(dy)/(dt)=y+1" and "y(0)=3". " What is t when y=1 ? Choose 1 answer: (A) t=0 (B) t=ln 2 (C) t=-ln 2 (D) t=1 (E) t=ln 4

Solve Differential Equation: First, we need to solve the differential equation (dy)/(dt) = y + 1 with the initial condition y(0) = 3. We can separate variables and integrate.Separate Variables: Separate the variables: (dy)/(y+1) = dt.Integrate Both Sides: Integrate both sides: ∫(1/(y+1)) dy = ∫ dt.Find Constant of Integration: The integral of 1/(y+1) with respect to y is ln|y+1|, and the integral of dt is t. So we have ln|y+1| = t + C, where C is the constant of integration.Exponentiate to Solve for y: Using the initial condition y(0) = 3, we can find C. Plugging in the values, we get ln|3+1| = 0 + C, which simplifies to ln(4) = C.Substitute y = 1: Now we have the particular solution ln|y+1| = t + ln(4). We can solve for y by exponentiating both sides to get rid of the natural logarithm.Solve for e^t: Exponentiate both sides: e^(ln|y+1|) = e^(t + ln(4)), which simplifies to |y+1| = e^t * 4.Take Natural Logarithm: Since we are looking for when y = 1, we substitute 1 into the equation: |1+1| = e^t * 4, which simplifies to 2 = e^t * 4.Final Solution: Divide both sides by 4 to solve for e^t: e^t = 2/4, which simplifies to e^t = 1/2.Take the natural logarithm of both sides to solve for t: ln(e^t) = ln(1/2), which simplifies to t = ln(1/2).We know that ln(1/2) is the same as -ln(2), so t = -ln(2).

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(dy)/(dt)=y+1" and "y(0)=3". "
What is
t when
y=1 ?
Choose 1 answer:
(A)
t=0
(B)
t=ln 2
(C)
t=-ln 2
(D)
t=1
(E)
t=ln 4

$\frac{d y}{d t}=y+1 \text { and } y(0)=3 \text {. }$ $\newline$ What is $t$ when $y=1$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $t=0$ $\newline$ (B) $t=\ln 2$ $\newline$ (C) $t=-\ln 2$ $\newline$ (D) $t=1$ $\newline$ (E) $t=\ln 4$

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Full solution

\frac{d y}{d t}=y+1 \text { and } y(0)=3 \text {. }

\newline

What is

t

when

y=1

\newline

Choose

1

answer:

\newline

(A)

t=0

\newline

(B)

t=\ln 2

\newline

(C)

t=-\ln 2

\newline

(D)

t=1

\newline

(E)

t=\ln 4

Solve Differential Equation: First, we need to solve the differential equation $\frac{dy}{dt} = y + 1$ with the initial condition $y(0) = 3$ . We can separate variables and integrate.
Separate Variables: Separate the variables: (\frac{dy}{y+$1$} = dt)\.
Integrate Both Sides: Integrate both sides: \(\int\frac{1}{y+1}\, dy = \int dt.
Find Constant of Integration: The integral of $\frac{1}{y+1}$ with respect to $y$ is $\ln|y+1|$ , and the integral of $dt$ is $t$ . So we have $\ln|y+1| = t + C$ , where $C$ is the constant of integration.
Exponentiate to Solve for y: Using the initial condition $y(0) = 3$ , we can find $C$ . Plugging in the values, we get $\ln|3+1| = 0 + C$ , which simplifies to $\ln(4) = C$ .
Substitute $y = 1$ : Now we have the particular solution $\ln|y+1| = t + \ln(4)$ . We can solve for $y$ by exponentiating both sides to get rid of the natural logarithm.
Solve for $e^t$ : Exponentiate both sides: $e^{\ln|y+1|} = e^{t + \ln(4)}$ , which simplifies to $|y+1| = e^t \cdot 4$ .
Take Natural Logarithm: Since we are looking for when $y = 1$ , we substitute $1$ into the equation: $|1+1| = e^t \times 4$ , which simplifies to $2 = e^t \times 4$ .
Final Solution: Divide both sides by $4$ to solve for $e^t$ : $e^t = \frac{2}{4}$ , which simplifies to $e^t = \frac{1}{2}$ .
Final Solution: Divide both sides by $4$ to solve for $e^t$ : $e^t = \frac{2}{4}$ , which simplifies to $e^t = \frac{1}{2}$ .Take the natural logarithm of both sides to solve for $t$ : $\ln(e^t) = \ln(\frac{1}{2})$ , which simplifies to $t = \ln(\frac{1}{2})$ .
Final Solution: Divide both sides by $4$ to solve for $e^t$ : $e^t = \frac{2}{4}$ , which simplifies to $e^t = \frac{1}{2}$ .Take the natural logarithm of both sides to solve for $t$ : $\ln(e^t) = \ln(\frac{1}{2})$ , which simplifies to $t = \ln(\frac{1}{2})$ .We know that $\ln(\frac{1}{2})$ is the same as $-\ln(2)$ , so $t = -\ln(2)$ .