Mathematics, 10.03.2021 21:20 shamiya15

Find the particular solution of the differential equation dy/dx=y−5/x−8 using the given boundary condition: y=7 when x=9. (Use symbolic notation and fractions where needed.)

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Answer from: angellong94

$\displaystyle y - 5 = 2(x - 8)$

General Formulas and Concepts:

Symbols

e (Euler's number) ≈ 2.7182

Pre-Algebra

Order of Operations: BPEMDAS

BracketsParenthesisExponentsMultiplicationDivisionAdditionSubtractionLeft to Right

Distributive Property

Equality Properties

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

Algebra I

Functions

Point-Slope Form: y - y₁ = m(x - x₁)

x₁ - x coordinatey₁ - y coordinatem - slope

Slope-Intercept Form: y = mx + b

m - slope b - y-intercept

Algebra II

Logarithms - ln and e

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

Slope Fields

Solving Differentials - Integrals

Integration Constant C

U-Substitution

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

Step-by-step explanation:

*Note:

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

Step 1: Define

$\displaystyle \frac{dy}{dx} = \frac{y - 5}{x - 8}$

x = 9, y = 7

Step 2: Rewrite Differential

Rewrite Leibniz Notation using Separation of Variables.

[Separation of Variables] Isolate x's together: $\displaystyle dy = \frac{y - 5}{x - 8}dx$ [Separation of Variables] Isolate y's together: $\displaystyle \frac{1}{y - 5}dy = \frac{1}{x - 8}dx$

Step 3: Integrate Pt. 1

Solving general form of differential using integration.

[Differential] Integrate both sides: $\displaystyle \int {\frac{1}{y - 5}} \, dy = \int {\frac{1}{x - 8}} \, dx$

Step 4: Identify Variables

Set up u-substitution for right integral.

Integral w/ respect to y

u = y - 5

du = dy

Integral w/ respect to x

z = x - 8

dz = dx

Step 5: Integrate Pt. 2

[Integrals] U-Substitution: $\displaystyle \int {\frac{1}{u}} \, du = \int {\frac{1}{z}} \, dz$ [Integrals] ln Integration: $\displaystyle ln|u| = ln|z| + C$ Back-Substitute: $\displaystyle ln|y - 5| = ln|x - 8| + C$ [Equality Property] Raise e on both sides: $\displaystyle e^{ln|y - 5|} = e^{ln|x - 8| + C}$ Simplify: $\displaystyle |y - 5| = C|x - 8|$

General Form: $\displaystyle |y - 5| = C|x - 8|$

Step 6: Solve Particular Solution

Since both sides have absolute value, assume that the particular solution will be positive.

Substitute in variables [General Form]: $\displaystyle |7 - 5| = C|9 - 8|$ [Particular] |Absolute Value| Subtract: $\displaystyle |2| = C|1|$ [Particular] Evaluate absolute values: $\displaystyle 2 = C(1)$ [Particular] Multiply: $\displaystyle 2 = C$ [Particular] Rewrite: $\displaystyle C = 2$

Substituting integration constant C into the general form:

Particular Solution (in Point-Slope Form): $\displaystyle y - 5 = 2(x - 8)$

Particular Solution (in Slope-Intercept Form): $\displaystyle y = 2x - 11$

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

Unit: Differentials and Slope Fields

Book: College Calculus 10e

Answer from: Quest

so firstly, we need to isolate the y variables to be able to solve these inequalities. firstly, add 0.5x on both sides of the first inequality and subtract 1.5x on both sides of the second inequality:

now since the slope is positive for the first inequality, this means that the line going upwards belongs to the first inequality, and the line going downwards belongs to the second inequality.

next, since y ≥ in the first inequality, this means that the shaded region will be above the first inequality's line, thus shading regions a and b.

next, since y ≤ in the second inequality, this means that the shaded region is going to be below this line, thus shading regions a and c.

since both lines shade region a, this means that region a is the solution.

step-by-step explanation:

mark brainliest!

Answer from: Quest

a is 8.b is 10.c is 10.5. d is -.5

step-by-step explanation:

just minus the smallest number from the biggest.

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