Mathematics, 27.05.2021 06:10 FoxodyGirl7134

Find the solution of the differential equation f' (t) = t^4+91-3/t having the boundary condition f(1) =1/4

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Answer from: campilk5

$\displaystyle f(t) = \frac{t^5}{5} + 91t - 3ln|t| - \frac{1819}{20}$

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

FunctionsFunction Notation

Calculus

Derivatives

Derivative Notation

Antiderivatives - Integrals

Integration Constant C

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

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

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle f'(t) = t^4 + 91 - \frac{3}{t}$

$\displaystyle f(1) = \frac{1}{4}$

Step 2: Integration

Integrate the derivative to find function.

[Derivative] Integrate: $\displaystyle \int {f'(t)} \, dt = \int {t^4 + 91 - \frac{3}{t}} \, dt$ Simplify: $\displaystyle f(t) = \int {t^4 + 91 - \frac{3}{t}} \, dt$ Rewrite [Integration Property - Addition/Subtraction]: $\displaystyle f(t) = \int {t^4} \, dt + \int {91} \, dt - \int {\frac{3}{t}} \, dt$ [1st Integral] Integrate [Integral Rule - Reverse Power Rule]: $\displaystyle f(t) = \frac{t^5}{5} + \int {91} \, dt - \int {\frac{3}{t}} \, dt$ [2nd Integral] Integrate [Integral Rule - Reverse Power Rule]: $\displaystyle f(t) = \frac{t^5}{5} + 91t - \int {\frac{3}{t}} \, dt$ [3rd Integral] Rewrite [Integral Property - Multiplied Constant]: $\displaystyle f(t) = \frac{t^5}{5} + 91t - 3\int {\frac{1}{t}} \, dt$ [3rd Integral] Integrate: $\displaystyle f(t) = \frac{t^5}{5} + 91t - 3ln|t| + C$

Our general solution is $\displaystyle f(t) = \frac{t^5}{5} + 91t - 3ln|t| + C$ .

Step 3: Find Particular Solution

Find Integration Constant C for function using given condition.

Substitute in condition [Function]: $\displaystyle f(1) = \frac{1^5}{5} + 91(1) - 3ln|1| + C$ Substitute in function value: $\displaystyle \frac{1}{4} = \frac{1^5}{5} + 91(1) - 3ln|1| + C$ Evaluate exponents: $\displaystyle \frac{1}{4} = \frac{1}{5} + 91(1) - 3ln|1| + C$ Evaluate natural log: $\displaystyle \frac{1}{4} = \frac{1}{5} + 91(1) - 3(0) + C$ Multiply: $\displaystyle \frac{1}{4} = \frac{1}{5} + 91 - 0 + C$ Add: $\displaystyle \frac{1}{4} = \frac{456}{5} - 0 + C$ Simplify: $\displaystyle \frac{1}{4} = \frac{456}{5} + C$ [Subtraction Property of Equality] Isolate C: $\displaystyle -\frac{1819}{20} = C$ Rewrite: $\displaystyle C = -\frac{1819}{20}$ Substitute in C [Function]: $\displaystyle f(t) = \frac{t^5}{5} + 91t - 3ln|t| - \frac{1819}{20}$

∴ Our particular solution to the differential equation is $\displaystyle f(t) = \frac{t^5}{5} + 91t - 3ln|t| - \frac{1819}{20}$ .

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

Unit: Integration

Book: College Calculus 10e

Answer from: Quest

a. turk and mcalister

b. howard, mission, and market

c. yes. since streets end some do not cross yet are not parallel to others. example: mcalister and 5th

step-by-step explanation:

parallel lines are lines which do not intersect ever and look a lot like railroad tracks. turk, mcalister and golden gate are examples of these.

intersect are streets which cross each other and a car on one can reach the other. howard, mission, and market intersect with 6th.

skew lines are lines which neither intersect nor are parallel. these lines exist here because some streets end and never intersect another like mcalister and 5th.

A. name the streets that are parallel to golden gate. b. name the streets that intersect 6th street.

Answer from: Quest

Where’s the

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