(dy)/(dx)=-3y, and y=2 when x=0. Solve the equation. Choose 1 answer: (A) y=2+e^(-3x) (B) y=1+e^(-3x) (C) y=2e^(-3x) () y=e^(-3x)

Recognize the Equation Type: Recognize the type of differential equation. The given differential equation is a first-order linear homogeneous differential equation.Solve the Equation: Solve the differential equation. To solve (dy)/(dx) = -3y, we can use separation of variables. Rearrange the equation to separate y and x: (dy)/y = -3 dxIntegrate Both Sides: Integrate both sides. Integrate the left side with respect to y and the right side with respect to x: ∫(1/y) dy = ∫-3 dx ln|y| = -3x + C, where C is the constant of integration.Solve for y: Solve for y. Exponentiate both sides to solve for y: e^(ln|y|) = e^(-3x + C) y = e^C * e^(-3x) Since e^C is a constant, we can denote it as A, where A = e^C. So, y = Ae^(-3x)Use Initial Condition: Use the initial condition to find A. Given that y=2 when x=0, we can substitute these values into y = Ae^(-3x) to find A: 2 = Ae^(0) 2 = AWrite Final Solution: Write the final solution. Now that we have found A, we can write the final solution: y = 2e^(-3x)

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Algebra 1

Transformations of quadratic functions

Full solution

\frac{d y}{d x}=-3 y

, and

y=2

when

x=0

\newline

Solve the equation.

\newline

Choose

1

answer:

\newline

(A)

y=2+e^{-3 x}

\newline

(B)

y=1+e^{-3 x}

\newline

(C)

y=2 e^{-3 x}

\newline

(D)

y=e^{-3 x}

Recognize the Equation Type: Recognize the type of differential equation. The given differential equation is a first-order linear homogeneous differential equation.
Solve the Equation: Solve the differential equation. $\newline$ To solve $(\frac{dy}{dx}) = -3y$ , we can use separation of variables. Rearrange the equation to separate $y$ and $x$ : $\newline$ $(\frac{dy}{y}) = -3 \, dx$
Integrate Both Sides: Integrate both sides. $\newline$ Integrate the left side with respect to $y$ and the right side with respect to $x$ : $\newline$ $\int(\frac{1}{y}) dy = \int-3 dx$ $\newline$ $\ln|y| = -3x + C$ , where $C$ is the constant of integration.
Solve for $y$ : Solve for $y$ . $\newline$ Exponentiate both sides to solve for $y$ : $\newline$ $e^{\ln|y|} = e^{(-3x + C)}$ $\newline$ $y = e^C \cdot e^{(-3x)}$ $\newline$ Since $e^C$ is a constant, we can denote it as $A$ , where $A = e^C$ . $\newline$ So, $y = Ae^{(-3x)}$
Use Initial Condition: Use the initial condition to find $A$ . Given that $y=2$ when $x=0$ , we can substitute these values into $y = Ae^{-3x}$ to find $A$ : $2 = Ae^{0}$ $2 = A$
Write Final Solution: Write the final solution. $\newline$ Now that we have found $A$ , we can write the final solution: $\newline$ $y = 2e^{-3x}$