(dy)/(dt)=3t and y(2)=3. What is t when y=6 ? Choose all answers that apply: A t=-2 B t=sqrt6 c t=6 D t=-sqrt6 E t=2

Integrate to find general solution: To solve for t when y equals 6, we first need to integrate the differential equation dy/dt = 3t to find the general solution for y in terms of t. We integrate 3t with respect to t to get y(t). The integral of 3t dt is (3/2)t^2 + C, where C is the constant of integration.Use initial condition to find C: Next, we use the initial condition y(2) = 3 to find the value of the constant C. We substitute t = 2 and y = 3 into the equation y(t) = (3/2)t^2 + C. 3 = (3/2)(2)^2 + C simplifies to 3 = (3/2)(4) + C.Determine particular solution: Solving for C, we get 3 = 6 + C, which means C = 3 - 6 = -3. Now we have the particular solution y(t) = (3/2)t^2 - 3.Find t when y equals 6: We want to find the value of t when y equals 6. We set y(t) = 6 and solve for t: 6 = (3/2)t^2 - 3.Isolate t in equation: Adding 3 to both sides of the equation, we get 6 + 3 = (3/2)t^2. This simplifies to 9 = (3/2)t^2.Solve for t: To isolate t^2, we multiply both sides by 2/3: (2/3) * 9 = t^2. This simplifies to 6 = t^2.Check validity of solutions: Taking the square root of both sides, we find that t = sqrt(6) or t = -sqrt(6).We check the interval for t given in the question prompt. Since we are not restricted by an interval for t, both t = sqrt(6) and t = -sqrt(6) are valid solutions.

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Full solution

\frac{d y}{d t}=3 t

and

y(2)=3

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What is

t

when

y=6

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Choose all answers that apply:

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(A)

t=-2

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(B)

t=\sqrt{6}

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(C)

t=6

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(D)

t=-\sqrt{6}

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(E)

t=2

Integrate to find general solution: To solve for $t$ when $y$ equals $6$ , we first need to integrate the differential equation $\frac{dy}{dt} = 3t$ to find the general solution for $y$ in terms of $t$ . We integrate $3t$ with respect to $t$ to get $y(t)$ . The integral of $3t$ dt is $y$ $0$ , where $y$ $1$ is the constant of integration.
Use initial condition to find $C$ : Next, we use the initial condition $y(2) = 3$ to find the value of the constant $C$ . We substitute $t = 2$ and $y = 3$ into the equation $y(t) = \frac{3}{2}t^2 + C$ . $3 = \frac{3}{2}(2)^2 + C$ simplifies to $3 = \frac{3}{2}(4) + C$ .
Determine particular solution: Solving for $C$ , we get $3 = 6 + C$ , which means $C = 3 - 6 = -3$ . Now we have the particular solution $y(t) = \frac{3}{2}t^2 - 3$ .
Find $t$ when $y$ equals $6$ : We want to find the value of $t$ when $y$ equals $6$ . We set $y(t) = 6$ and solve for $t$ : $6 = \frac{3}{2}t^2 - 3$ .
Isolate t in equation: Adding $3$ to both sides of the equation, we get $6 + 3 = \left(\frac{3}{2}\right)t^2$ . This simplifies to $9 = \left(\frac{3}{2}\right)t^2$ .
Solve for $t$ : To isolate $t^2$ , we multiply both sides by $\frac{2}{3}$ : $\left(\frac{2}{3}\right) \times 9 = t^2$ . This simplifies to $6 = t^2$ .
Check validity of solutions: Taking the square root of both sides, we find that $t = \sqrt{6}$ or $t = -\sqrt{6}$ .
Check validity of solutions: Taking the square root of both sides, we find that $t = \sqrt{6}$ or $t = -\sqrt{6}$ .We check the interval for $t$ given in the question prompt. Since we are not restricted by an interval for $t$ , both $t = \sqrt{6}$ and $t = -\sqrt{6}$ are valid solutions.