(dy)/(dt)=y, and y=1 when t=4. Solve the equation. Choose 1 answer: (A) y=e^(t-4) (B) y=4e^(t-1) (C) y=e^(4t) (D) y=4e^(t)

Identify Differential Equation: The given differential equation is a first-order linear differential equation which can be written as: (dy)/(dt) = y This is a separable differential equation, and we can solve it by separating the variables y and t.Separate Variables: To separate the variables, we divide both sides by y and multiply both sides by dt to get: (1/y) dy = dt Now we can integrate both sides to find the general solution.Integrate to Find General Solution: Integrating the left side with respect to y and the right side with respect to t, we get: ∫(1/y) dy = ∫ dt ln|y| = t + C where C is the constant of integration.Exponentiate to Solve for y: To solve for y, we exponentiate both sides to get rid of the natural logarithm: e^(ln|y|) = e^(t + C) y = e^t * e^C Since e^C is just another constant, we can rename it as C': y = C'e^tApply Initial Condition: Now we use the initial condition y=1 when t=4 to find the specific value of C': 1 = C'e^4 C' = 1/e^4Find Particular Solution: Substitute C' back into the general solution to get the particular solution: y = (1/e^4)e^t y = e^(t-4) This matches answer choice (A).

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(dy)/(dt)=y, and
y=1 when
t=4.
Solve the equation.
Choose 1 answer:
(A)
y=e^(t-4)
(B)
y=4e^(t-1)
(C)
y=e^(4t)
(D)
y=4e^(t)

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

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Algebra 2

Composition of linear and quadratic functions: find an equation

Full solution

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

, and

y=1

when

t=4

\newline

Solve the equation.

\newline

Choose

1

answer:

\newline

(A)

y=e^{t-4}

\newline

(B)

y=4 e^{t-1}

\newline

(C)

y=e^{4 t}

\newline

(D)

y=4 e^{t}

Identify Differential Equation: The given differential equation is a first-order linear differential equation which can be written as: $\newline$ $\frac{dy}{dt} = y$ $\newline$ This is a separable differential equation, and we can solve it by separating the variables $y$ and $t$ .
Separate Variables: To separate the variables, we divide both sides by $y$ and multiply both sides by $dt$ to get: $\newline$ $(\frac{1}{y}) dy = dt$ $\newline$ Now we can integrate both sides to find the general solution.
Integrate to Find General Solution: Integrating the left side with respect to $y$ and the right side with respect to $t$ , we get: $\newline$ $\int(\frac{1}{y}) \, dy = \int dt$ $\newline$ $\ln|y| = t + C$ $\newline$ where $C$ is the constant of integration.
Exponentiate to Solve for $y$ : To solve for $y$ , we exponentiate both sides to get rid of the natural logarithm: $\newline$ $e^{\ln|y|} = e^{t + C}$ $\newline$ $y = e^t \cdot e^C$ $\newline$ Since $e^C$ is just another constant, we can rename it as $C'$ : $\newline$ $y = C'e^t$
Apply Initial Condition: Now we use the initial condition $y=1$ when $t=4$ to find the specific value of $C'$ : $\newline$ $1 = C'e^{4}$ $\newline$ $C' = \frac{1}{e^{4}}$
Find Particular Solution: Substitute $C'$ back into the general solution to get the particular solution: $\newline$ $y = (1/e^4)e^t$ $\newline$ $y = e^{t-4}$ $\newline$ This matches answer choice $(A)$ .