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

Recognize type: Recognize the type of differential equation. The given differential equation (dy)/(dx)=-4y is a first-order linear homogeneous differential equation.Solve equation: Solve the differential equation. To solve this equation, we can use separation of variables. We rearrange the terms to separate the variables y and x: (dy)/y = -4 dx Now, we integrate both sides: ∫(1/y) dy = ∫-4 dx The integral of (1/y) dy is ln|y|, and the integral of -4 dx is -4x. So we have: ln|y| = -4x + C, where C is the constant of integration.Find 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)Use initial condition: Use the initial condition to find the constant 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) Now, we solve for C': C' = 3/e^(-8) C' = 3e^(8)Find constant: Write the final solution. Now that we have the value of C', we can write the final solution: y = 3e^(8)e^(-4x) We can combine the exponents since they have the same base: y = 3e^(8-4x) This matches answer choice (A).

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

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

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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^{8-4 x}

\newline

(B)

y=3 e^{-4 x}

\newline

(C)

y=3 e^{4-4 x}

\newline

(D)

y=6 e^{-4 x}

Recognize type: Recognize the type of differential equation. $\newline$ The given differential equation $\frac{dy}{dx}=-4y$ is a first-order linear homogeneous differential equation.
Solve equation: Solve the differential equation. $\newline$ To solve this equation, we can use separation of variables. We rearrange the terms to separate the variables $y$ and $x$ : $\newline$ $\frac{dy}{y} = -4 dx$ $\newline$ Now, we integrate both sides: $\newline$ $\int(\frac{1}{y}) dy = \int-4 dx$ $\newline$ The integral of $\frac{1}{y}$ dy is $\ln|y|$ , and the integral of $-4 dx$ is $-4x$ . So we have: $\newline$ $\ln|y| = -4x + C$ , where $C$ is the constant of integration.
Find $y$ : 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 $C'$ : $\newline$ $y = C'e^{(-4x)}$
Use initial condition: Use the initial condition to find the constant $C'$ . We are given that $y=3$ when $x=2$ . We substitute these values into the equation to find $C'$ : $3 = C'e^{(-4*2)}$ $3 = C'e^{(-8)}$ Now, we solve for $C'$ : $C' = \frac{3}{e^{(-8)}}$ $C' = 3e^{8}$
Find constant: Write the final solution. $\newline$ Now that we have the value of $C'$ , we can write the final solution: $\newline$ $y = 3e^{8}e^{-4x}$ $\newline$ We can combine the exponents since they have the same base: $\newline$ $y = 3e^{8-4x}$ $\newline$ This matches answer choice $(A)$ .