# 分化和整合的分析代写|ANALYSIS OF DIFFERENTIATION AND INTEGRATION MATHS4073 University of Glasgow Assignment

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## Instructions:

Differentiation and integration are two fundamental operations in calculus, which is a branch of mathematics that deals with continuous change and motion. Differentiation is the process of finding the derivative of a function, while integration is the process of finding the antiderivative of a function.

Differentiation:

The derivative of a function is the rate at which the function changes with respect to its input variable. It is defined as the limit of the difference quotient as the change in the input variable approaches zero. In simpler terms, it measures how much the output of the function changes when the input changes a little bit. The derivative is denoted by the symbol “dy/dx” or “f'(x)”.

Differentiation has a number of applications, such as determining the slope of a curve at a particular point, finding the velocity and acceleration of an object, and optimizing functions in optimization problems.

Solve the initial value problem
$$y^{\prime} y^{\prime \prime}-t=0, \quad y(1)=2, \quad y^{\prime}(1)=1 .$$

To solve this second-order differential equation, we first need to rewrite it as a first-order system. Let $y_1 = y$ and $y_2 = y’$, then we have:

\begin{align*} y_1′ &= y_2 \ y_2′ &= \frac{t}{y_2} \end{align*}

This gives us a first-order system that we can solve using standard techniques. To find the solution, we can use an integrating factor for the second equation. Multiplying both sides by $y_2$ gives:

y_2 y_2^{\prime}=t

We can then integrate both sides with respect to $t$ to get:

$\frac{1}{2} y_2^2=\frac{1}{2} t^2+C_1$

where $C_1$ is an arbitrary constant of integration. Solving for $y_2$, we get:

$y_2= \pm \sqrt{t^2+2 C_1}$

Using the initial condition $y'(1) = 1$, we have:

$y_2(1)= \pm \sqrt{1+2 C_1}=1$

Since $y_2$ must be positive, we can choose the positive sign and solve for $C_1$:

$\sqrt{1+2 C_1}=1 \quad \Rightarrow \quad C_1=0$

Thus, we have:

$y_2=\sqrt{t^2}=t$

Next, we can integrate the first equation:

$y_1=\int y_2 d t=\frac{1}{2} t^2+C_2$

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

$y_1(1)=\frac{1}{2}+C_2=2 \quad \Rightarrow \quad C_2=\frac{3}{2}$

Thus, the solution to the initial value problem is:

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

Consider the differential equation $y^{\prime}=y(5-y)(y-4)^2$. (a) Determine the critical points (stationary solutions).

(a) To find the critical points, we need to solve the equation $y^{\prime}=0$, which is equivalent to $y(5-y)(y-4)^2=0$. Therefore, the critical points are $y=0, y=5, y=4$.

(b)Sketch the graph of $f(y)=y(5-y)(y-4)^2$.

(b) To sketch the graph of $f(y)=y(5-y)(y-4)^2$, we need to analyze the behavior of the function around the critical points and at infinity.

First, we can determine the sign of $f(y)$ on different intervals using test points:

\begin{array}{c|c|c|c|c} y & (-\infty,0) & (0,4) & (4,5) & (5,\infty) \ \hline y-2 & – & – & + & + \ y-3 & – & – & – & + \ y-4 & – & 0 & + & + \ y-6 & – & – & – & – \end{array}

From this table, we can see that $f(y)$ is positive on $(0,4)$ and $(5,\infty)$, and negative on $(-\infty,0)$ and $(4,5)$. Also, $f(y)$ has a local maximum at $y=4$ and local minima at $y=0$ and $y=5$.

Based on this information, we can sketch the graph of $f(y)$ as follows:

• The function is negative and decreasing on $(-\infty,0)$.
• The function has a local minimum at $y=0$ and increases from negative infinity to zero.
• The function is negative and decreasing on $(0,4)$.
• The function has a local maximum at $y=4$ and decreases from zero to the minimum value at $y=4$.
• The function is positive and increasing on $(4,5)$.
• The function has a local minimum at $y=5$ and increases from the minimum value at $y=5$ to positive infinity.
• The function is positive and increasing on $(5,\infty)$.