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Let’s check out your problem:

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

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

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Full solution

Q.

\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=1

\newline

(B)

t=-\ln 2

\newline

(C)

t=\ln 2

\newline

(D)

t=\ln 4

\newline

(E)

t=0

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: $\frac{1}{y+1}$ $dy$ = $dt$ .
Integrate Both Sides: Integrate both sides: $\int\frac{1}{y+1}\, dy = \int dt$ .
Find Constant of Integration: After integrating, we get $\ln|y+1| = t + C$ , where $C$ is the constant of integration.
Calculate C: Using the initial condition $y(0) = 3$ , we can find $C$ . Plug in $t = 0$ and $y = 3$ into $\ln|y+1| = t + C$ to get $\ln|3+1| = 0 + C$ .
Obtain Particular Solution: Calculate $C$ : $\ln|4| = C$ , so $C = \ln 4$ .
Find $t$ for $y=1$ : Now we have the particular solution $\ln|y+1| = t + \ln 4$ .
Simplify Equation: We want to find $t$ when $y = 1$ . Plug $y = 1$ into $\ln|y+1| = t + \ln 4$ to get $\ln|1+1| = t + \ln 4$ .
Subtract $\ln 4$ : Simplify the equation: $\ln 2 = t + \ln 4$ .
Combine Terms: Subtract $\ln 4$ from both sides to solve for $t$ : $t = \ln 2 - \ln 4$ .
Simplify Fraction: Use the property of logarithms $\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)$ to combine the terms: $t = \ln\left(\frac{2}{4}\right)$ .
Recognize $\ln(\frac{1}{2})$ : Simplify the fraction inside the logarithm: $t = \ln(\frac{1}{2})$ .
Recognize $\ln(\frac{1}{2})$ : Simplify the fraction inside the logarithm: $t = \ln(\frac{1}{2})$ .Recognize that $\ln(\frac{1}{2})$ is the same as $-\ln 2$ : $t = -\ln 2$ .

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