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timurjin [86]
2 years ago
5

Find the particular solution of the differential equation? /=5^3+9^2, when =1, =8

Mathematics
1 answer:
kipiarov [429]2 years ago
3 0

Answer:

\displaystyle s = \frac{5t^4}{4} + \frac{9}{t} - \frac{9}{4}

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality

<u>Algebra I</u>

  • Exponential Rule [Rewrite]:                                                                           \displaystyle b^{-m} = \frac{1}{b^m}

<u>Calculus</u>

Derivatives

Derivative Notation

Solving Differentials - Integrals

Integration Constant C

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

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

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

Step-by-step explanation:

*Note:

Ignore the Integration Constant C on the left hand side of the differential equation when integrating.

<u>Step 1: Define</u>

\displaystyle \frac{ds}{dt} = 5t^3 + \frac{9}{t^2}

t = 1

s = 8

<u>Step 2: Integrate</u>

  1. [Derivative] Rewrite [Leibniz's Notation]:                                                     \displaystyle ds = (5t^3 + \frac{9}{t^2})dt
  2. [Equality Property] Integrate both sides:                                                     \displaystyle \int {} \, ds = \int {(5t^3 + \frac{9}{t^2})} \, dt
  3. [Left Integral] Reverse Power Rule:                                                             \displaystyle s = \int {(5t^3 + \frac{9}{t^2})} \, dt
  4. [Right Integral] Rewrite [Integration Property - Addition]:                           \displaystyle s = \int {5t^3} \, dt + \int {\frac{9}{t^2}} \, dt
  5. [Right Integrals] Rewrite [Integration Property - Multiplied Constant]:     \displaystyle s = 5\int {t^3} \, dt + 9\int {\frac{1}{t^2}} \, dt
  6. [Right Integrals] Rewrite [Exponential Rule - Rewrite]:                               \displaystyle s = 5\int {t^3} \, dt + 9\int {t^{-2}} \, dt
  7. [Right Integrals] Reverse Power Rule:                                                         \displaystyle s = 5(\frac{t^4}{4}) + 9(\frac{t^{-1}}{-1}) + C
  8. [Right Integrals] Rewrite [Exponential Rule - Rewrite]:                               \displaystyle s = 5(\frac{t^4}{4}) + 9(\frac{1}{t}) + C
  9. Multiply:                                                                                                         \displaystyle s = \frac{5t^4}{4} + \frac{9}{t} + C

<u>Step 3: Solve</u>

  1. Substitute in variables:                                                                                 \displaystyle 8 = \frac{5(1)^4}{4} + \frac{9}{1} + C
  2. Evaluate exponents:                                                                                     \displaystyle 8 = \frac{5}{4} + \frac{9}{1} + C
  3. Divide:                                                                                                           \displaystyle 8 = \frac{5}{4} + 9 + C
  4. Add:                                                                                                               \displaystyle 8 = \frac{41}{4} + C
  5. [Subtraction Property of Equality] Isolate <em>C</em>:                                               \displaystyle \frac{-9}{4} = C
  6. Rewrite:                                                                                                          \displaystyle C = \frac{-9}{4}

Particular Solution: \displaystyle s = \frac{5t^4}{4} + \frac{9}{t} - \frac{9}{4}

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

Unit: Differentials Equations and Slope Fields

Book: College Calculus 10e

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