -x-ln4
B) y=x-ln4
C) y = -ln(-e^x+5)
D) y = -ln(e^x+3)
E) y = ln(e^x+3)

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Answer from: hernan99961

C) y = -ln(-eˣ + 5)

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

BracketsParenthesisExponentsMultiplicationDivisionAdditionSubtractionLeft to Right

Equality Properties

Multiplication Property of EqualityDivision Property of EqualityAddition Property of EqualitySubtraction Property of Equality

Algebra I

Function NotationExponential Rule [Multiplying]: $\displaystyle b^m \cdot b^n = b^{m + n}$ Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$

Algebra II

Log PropertiesNatural log ln(x) and eˣ

Calculus

Antiderivatives - Integrals

Integration Constant C

U-Substitution

Slope Fields

Solving DifferentialsSeparation of Variables

Explanation:

Step 1: Define

$\displaystyle \frac{dy}{dx} = e^{y + x} \\y(0) = -ln4$

Step 2: Rewrite

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

[Differential Equation] Rewrite [Exponential Rule - Multiplying]: $\displaystyle \frac{dy}{dx} = e^y \cdot e^x$ [Diff Eq] Isolate x terms together [Multiplication Property of Equality]: $\displaystyle dy = e^y \cdot e^x dx$ [Diff Eq] Isolate y terms together [Division Property of Equality]: $\displaystyle \frac{dy}{e^y} = e^x dx$ [Diff Eq] Rewrite: $\displaystyle \frac{1}{e^y} dy = e^x dx$ [Diff Eq] Rewrite y [Exponential Rule - Rewrite]: $\displaystyle e^{-y} dy = e^x dx$

Step 3: Integrate Pt. 1

[Diff Eq] Integrate both sides [Equality Property]: $\displaystyle \int {e^{-y}} \, dy = \int {e^x} \, dx$

Step 4: Identify Variables for U-Substitution

Set variables for u-sub for y.

u = -y

du = -dy

Step 5: Integrate Pt. 2

[Integrals] Rewrite: $\displaystyle -\int {-e^{-y}} \, dy = \int {e^x} \, dx$ [Integrals] U-Substitution: $\displaystyle -\int {e^u} \, du = \int {e^x} \, dx$ [Integrals] eˣ integration: $\displaystyle -e^u = e^x + C$ [Integral Expression] Back-substitution: $\displaystyle -e^{-y} = e^x + C$

Step 6: Solve Differential Equation Pt. 1

[Int Exp] Divide -1 on both sides [Division Property of Equality]: $\displaystyle e^{-y} = -e^x - C$ [Int Exp] Natural log both sides (isolate y term) [Equality Property]: $\displaystyle -y = ln(-e^x - C)$ [Int Exp] Divide -1 on both sides [Division Property of Equality]: $\displaystyle y = -ln(-e^x - C)$

This is our differential function.

Step 7: Solve Differential Equation Pt. 2

[Diff Function] Substitute in given point: $\displaystyle -ln4 = -ln(-e^0 - C)$ [Diff Function] Evaluate exponent: $\displaystyle -ln4 = -ln(-1 - C)$ [Diff Function] Divide -1 on both sides [Division Property of Equality]: $\displaystyle ln4 = ln(-1 - C)$ [Diff Function] e both sides [Equality Property]: $\displaystyle 4 = -1 - C$ [Diff Function] Add 1 on both sides [Addition Property of Equality]: $\displaystyle 5 = -C$ [Diff Function] Divide -1 on both sides [Division Property of Equality]: $\displaystyle -5 = C$ [Diff Function] Rewrite: $\displaystyle C = -5$ [Diff Function] Substitute in Integration Constant C: $\displaystyle y = -ln(-e^x - -5)$ [Diff Function] Simplify: $\displaystyle y = -ln(-e^x + 5)$

Topic: Calculus

Unit: Slope Fields

Book: College Calculus 10e

Answer from: Quest

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