Question

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

Answer 1

Answer:

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

General Formulas and Concepts:

Pre-Algebra

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

Algebra I

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

Calculus

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.

Step 1: Define

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

t = 1

s = 8

Step 2: Integrate

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:                                                                                                         $\displaystyle s = \frac{5t^4}{4} + \frac{9}{t} + C$

Step 3: Solve

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 C:                                               $\displaystyle \frac{-9}{4} = C$
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