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Chapter 5: Applications of Integration
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Section 5.6: Differential Equations

Essentials


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An algebraic equation is an open statement that is true when the "openings" in the statement are filled with the appropriate algebraic expressions. Thus, $2xplus;3equals;5$ becomes true when $x$, the "opening," is replaced with the number 1.
A differential equation is an open statement in which the openings are are a function and at least one of its derivatives. Table 5.6.1 lists several examples of differential equations.
Differential Equation

Comments

$y\prime \left(x\right)\=1$

Immediately integrable to the general solution $y\left(x\right)\=x\+c$

$y\prime \left(x\right)\=y\left(x\right)$

Knowledge of the exponential function suggests the general solution $y\left(x\right)\=c{e}^{x}$, but this equation is actually separable.

$y\prime \left(x\right)\+2y\left(x\right)equals;x$

The equation is firstorder, linear, and yields to the "recipe" $y\={e}^{\int 2\mathit{DifferentialD;}x}\left(\int x{e}^{\int 2\mathit{DifferentialD;}x}\mathit{DifferentialD;}xplus;c\right)$.
The general firstorder, linear, equation is $p\left(x\right)y\prime plus;q\left(x\right)yequals;r\left(x\right)$.

$\frac{\mathrm{dy}}{\mathrm{dx}}\=\frac{{x}^{2}xyplus;{y}^{2}}{{x}^{2}{y}^{2}}$

The equation is homogeneous, and becomes firstorder, linear, under the change of variables $v\left(x\right)\=y\left(x\right)\/x$.

$y\prime \left(x\right)\+{y}^{2}\left(x\right)\=x$

The equation is nonlinear. If functions of either $y$ or $y\prime$ appear in the equation, it is no longer linear.

$y\prime \left(x\right)\+2y\left(x\right)equals;{y}^{3}\left(x\right)$

This is a Bernoulli equation, and becomes firstorder, linear, under the change of variables $y\left(x\right)\={z}^{k}\left(x\right)$ for some special value of $k$.

$y\prime \left(x\right)f\left(y\right)equals;g\left(x\right)$

This is the general separable equation whose solution is given implicitly by direct integration: $\int f\left(y\right)\mathit{DifferentialD;}yequals;\int g\left(x\right)\mathit{DifferentialD;}xplus;c$.

Table 5.6.1 Examples of differential equations



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The separable differential equation affords great opportunity to practice evaluating both indefinite and definite integrals. It will be the main focus of this chapter.
Since the general solution of the firstorder differential equation $f\left(y\,y\prime \,x\right)\=0$ contains one arbitrary constant of integration, the solution of such an equation represents a family of curves. One unique member of this family can be distinguished by imposing one algebraic condition of the form $y\left({x}_{0}\right)\={y}_{0}$, called an initial condition. A differential equation and its associated initial condition(s) is called an initial value problem (IVP).
An implicit solution of the IVP $y\prime \left(x\right)f\left(y\right)equals;g\left(x\right)comma;y\left({x}_{0}\right)equals;{y}_{0}$, is ${\int}_{{y}_{0}}^{y}f\left(s\right)DifferentialD;sequals;{\int}_{{x}_{0}}^{x}g\left(s\right)DifferentialD;s$. The differential equation is separable; and the variable of integration on each side is immaterial, as long as it does not duplicate one of the limits.


Examples


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Example 5.6.1

Obtain the general solution of the differential equation $y\prime \left(x\right)\=xy\left(x\right)$.

Example 5.6.2

Obtain the general solution to the differential equation $x\mathrm{dx}2y\sqrt{1plus;{x}^{2}}\mathrm{dy}equals;0$.

Example 5.6.3

Solve the initialvalue problem consisting of the differential equation $x2y\sqrt{1plus;{x}^{2}}\cdot y\prime equals;0$ and the initial condition $y\left(1\right)\=2$. Graph the solution.

Example 5.6.4

Graph the solution of the initialvalue problem consisting of the differential equation $y\prime \=4x{y}^{2}\+8{y}^{2}\+x\+2$, and the initial condition $y\left(1\right)\=0$.

Example 5.6.5

A tank contains 50 lbs of salt dissolved in 1000 gallons of water. Brine containing $1\/100$ lb of salt per gal of water enters the tank at a rate of 40 gal per minute, mixes instantaneously, and drains at the same rate. Determine the tank's salt content 15 minutes later.

Example 5.6.6

If $y\>0$, solve the initialvalue problem consisting of the differential equation $y\prime \left(t\right)\=ky\left(t\right)$ and the initial condition $y\left(0\right)\={y}_{0}$.

Example 5.6.7

Solve the initialvalue problem consisting of the logistic differential equation $\stackrel{\.}{y}\/y\=k\left(cy\right)$, and the initial condition $y\left(0\right)\={y}_{0}$.

Example 5.6.8

A species undergoes logistic growth, governed by the formula developed in Example 5.6.7. Observation yields the following three data points.
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$\[\begin{array}{cc}\mathrm{Time\; in\; years}& \mathrm{Population\; Size}\\ 1& 1300\\ 3& 1870\\ 4& 2070\end{array}\]$ ${}$
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Determine the carrying capacity $c$, the initial population ${y}_{0}$, and the rate constant $k$, if it is known that $k\>0$.

Example 5.6.9

Obtain the general solution of the differential equation $\stackrel{\.}{u}\=k\left(u{u}_{s}\right)$, where $u\left(t\right)$ represents the temperature of a body in thermal contact with its surroundings at a fixed temperature ${u}_{s}$. This equation, sometimes called Newton's law of cooling, simply states that the rate of change of temperature of the body is proportional to the difference in temperature between the body and its surroundings.

Example 5.6.10

A medical examiner (M.E.) notes the temperature of the body of a deceased person is $85$°F, and the environment in which the body has been located is 65°F. Being careful not to alter the surrounding temperature, the M.E. waits 15 minutes and again checks the body's temperature, finding it to be $80$°F. Using Newton's law of cooling, what estimate can the M.E. make for the time of death of the deceased?



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