if dy/dx = ysec^2(x) and y= 5 when x = 0, then y=?

А. e^tanx +4
B. e^tanx +5
C. 5e^tanx
D. tanx + 5
E. tanx + 5e^x

Answers

Answer 1

Answer:

C. [tex]\displaystyle y = 5e^{tan(x)}[/tex]

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|Absolute Values|Anything to the 0th power is 1

Algebra II

Logarithms and Natural logsEuler's number e

Calculus

Derivatives

Derivative Notation

Trig Derivatives

Differential Equations

Separation of variablesGeneral and particular solutions

Antiderivatives - Integration

Integration Constant C

Trig Integration

Logarithmic Integration

Step-by-step explanation:

Step 1: Define

Identify

[tex]\displaystyle \frac{dy}{dx} = ysec^2(x)[/tex]

x = 0, y = 5

Step 2: Rewrite Differential

Separation of variables

[Division Property of Equality] Isolate y terms together:                             [tex]\displaystyle \frac{1}{y}\frac{dy}{dx} = sec^2(x)[/tex][Multiplication Property of Equality] Isolate x terms together:                   [tex]\displaystyle \frac{1}{y}dy = sec^2(x)dx[/tex]

Step 3: Find General Solution

[Equality Property] Integrate both sides:                                                     [tex]\displaystyle \int {\frac{1}{y}} \, dy = \int {sec^2(x)} \, dx[/tex][1st Integral] Integrate [Logarithmic Integration]:                                         [tex]\displaystyle ln|y| = \int {sec^2(x)} \, dx[/tex][2nd Integral] Integrate [Trig Integration]:                                                   [tex]\displaystyle ln|y| = tan(x) + C[/tex][Equality Property] e both sides:                                                                   [tex]\displaystyle e^{ln|y|} = e^{tan(x) + C}[/tex]Simplify:                                                                                                         [tex]\displaystyle |y| = Ce^{tan(x)}[/tex]Rewrite:                                                                                                         [tex]\displaystyle y = \pm Ce^{tan(x)}[/tex]

Step 4: Find Particular Solution

Substitute in variables [Function]:                                                               [tex]\displaystyle |5| = Ce^{tan(0)}[/tex]Evaluate absolute value:                                                                               [tex]\displaystyle 5 = Ce^{tan(0)}[/tex]Evaluate trig:                                                                                                 [tex]\displaystyle 5 = Ce^0[/tex]Evaluate exponent:                                                                                       [tex]\displaystyle 5 = C(1)[/tex]Multiply:                                                                                                             [tex]\displaystyle 5 = C[/tex]Rewrite:                                                                                                         [tex]\displaystyle C =5[/tex]Substitute in C [General Solution]:                                                               [tex]\displaystyle y = 5e^{tan(x)}[/tex]

∴ Our answer is C.

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

Unit: Differential Equations

Book: College Calculus 10e


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