Unit 7: Differential Equations

Suppose a population $P(t)$ grows at a rate proportional to its current size. “Rate” gives $dP/dt$, “proportional to its current size” gives $kP$, and “grows” requires a positive sign. The model is $dP/dt=kP$ with $k>0$. If $t$ is measured in years, $dP/dt$ has units of individuals per year and $k$ has units of $1/$year.

Verify $y=3e^{2x}$ for $dy/dx=2y$. Differentiation gives $dy/dx=6e^{2x}$, while substituting the candidate on the right gives $2y=2(3e^{2x})=6e^{2x}$. Both sides agree for every real $x$, so the candidate is a solution.

More generally, $y=Ce^{2x}$ gives $dy/dx=2Ce^{2x}=2y$ for every constant $C$. Thus the same differential equation has infinitely many solutions. An additional condition such as $y(0)=3$ selects $C=3$ from that family.

The zero-slope locations satisfy $x-y=0$, or $y=x$. Above that line, $x-y<0$, so solution curves locally fall; below it, $x-y>0$, so they locally rise. This pattern estimates how solutions bend as they move across the plane.

For $dy/dx=y$, segments above the $x$-axis have positive slope, those on the axis are horizontal, and those below have negative slope. A solution through $(0,1)$ therefore rises as it follows the upper field; a solution through $(0,-1)$ falls in the lower field; $y=0$ itself follows the horizontal segments. Together these curves illustrate a family of solutions.

To estimate a solution through $(x_0,y_0)$, begin exactly there and sketch smoothly in both directions, continually matching nearby segment slopes. The initial point narrows the field's whole family to the solution curve or curves compatible with that condition.

For $dy/dx=x+y$, $y(0)=1$, and $h=0.1$: from $(0,1)$ the slope is $1$, so $(x_1,y_1)=(0.1,1+0.1(1))=(0.1,1.1)$. Now the slope is $0.1+1.1=1.2$, so $(x_2,y_2)=(0.2,1.1+0.1(1.2))=(0.2,1.22)$. Thus $y(0.2)\approx1.22$.

For $dy/dx=xy$, first assume $y\ne0$ and write $dy/y=x\,dx$. Then $\ln|y|=x^2/2+C$, so $y=Ce^{x^2/2}$ after absorbing sign and magnitude into $C$. Differentiation gives $y'=Cxe^{x^2/2}=xy$, confirming the family.

When $dy/dx=f(x)$, this formula automatically satisfies both requirements: the Fundamental Theorem of Calculus gives $F'(x)=f(x)$, and $F(a)=y_0$ because an integral with equal bounds is $0$. The integration variable $t$ is a placeholder, leaving $x$ as the input of $F$.

For $dy/dx=y^2$ with $y(0)=1$, separation gives $y^{-2}dy=dx$, so $-1/y=x+C$. Substituting $(0,1)$ gives $C=-1$, hence $y=1/(1-x)$. Differentiating yields $y'=1/(1-x)^2=y^2$, and $y(0)=1$, so both the equation and initial condition hold.

In the illustrative model $dy/dt=0.03y$, if $y=200$ units at a certain time, then $dy/dt=0.03(200)=6$ units per time unit at that instant. The constant $0.03$ is the relative growth rate; it is not a constant increase of $6$ forever.

At $t=0$, the general solution gives $y(0)=Ce^0=C$, so the initial condition forces $C=y_0$. Differentiating $y_0e^{kt}$ gives $ky_0e^{kt}=ky$, verifying both the differential equation and the initial value.

For the illustrative initial-value problem $dy/dt=0.2y$ and $y(0)=50$, the particular solution is $y=50e^{0.2t}$. At $t=3$, the model gives $y(3)=50e^{0.6}$ units. Because $k=0.2>0$, the amount grows; a negative $k$ would produce decay.

The initial condition tells which row applies. If $0<y(0)<a$, the solution increases but the factor $a-y$ shrinks, so growth slows and $y$ approaches $a$ as $t\to\infty$. If $y(0)=0$ or $y(0)=a$, the rate remains zero and the solution stays at that equilibrium.

For values between $0$ and $a$, the growth rate as a function of $y$ is $k(ay-y^2)$, a downward-opening quadratic. Its vertex occurs at $y=a/2$, so the quantity is increasing fastest when it reaches half the carrying capacity.

In the illustrative model $dy/dt=ky(1000-y)$ with $k>0$ and $y(0)=100$, the quantity initially grows, approaches the carrying capacity $1000$, and has its greatest growth rate when $y=500$. None of these conclusions requires solving for $y(t)$.