We can begin by using the method of integration by parts to integrate $y’y”$. Let $u = y’$ and $v’ = y”$, so that $u’ = y”$ and $v = y’$. Then we have

$y^{\prime} y^{\prime \prime}=u v^{\prime}=\frac{1}{2}\left(u^2\right)^{\prime}$.

Using this, we can rewrite the differential equation as

$\frac{1}{2}\left(y^{\prime 2}\right)^{\prime}-t=0$

Integrating once with respect to $t$, we get

$\frac{1}{2} y^{\prime 2}-\frac{1}{2} t^2=C_1$,

where $C_1$ is a constant of integration. We can then solve for $y’$ to get

$y^{\prime}= \pm \sqrt{t^2+2 C_1}$.

Using the initial condition $y(1) = 2$, we get

$y^{\prime}(1)= \pm \sqrt{1+2 C_1}=1$

Solving for $C_1$, we find that $C_1 = 0$. Thus, we have

$y^{\prime}= \pm \sqrt{t^2}, \quad y(1)=2$.

Since $y$ is increasing, we take the positive sign, giving

$y^{\prime}=t, \quad y(1)=2$.

Integrating with respect to $t$, we get

$y=\frac{1}{2} t^2+C_2$.

Using the initial condition $y(1) = 2$, we get $C_2 = \frac{3}{2}$, so the solution to the initial value problem is

$y=\frac{1}{2} t^2+\frac{3}{2}$

To determine the critical points, we need to find the values of $y$ for which $y’=0$. Thus, we solve the equation:

$y^{\prime}=y(5-y)(y-4)^2=0$

This equation is satisfied when $y=0,5$, or $4$. Therefore, the critical points are $y=0,5,$ and $4$.

To sketch the graph of $f(y)=y(5-y)(y-4)^2$, we can use the critical points found in part (a) and the behavior of $f(y)$ as $y$ approaches positive or negative infinity.

First, we note that $f(y)$ changes sign at $y=0,4,$ and $5$. We can determine the sign of $f(y)$ on each interval by testing a point in each interval. For example, for $y<0$, we can test $y=-1$, and we get:

$f(-1)=(-1)(5-(-1))(4-(-1))^2=-120<0$.

Therefore, $f(y)$ is negative on $(-\infty,0)$.

Next, we can consider the behavior of $f(y)$ as $y$ approaches infinity or negative infinity. Since the highest power of $y$ in $f(y)$ is $y^4$, we know that $f(y)$ approaches infinity as $y$ approaches positive or negative infinity.

Putting all of this together, we can sketch the graph of $f(y)$ as follows:

The graph has a local maximum at $y=4$ and a local minimum at $y=5$. The critical point $y=0$ is a double root.

To derive the acceleration of an individual car, we can start with the definition of acceleration, which is the rate of change of velocity with respect to time, i.e.,

$a=\frac{d u}{d t}$

Using the chain rule of differentiation, we can express this as

$a=\frac{\partial u}{\partial t}+\frac{\partial u}{\partial x} \frac{d x}{d t}$.

Here, the second term on the right-hand side represents the change in velocity due to a change in position, which is given by the product of the velocity gradient $\partial u / \partial x$ and the car’s velocity $dx/dt$. We can rewrite this in terms of the velocity field $u(x,t)$ by using the chain rule again:

$\frac{d x}{d t}=u(x, t)$

Substituting this into the previous equation gives

$a=\frac{\partial u}{\partial t}+u(x, t) \frac{\partial u}{\partial x}$.

This is the desired expression for the acceleration of an individual car in terms of the velocity field.

The number of ways a positive integer $n$ can be written as a sum of positive integers, taking order into account, is given by the partition function $p(n)$. Unfortunately, there is no known closed-form expression for $p(n)$.

However, we can compute $p(n)$ recursively using the following recurrence relation:

$p(n)=\sum_{k=1}^n p(n-k)$

where the sum is taken over all positive integers $k$ such that $k \leq n$.

The base case for this recurrence relation is $p(0) = 1$, which represents the empty sum that adds up to $0$.

Using this recurrence relation, we can compute $p(n)$ for small values of $n$:

\begin{align*} p(1) &= 1 \ p(2) &= 2 \ p(3) &= 3 \ p(4) &= 5 \ p(5) &= 7 \ p(6) &= 11 \ p(7) &= 15 \ p(8) &= 22 \ p(9) &= 30 \ p(10) &= 42 \end{align*}

and so on.

For larger values of $n$, computing $p(n)$ using this recurrence relation becomes computationally expensive. However, there are more efficient algorithms for computing $p(n)$, such as the pentagonal number theorem and the Hardy-Ramanujan-Rademacher formula.

Let $S_k$ denote the number of $k$-element subsets of ${1,2,\ldots,p}$ that have a sum congruent to $0$ modulo $p$. We will derive a recursive formula for $S_k$.

Consider a fixed integer $a_1$. Then the remaining $k-1$ integers must sum to $-a_1$ modulo $p$. If $a_1=0$, then we have $S_{k-1}$ choices for the remaining $k-1$ integers. Otherwise, there are $S_{k-1}$ choices for the remaining $k-1$ integers when we consider the $k-1$-element subset of ${1,2,\ldots,p}\setminus{a_1}$ that sums to $-a_1$ modulo $p$. Thus, we have the recursion

$S_k=S_{k-1}+(p-1) S_{k-1}=p S_{k-1}$

where the factor of $p-1$ accounts for the fact that there are $p-1$ choices for $a_1$.

Since $S_1 = p$, we have $S_k = p^k$ for all positive integers $k$. Therefore, there are $p^k$ $k$-element subsets of ${1,2,\ldots,p}$ that have a sum congruent to $0$ modulo $p$.