(dy)/(dx)=-4y, and y=3 when x=2. Solve the equation. Choose 1 answer: (A) y=3e^(-4x) (B) y=6e^(-4x) (C) y=3e^(8-4x) (D) y=3e^(4-4x)

Recognize Type of Differential Equation: Recognize the type of differential equation. The given differential equation (dy)/(dx) = -4y is a first-order linear homogeneous differential equation with a constant coefficient.Solve the Equation: Solve the differential equation. To solve this equation, we can use separation of variables. We rearrange the terms to get dy/y = -4 dx and then integrate both sides. ∫(1/y) dy = ∫-4 dxPerform the Integration: Perform the integration. The integral of 1/y with respect to y is ln|y|, and the integral of -4 with respect to x is -4x. So we have: ln|y| = -4x + C, where C is the constant of integration.Solve for y: Solve for y. To solve for y, we exponentiate both sides to get rid of the natural logarithm: e^(ln|y|) = e^(-4x + C) y = e^(-4x) * e^C Since e^C is just a constant, we can rename it as C': y = C'e^(-4x)Apply Initial Condition: Apply the initial condition to find C'. We are given that y = 3 when x = 2. We substitute these values into the equation to find C': 3 = C'e^(-4*2) 3 = C'e^(-8) C' = 3/e^(-8)Calculate C': Calculate the value of C'. To find C', we calculate 3/e^(-8): C' = 3 * e^8 This is incorrect because we should have divided by e^(-8), not multiplied. This is a math error.

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(dy)/(dx)=-4y, and
y=3 when
x=2.
Solve the equation.
Choose 1 answer:
(A)
y=3e^(-4x)
(B)
y=6e^(-4x)
(C)
y=3e^(8-4x)
(D)
y=3e^(4-4x)

$\frac{d y}{d x}=-4 y$ , and $y=3$ when $x=2$ . $\newline$ Solve the equation. $\newline$ Choose $1$ answer: $\newline$ (A) $y=3 e^{-4 x}$ $\newline$ (B) $y=6 e^{-4 x}$ $\newline$ (C) $y=3 e^{8-4 x}$ $\newline$ (D) $y=3 e^{4-4 x}$

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Math Problems

Algebra 1

Transformations of quadratic functions

Full solution

\frac{d y}{d x}=-4 y

, and

y=3

when

x=2

\newline

Solve the equation.

\newline

Choose

1

answer:

\newline

(A)

y=3 e^{-4 x}

\newline

(B)

y=6 e^{-4 x}

\newline

(C)

y=3 e^{8-4 x}

\newline

(D)

y=3 e^{4-4 x}

Recognize Type of Differential Equation: Recognize the type of differential equation. The given differential equation $\frac{dy}{dx} = -4y$ is a first-order linear homogeneous differential equation with a constant coefficient.
Solve the Equation: Solve the differential equation. $\newline$ To solve this equation, we can use separation of variables. We rearrange the terms to get $\frac{dy}{y} = -4 dx$ and then integrate both sides. $\newline$ $\int \frac{1}{y} dy = \int -4 dx$
Perform the Integration: Perform the integration. $\newline$ The integral of $\frac{1}{y}$ with respect to $y$ is $\ln|y|$ , and the integral of $-4$ with respect to $x$ is $-4x$ . So we have: $\newline$ $\ln|y| = -4x + C$ , where $C$ is the constant of integration.
Solve for y: Solve for y. $\newline$ To solve for y, we exponentiate both sides to get rid of the natural logarithm: $\newline$ $e^{\ln|y|} = e^{-4x + C}$ $\newline$ $y = e^{-4x} \cdot e^C$ $\newline$ Since $e^C$ is just a constant, we can rename it as $C'$ : $\newline$ $y = C'e^{-4x}$
Apply Initial Condition: Apply the initial condition to find $C'$ . We are given that $y = 3$ when $x = 2$ . We substitute these values into the equation to find $C'$ : $3 = C'e^{(-4\cdot2)}$ $3 = C'e^{(-8)}$ $C' = \frac{3}{e^{-8}}$
Calculate $C'$ : Calculate the value of $C'$ . $\newline$ To find $C'$ , we calculate $\frac{3}{e^{-8}}$ : $\newline$ $C' = 3 \times e^8$ $\newline$ This is incorrect because we should have divided by $e^{-8}$ , not multiplied. This is a math error.