(dy)/(dt)=-7y, and y=3 when t=0. Solve the equation. Choose 1 answer: (A) y=3e^(-7t) (B) y=3+e^(-7t) (C) y=2+e^(-7t) (D) y=e^(-7t)

Identify Equation Type: The given differential equation is a first-order linear homogeneous differential equation. To solve it, we can use separation of variables.Separate Variables: Separate the variables y and t by dividing both sides by y and multiplying both sides by dt: (dy/y) = -7 dtIntegrate Both Sides: Integrate both sides of the equation: ∫(1/y) dy = ∫-7 dtFind Constant of Integration: The integral of 1/y with respect to y is ln|y|, and the integral of -7 with respect to t is -7t. So we have: ln|y| = -7t + C, where C is the constant of integration.Apply Initial Condition: To find the constant of integration C, we use the initial condition y=3 when t=0: ln|3| = -7(0) + C C = ln(3)Use Natural Logarithm: Now we have the equation with the constant C: ln|y| = -7t + ln(3)Exponentiate Both Sides: To solve for y, we exponentiate both sides of the equation to get rid of the natural logarithm: e^(ln|y|) = e^(-7t + ln(3))Simplify Exponents: Since e^(ln|x|) = |x| for any x, and since y is positive (as it is an exponential decay problem), we can drop the absolute value: y = e^(-7t) * e^(ln(3))Final Solution: Simplify the right side by combining the exponents: y = 3 * e^(-7t)The solution to the differential equation is y = 3 * e^(-7t), which corresponds to option (A).

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(dy)/(dt)=-7y, and
y=3 when
t=0.
Solve the equation.
Choose 1 answer:
(A)
y=3e^(-7t)
(B)
y=3+e^(-7t)
(C)
y=2+e^(-7t)
(D)
y=e^(-7t)

$\frac{d y}{d t}=-7 y$ , and $y=3$ when $t=0$ . $\newline$ Solve the equation. $\newline$ Choose $1$ answer: $\newline$ (A) $y=3 e^{-7 t}$ $\newline$ (B) $y=3+e^{-7 t}$ $\newline$ (C) $y=2+e^{-7 t}$ $\newline$ (D) $y=e^{-7 t}$

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Find derivatives of using multiple formulae

Full solution

\frac{d y}{d t}=-7 y

, and

y=3

when

t=0

\newline

Solve the equation.

\newline

Choose

1

answer:

\newline

(A)

y=3 e^{-7 t}

\newline

(B)

y=3+e^{-7 t}

\newline

(C)

y=2+e^{-7 t}

\newline

(D)

y=e^{-7 t}

Identify Equation Type: The given differential equation is a first-order linear homogeneous differential equation. To solve it, we can use separation of variables.
Separate Variables: Separate the variables $y$ and $t$ by dividing both sides by $y$ and multiplying both sides by $dt$ : $\left(\frac{dy}{y}\right) = -7 dt$
Integrate Both Sides: Integrate both sides of the equation: $\int(\frac{1}{y}) \, dy = \int -7 \, dt$
Find Constant of Integration: The integral of $\frac{1}{y}$ with respect to $y$ is $\ln|y|$ , and the integral of $-7$ with respect to $t$ is $-7t$ . So we have: $\newline$ $\ln|y| = -7t + C$ , where $C$ is the constant of integration.
Apply Initial Condition: To find the constant of integration $C$ , we use the initial condition $y=3$ when $t=0$ :
$\ln|3| = -7(0) + C$
$C = \ln(3)$
Use Natural Logarithm: Now we have the equation with the constant $C$ : $\ln|y| = -7t + \ln(3)$
Exponentiate Both Sides: To solve for $y$ , we exponentiate both sides of the equation to get rid of the natural logarithm: $e^{\ln|y|} = e^{-7t + \ln(3)}$
Simplify Exponents: Since $e^{\ln|x|} = |x|$ for any $x$ , and since $y$ is positive (as it is an exponential decay problem), we can drop the absolute value: $\newline$ $y = e^{-7t} \cdot e^{\ln(3)}$
Final Solution: Simplify the right side by combining the exponents: $y = 3 \times e^{(-7t)}$
Final Solution: Simplify the right side by combining the exponents: $\newline$ $y = 3 \cdot e^{(-7t)}$ The solution to the differential equation is $y = 3 \cdot e^{(-7t)}$ , which corresponds to option (A).