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(dy)/(dt)=2t+3 and y(1)=6.
What is t when y=0 ?
Choose all answers that apply:
(A) t=-3
(B) t=-1
(C) t=1
(D) t=0
(E) t=-2
(F) t=-4

$\frac{d y}{d t}=2 t+3 \text { and } y(1)=6 .$ $\newline$ What is $t$ when $y=0$ ? $\newline$ Choose all answers that apply: $\newline$ (A) $t=-3$ $\newline$ (B) $t=-1$ $\newline$ (C) $t=1$ $\newline$ (D) $t=0$ $\newline$ (E) $t=-2$ $\newline$ (F) $t=-4$

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Full solution

Q.

\frac{d y}{d t}=2 t+3 \text { and } y(1)=6 .

\newline

What is

t

when

y=0

?

\newline

Choose all answers that apply:

\newline

(A)

t=-3

\newline

(B)

t=-1

\newline

(C)

t=1

\newline

(D)

t=0

\newline

(E)

t=-2

\newline

(F)

t=-4

Given Differential Equation: Given the differential equation $\frac{dy}{dt} = 2t + 3$ and the initial condition $y(1) = 6$ , we want to find the value of $t$ when $y = 0$ . To do this, we will separate variables and integrate both sides of the differential equation.
Separate Variables and Integrate: Separate the variables by moving all terms involving $y$ to one side and all terms involving $t$ to the other side. Since there are no $y$ terms on the right side, we can directly integrate with respect to $t$ .
Use Initial Condition: Integrate both sides of the equation with respect to $t$ . The integral of $\frac{dy}{dt}$ with respect to $t$ is $y$ , and the integral of $2t + 3$ with respect to $t$ is $t^2 + 3t + C$ , where $C$ is the constant of integration. $\newline$ $\int(dy) = \int(2t + 3)dt$ $\newline$ $y = t^2 + 3t + C$
Find Constant C: Use the initial condition $y(1) = 6$ to find the value of the constant $C$ . $6 = (1)^2 + 3(1) + C$ $6 = 1 + 3 + C$ $C = 6 - 4$ $C = 2$
Write General Solution: Now that we have the constant $C$ , we can write the general solution to the differential equation as: $\newline$ $y = t^2 + 3t + 2$
Find Value of t: To find the value of $t$ when $y = 0$ , we set the equation $y = t^2 + 3t + 2$ equal to $0$ and solve for $t$ . $0 = t^2 + 3t + 2$
Factor Quadratic Equation: Factor the quadratic equation to find the values of $t$ . $0 = (t + 1)(t + 2)$ This gives us two possible solutions for $t$ : $t = -1$ and $t = -2$ .
Check Answer Choices: Check the answer choices to see which ones match our solutions. The correct answers are: $\newline$ B $t = -1$ $\newline$ E $t = -2$

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