 # Additional Exam 1 Problems

These are a few additional problems for Exam 1, but please note this review does not cover all sections on your Exam 1.
Directions. It is recommended you work the problems yourself, and then click "Answer" to check your answer. If you do not understand a problem, you can click "Video" to learn how to solve it.
1. For the initial value problem $$(t^2-4)y'+2ty=3t^2, \hskip.2in y(1)=-3$$
1. Determine an interval in which the solution to the initial value problem  is certain to exist.
2. Solve the initial value problem.​

1. $$I=(-2,2)$$
2. $$y=\dfrac{t^2-2t+4}{t-2}, \quad I.V.=(-\infty,2)$$

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2. A large tank initially contains 10 L of fresh water. A brine containing 20 g/L of salt flows into the tank at a rate of 3 L/min. The solution inside the tank is kept well stirred and flows out of the tank at a rate of 2 L/min. Determine the concentration of salt in the tank as a function of time.​

$$C(t)=20-20,000(t+10)^{-3}$$

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3. Given the differential equation $\dfrac{dy}{dt}=7y-y^2-10$
1. Find the equilibrium solutions.
2. Sketch the phase line and determine whether the equilibrium solutions are stable, unstable, or semistable.
3. Sketch the graph of some solutions.
4. Determine the behavior of $$y(t)$$ as $$t$$ increases for all possible values of $$y(0) = y_0$$.
5. Do any solutions admit a vertical asymptote?
6. Solve the equation.​​

1. Equilibrium solutions: $$y=2$$, $$y=5$$
2. $$y=2$$ is unstable and $$y=5$$ is stable. See the video for the sketch of the phase line.
3. See the video.
4. $$\begin{cases} y_0<2 & \Rightarrow & y\to -\infty \quad \text{as }t \text{ increases}\\ y_0=2 & \Rightarrow & y(t)=2 \\ 2< y_0<5 & \Rightarrow & y(t) \to 5\quad \text{as }t \text{ increases} \\ y_0=5 & \Rightarrow & y(t)= 5 \\ y_0>5 & \Rightarrow & y(t)\to 5\quad \text{as }t \text{ increases} \end{cases}$$
5. Yes. See video for explanation.
6. $$y=\dfrac{5-2Ce^{-3t}}{1-Ce^{-3t}}$$

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4. Find an integrating factor for the equation $(y^2+xy)+(x^2+3xy)y'=0$ and then solve the equation.

$$\dfrac{1}{2}x^2y^2+xy^3=C$$

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