Mathematics, 01.06.2021 04:10, hskdbdjd

The
exact solusion of the IVP

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answered: bstine6678

$\displaystyle y = \sqrt{2}e^{\frac{1}{2} ln|2 + x^2|}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right

Equality Properties

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

Algebra I

|Absolute Value|FunctionsFunction NotationExponential Rule [Multiplying]: $\displaystyle b^m \cdot b^n = b^{m + n}$

Algebra II

Logarithms and Natural LogsEuler's number e

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Differential Equations

Separation of Variables

Antiderivatives - Integrals

Integration Constant C

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

U-Substitution

Logarithmic Integration

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle y' = \frac{xy}{2 + x^2}$

$\displaystyle y(0) = 2$

Step 2: Rewrite

Separation of Variables

Rewrite Derivative Notation: $\displaystyle \frac{dy}{dx} = \frac{xy}{2 + x^2}$ [Division Property of Equality] Isolate y's: $\displaystyle \frac{1}{y} \frac{dy}{dx} = \frac{x}{2 + x^2}$ [Multiplication Property of Equality] Rewrite Derivative Notation: $\displaystyle \frac{1}{y} dy = \frac{x}{2 + x^2} dx$

Step 3: Find General Solution Pt. 1

Integration

[Equality Property] Integrate both sides: $\displaystyle \int {\frac{1}{y}} \, dy = \int {\frac{x}{2 + x^2}} \, dx$ [1st Integral] Integrate [Logarithmic Integration]: $\displaystyle ln|y| = \int {\frac{x}{2 + x^2}} \, dx$

Step 4: Identify Variables

Identify variables for u-substitution for 2nd integral.

u = 2 + x²

du = 2xdx

Step 5: Find General Solution Pt. 2

[2nd Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle ln|y| = \frac{1}{2}\int {\frac{2x}{2 + x^2}} \, dx$ [2nd Integral] U-Substitution: $\displaystyle ln|y| = \frac{1}{2}\int {\frac{1}{u}} \, du$ [2nd Integral] Integrate [Logarithmic Integration]: $\displaystyle ln|y| = \frac{1}{2} ln|u| + C$ [Equality Property] e both sides: $\displaystyle e^{ln|y|} = e^{\frac{1}{2} ln|u| + C}$ Simplify: $\displaystyle |y| = e^{\frac{1}{2} ln|u| + C}$ Rewrite [Exponential Rule - Multiplying]: $\displaystyle |y| = e^{\frac{1}{2} ln|u|} \cdot e^C$ Simplify: $\displaystyle |y| = Ce^{\frac{1}{2} ln|u|}$ Back-Substitute: $\displaystyle |y| = Ce^{\frac{1}{2} ln|2 + x^2|}$

Our general solution is $\displaystyle |y| = Ce^{\frac{1}{2} ln|2 + x^2|}$ .

Step 6: Find Particular Solution

Substitute in point: $\displaystyle |2| = Ce^{\frac{1}{2} ln|2 + 0^2|}$ Evaluate |Absolute Value|: $\displaystyle 2 = Ce^{\frac{1}{2} ln|2 + 0^2|}$ |Absolute Value| Evaluate exponents: $\displaystyle 2 = Ce^{\frac{1}{2} ln|2 + 0|}$ |Absolute Value| Add: $\displaystyle 2 = Ce^{\frac{1}{2} ln|2|}$ |Absolute Value| Evaluate: $\displaystyle 2 = Ce^{\frac{1}{2} ln(2)}$ [Division Property of Equality] Isolate C: $\displaystyle \sqrt{2} = C$ Rewrite: $\displaystyle C = \sqrt{2}$ Substitute in C [General Solution]: $\displaystyle y = \sqrt{2}e^{\frac{1}{2} ln|2 + x^2|}$

∴ Our particular solution is $\displaystyle y = \sqrt{2}e^{\frac{1}{2} ln|2 + x^2|}$ .

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

Unit: Differential Equations

Book: College Calculus 10e

answered: Guest

Yes i also think it is b

answered: Guest

is should be y = 5x

step-by-step explanation:

correct me if im wrong