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# 数学代写|微积分代写Calculus代考|Exponential Change

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## 数学代写|微积分代写Calculus代考|Exponential Change

In modeling many real-world situations, a quantity $y$ increases or decreases at a rate proportional to its size at a given time $t$. Examples of such quantities include the size of a population, the amount of a decaying radioactive material, and the temperature difference between a hot object and its surrounding medium. Such quantities are said to undergo exponential change.

If the amount present at time $t=0$ is called $y_0$, then we can find $y$ as a function of $t$ by solving the following initial value problem:
$$\text { Differential equation: } \quad \begin{array}{rlrl} \frac{d y}{d t} & =k y \ \text { Initial condition: } & y & =y_0 \text { when } t=0 . \end{array}$$
If $y$ is positive and increasing, then $k$ is positive, and we use Equation (1a) to say that the rate of growth is proportional to what has already been accumulated. If $y$ is positive and decreasing, then $k$ is negative, and we use Equation (1a) to say that the rate of decay is proportional to the amount still left.

We see right away that the constant function $y=0$ is a solution of Equation (1a) if $y_0=0$. To find the nonzero solutions, we divide Equation (1a) by $y$ :
\begin{aligned} \frac{1}{y} \cdot \frac{d y}{d t} & =k & & y \neq 0 \ \int \frac{1}{y} \frac{d y}{d t} d t & =\int k d t & & \text { Integrate with respect to } t ; \ \ln |y| & =k t+C & & \int(1 / u) d u=\ln |u|+C . \ |y| & =e^{k t+C} & & \text { Exponentiate. } \ |y| & =e^C \cdot e^{k t} & & e^{a+b}=e^a \cdot e^b \ y & = \pm e^C e^{k t} & & \text { If }|y|=r, \text { then } y= \pm r . \ y & =A e^{k t .} & & A \text { is a shorter name for } \pm e^C . \end{aligned}
By allowing $A$ to take on the value 0 in addition to all possible values $\pm e^C$, we can include the solution $y=0$ in the formula.

We find the value of $A$ for the initial value problem by solving for $A$ when $y=y_0$ and $t=0$ :
$$y_0=A e^{k \cdot 0}=A$$

## 数学代写|微积分代写Calculus代考|Separable Differential Equations

Exponential change is modeled by a differential equation of the form $d y / d x=k y$, where $k$ is a nonzero constant. More generally, suppose we have a differential equation of the form
$$\frac{d y}{d x}=f(x, y),$$
where $f$ is a function of both the independent and dependent variables. A solution of the equation is a differentiable function $y=y(x)$ defined on an interval of $x$-values (perhaps infinite) such that
$$\frac{d}{d x} y(x)=f(x, y(x))$$

on that interval. That is, when $y(x)$ and its derivative $y^{\prime}(x)$ are substituted into the differential equation, the resulting equation is true for all $x$ in the solution interval. The general solution is a solution $y(x)$ that contains all possible solutions and it always contains an arbitrary constant.

Equation (3) is separable if $f$ can be expressed as a product of a function of $x$ and a function of $y$. The differential equation then has the form
\frac{d y}{d x}=g(x) h(y) . \quad \begin{aligned} & g \text { is a function of } x \ & h \text { is a function of } y . \end{aligned}
Then collect all $y$ terms with $d y$ and all $x$ terms with $d x$ :
$$\frac{1}{h(y)} d y=g(x) d x$$
Now we simply integrate both sides of this equation:
$$\int \frac{1}{h(y)} d y=\int g(x) d x .$$
After completing the integrations, we obtain the solution $y$ defined implicitly as a function of $x$.

The justification that we can integrate both sides in Equation (4) in this way is based on the Substitution Rule (Section 5.5):
\begin{aligned} \int \frac{1}{h(y)} d y & =\int \frac{1}{h(y(x))} \frac{d y}{d x} d x \ & =\int \frac{1}{h(y(x))} h(y(x)) g(x) d x \quad \frac{d y}{d x}=h(y(x)) g(x) \ & =\int g(x) d x \end{aligned}

## 数学代写|微积分代写Calculus代考|Exponential Change

$$\text { Differential equation: } \quad \begin{array}{rlrl} \frac{d y}{d t} & =k y \ \text { Initial condition: } & y & =y_0 \text { when } t=0 . \end{array}$$

\begin{aligned} \frac{1}{y} \cdot \frac{d y}{d t} & =k & & y \neq 0 \ \int \frac{1}{y} \frac{d y}{d t} d t & =\int k d t & & \text { Integrate with respect to } t ; \ \ln |y| & =k t+C & & \int(1 / u) d u=\ln |u|+C . \ |y| & =e^{k t+C} & & \text { Exponentiate. } \ |y| & =e^C \cdot e^{k t} & & e^{a+b}=e^a \cdot e^b \ y & = \pm e^C e^{k t} & & \text { If }|y|=r, \text { then } y= \pm r . \ y & =A e^{k t .} & & A \text { is a shorter name for } \pm e^C . \end{aligned}

$$y_0=A e^{k \cdot 0}=A$$

## 数学代写|微积分代写Calculus代考|Separable Differential Equations

$$\frac{d y}{d x}=f(x, y),$$

$$\frac{d}{d x} y(x)=f(x, y(x))$$

\frac{d y}{d x}=g(x) h(y) . \quad \begin{aligned} & g \text { is a function of } x \ & h \text { is a function of } y . \end{aligned}

$$\frac{1}{h(y)} d y=g(x) d x$$

$$\int \frac{1}{h(y)} d y=\int g(x) d x .$$

\begin{aligned} \int \frac{1}{h(y)} d y & =\int \frac{1}{h(y(x))} \frac{d y}{d x} d x \ & =\int \frac{1}{h(y(x))} h(y(x)) g(x) d x \quad \frac{d y}{d x}=h(y(x)) g(x) \ & =\int g(x) d x \end{aligned}

## MATLAB代写

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