Physics involves some first-order differential equations. For completeness (and review) it seems desirable to touch on them briefly. We consider here differential equations of the
general form
dy
dx =f (x, y)= −P (x, y)
Q(x, y). (9.16)
Equation (9.16) is clearly a first-order, ordinary differential equation. It is first order be- cause it contains the first and no higher derivatives. It is ordinary because the only deriva- tive,dy/dx, is an ordinary, or total, derivative. Equation (9.16) may or may not be linear, although we shall treat the linear case explicitly later, Eq. (9.25).
Separable Variables
Frequently Eq. (9.16) will have the special form dy
dx =f (x, y)= −P (x)
Q(y). (9.17)
Then it may be rewritten as
P (x) dx+Q(y) dy=0.
Integrating from(x0, y0)to(x, y)yields x
x0
P (x) dx+ y
y0
Q(y) dy=0.
Since the lower limits,x0andy0, contribute constants, we may ignore the lower limits of integration and simply add a constant of integration. Note that this separation of variables technique does not require that the differential equation be linear.
Example 9.2.1 PARACHUTIST
We want to find the velocity of the falling parachutist as a function of time and are partic- ularly interested in the constant limiting velocity,v0, that comes about by air drag, taken, to be quadratic,−bv2, and opposing the force of the gravitational attraction,mg, of the Earth. We choose a coordinate system in which the positive direction is downward so that the gravitational force is positive. For simplicity we assume that the parachute opens im- mediately, that is, at timet=0, wherev(t=0)=0, our initial condition. Newton’s law applied to the falling parachutist gives
mv˙=mg−bv2, wheremincludes the mass of the parachute.
The terminal velocity,v0, can be found from the equation of motion ast→ ∞; when there is no acceleration,v˙=0, so
bv02=mg, or v0= )mg
b .
The variablest andvseparate
dv
g−mbv2=dt,
which we integrate by decomposing the denominator into partial fractions. The roots of the denominator are atv= ±v0.Hence
g− b
mv2 −1
= m 2v0b
1
v+v0− 1 v−v0
. Integrating both terms yields
v dV g−mbV2 =1
2 )m
gblnv0+v v0−v =t.
Solving for the velocity yields v=e2t /T −1
e2t /T +1v0=v0sinhTt
coshTt =v0tanh t T, whereT =
m
gb is the time constant governing the asymptotic approach of the velocity to the limiting velocity,v0.
Putting in numerical values,g=9.8 m/s2and takingb=700 kg/m,m=70 kg, gives v0=√
9.8/10∼1 m/s∼3.6 km/h∼2.23 mi/h, the walking speed of a pedestrian at landing, and T =
m
bg =1/√
10ã9.8∼0.1 s. Thus, the constant speed v0 is reached within a second. Finally, because it is always important to check the solution, we verify that our solution satisfies
˙
v=cosht /T cosht /T
v0
T − sinh2t /T cosh2t /T
v0
T =v0
T − v2
T v0=g− b mv2,
that is, Newton’s equation of motion. The more realistic case, where the parachutist is in free fall with an initial speedvi =v(0) >0 before the parachute opens, is addressed in
Exercise 9.2.18.
Exact Differential Equations
We rewrite Eq. (9.16) as
P (x, y) dx+Q(x, y) dy=0. (9.18) This equation is said to be exact if we can match the left-hand side of it to a differentialdϕ,
dϕ=∂ϕ
∂xdx+∂ϕ
∂ydy. (9.19)
Since Eq. (9.18) has a zero on the right, we look for an unknown function ϕ(x, y)= constant anddϕ=0.
We have (if such a functionϕ(x, y)exists)
P (x, y) dx+Q(x, y) dy=∂ϕ
∂xdx+∂ϕ
∂ydy (9.20a)
and
∂ϕ
∂x =P (x, y), ∂ϕ
∂y =Q(x, y). (9.20b)
The necessary and sufficient condition for our equation to be exact is that the second, mixed partial derivatives ofϕ(x, y)(assumed continuous) are independent of the order of differentiation:
∂2ϕ
∂y∂x =∂P (x, y)
∂y =∂Q(x, y)
∂x = ∂2ϕ
∂x∂y. (9.21)
Note the resemblance to Eqs. (1.133a) of Section 1.13, “Potential Theory.” If Eq. (9.18) corresponds to a curl (equal to zero), then a potential,ϕ(x, y), must exist.
Ifϕ(x, y)exists, then from Eqs. (9.18) and (9.20a) our solution is ϕ(x, y)=C.
We may construct ϕ(x, y)from its partial derivatives just as we constructed a magnetic vector potential in Section 1.13 from its curl. See Exercises 9.2.7 and 9.2.8.
It may well turn out that Eq. (9.18) is not exact and that Eq. (9.21) is not satisfied.
However, there always exists at least one and perhaps an infinity of integrating factors α(x, y)such that
α(x, y)P (x, y) dx+α(x, y)Q(x, y) dy=0
is exact. Unfortunately, an integrating factor is not always obvious or easy to find. Unlike the case of the linear first-order differential equation to be considered next, there is no systematic way to develop an integrating factor for Eq. (9.18).
A differential equation in which the variables have been separated is automatically exact.
An exact differential equation is not necessarily separable.
The wave front method of Section 9.1 also works for a first-order PDE:
a(x, y)∂ψ
∂x +b(x, y)∂ψ
∂y =0. (9.22a)
We look for a solution of the formψ=F (ξ ), whereξ(x, y)=constant for varyingxand ydefines the wave front. Hence
dξ=∂ξ
∂xdx+∂ξ
∂ydy=0, (9.22b)
while the PDE yields
a∂ξ
∂x +b∂ξ
∂y dF
dξ =0 (9.23a)
withdF /dξ=0 in general. Comparing Eqs. (9.22b) and (9.23a) yields dx
a =dy
b , (9.23b)
which reduces the PDE to a first-order ODE for the tangent dy/dx of the wave front functionξ(x, y).
When there is an additional source term in the PDE, a∂ψ
∂x +b∂ψ
∂y +cψ=0, (9.23c)
then we use the Ansatzψ=ψ0(x, y)F (ξ ), which converts our PDE to F
a∂ψ0
∂x +b∂ψ0
∂y +cψ0
+ψ0dF dξ
a∂ξ
∂x +b∂ξ
∂y
=0. (9.24)
If we can guess a solution ψ0 of Eq. (9.23c), then Eq. (9.24) reduces to our previous equation, Eq. (9.23a), from which the ODE of Eq. (9.23b) follows.
Linear First-Order ODEs
Iff (x, y)in Eq. (9.16) has the form−p(x)y+q(x), then Eq. (9.16) becomes dy
dx+p(x)y=q(x). (9.25)
Equation (9.25) is the most general linear first-order ODE. Ifq(x)=0, Eq. (9.25) is homogeneous (iny). A nonzeroq(x)may represent a source or a driving term. Equa- tion (9.25) is linear; each term is linear iny ordy/dx. There are no higher powers, that is,y2, and no products,y(dy/dx). Note that the linearity refers to theyanddy/dx;p(x) andq(x)need not be linear inx. Equation (9.25), the most important of these first-order ODEs for physics, may be solved exactly.
Let us look for an integrating factorα(x)so that α(x)dy
dx +α(x)p(x)y=α(x)q(x) (9.26)
may be rewritten as
d dx
α(x)y =α(x)q(x). (9.27)
The purpose of this is to make the left-hand side of Eq. (9.25) a derivative so that it can be integrated — by inspection. It also, incidentally, makes Eq. (9.25) exact. Expanding Eq. (9.27), we obtain
α(x)dy dx+dα
dxy=α(x)q(x).
Comparison with Eq. (9.26) shows that we must require dα
dx =α(x)p(x). (9.28)
Here is a differential equation forα(x), with the variablesαandxseparable. We separate variables, integrate, and obtain
α(x)=exp x
p(x) dx
(9.29)
as our integrating factor.
Withα(x)known we proceed to integrate Eq. (9.27). This, of course, was the point of introducingαin the first place. We have
x d dx
α(x)y(x) dx= x
α(x)q(x) dx.
Now integrating by inspection, we have α(x)y(x)=
x
α(x)q(x) dx+C.
The constants from a constant lower limit of integration are lumped into the constantC.
Dividing byα(x), we obtain y(x)=
α(x) −1 x
α(x)q(x) dx+C
. Finally, substituting in Eq. (9.29) forαyields
y(x)=exp
− x
p(t ) dt x
exp s
p(t ) dt
q(s) ds+C
. (9.30)
Here the (dummy) variables of integration have been rewritten to make them unambigu- ous. Equation (9.30) is the complete general solution of the linear, first-order differential equation, Eq. (9.25). The portion
y1(x)=Cexp
− x
p(t ) dt
(9.31) corresponds to the caseq(x)=0 and is a general solution of the homogeneous differential equation. The other term in Eq. (9.30),
y2(x)=exp
− x
p(t ) dt
x
exp s
p(t ) dt
q(s) ds, (9.32) is a particular solution corresponding to the specific source termq(x).
Note that if our linear first-order differential equation is homogeneous(q=0), then it is separable. Otherwise, apart from special cases such asp=constant,q=constant, and q(x)=ap(x), Eq. (9.25) is not separable.
Let us summarize this solution of the inhomogeneous ODE in terms of a method called variation of the constant as follows. In the first step, we solve the homogeneous ODE by separation of variables as before, giving
y′
y = −p, lny= − x
p(X) dX+lnC, y(x)=Ce−
x p(X) dX.
In the second step, we let the integration constant becomex-dependent, that is,C→C(x).
This is the “variation of the constant” used to solve the inhomogeneous ODE. Differenti- atingy(x)we obtain
y′= −pCe−
p(x) dx
+C′(x)e−
p(x) dx
= −py(x)+C′(x)e−
p(x) dx.
Comparing with the inhomogeneous ODE we find the ODE forC:
C′e−
p(x) dx
=q, or C(x)= x
e
X
p(Y ) dYq(X) dX.
Substituting thisCintoy=C(x)e−xp(X) dXreproduces Eq. (9.32).
Example 9.2.2 RL CIRCUIT
For a resistance-inductance circuit Kirchhoff’s law leads to LdI (t )
dt +RI (t )=V (t )
for the currentI (t ), whereLis the inductance andRis the resistance, both constant.V (t ) is the time-dependent input voltage.
From Eq. (9.29) our integrating factorα(t )is α(t )=exp
t R
Ldt=eRt /L. Then by Eq. (9.30),
I (t )=e−Rt /L t
eRt /LV (t ) L dt+C
,
with the constantCto be determined by an initial condition (a boundary condition).
For the special caseV (t )=V0, a constant, I (t )=e−Rt /L
V0 L ãL
ReRt /L+C
=V0
R +Ce−Rt /L. If the initial condition isI (0)=0, thenC= −V0/Rand
I (t )=V0 R
1−e−Rt /L .
Now we prove the theorem that the solution of the inhomogeneous ODE is unique up to an arbitrary multiple of the solution of the homogeneous ODE.
To show this, supposey1, y2both solve the inhomogeneous ODE, Eq. (9.25); then y1′ −y2′+p(x)(y1−y2)=0
follows by subtracting the ODEs and says thaty1−y2is a solution of the homogeneous ODE. The solution of the homogeneous ODE can always be multiplied by an arbitrary constant.
We also prove the theorem that a first-order homogeneous ODE has only one linearly independent solution. This is meant in the following sense. If two solutions are linearly dependent, by definition they satisfyay1(x)+by2(x)=0 with nonzero constantsa, bfor all values ofx.If the only solution of this linear relation isa=0=b, then our solutions y1andy2are said to be linearly independent.
To prove this theorem, supposey1, y2both solve the homogeneous ODE. Then y1′
y1= −p(x)=y2′ y2
implies W (x)≡y1′y2−y1y′2≡0. (9.33) The functional determinantW is called the Wronskian of the pairy1,y2.We now show thatW≡0 is the condition for them to be linearly dependent. Assuming linear dependence, that is,
ay1(x)+by2(x)=0
with nonzero constantsa, bfor all values ofx, we differentiate this linear relation to get another linear relation,
ay1′(x)+by2′(x)=0.
The condition for these two homogeneous linear equations in the unknownsa,bto have a nontrivial solution is that their determinant be zero, which isW=0.
Conversely, fromW=0, there follows linear dependence, because we can find a non- trivial solution of the relation
y1′ y1=y2′
y2
by integration, which gives
lny1=lny2+lnC, or y1=Cy2.
Linear dependence and the Wronskian are generalized to three or more functions in Sec- tion 9.6.
Exercises
9.2.1 From Kirchhoff’s law the currentI in anRC(resistance–capacitance) circuit (Fig. 9.1) obeys the equation
RdI dt + 1
CI=0.
(a) FindI (t ).
(b) For a capacitance of 10,000 àF charged to 100 V and discharging through a resis- tance of 1 M, find the currentI fort=0 and fort=100 seconds.
Note. The initial voltage isI0RorQ/C, whereQ=∞
0 I (t ) dt.
9.2.2 The Laplace transform of Bessel’s equation(n=0)leads to s2+1
f′(s)+sf (s)=0.
Solve forf (s).
FIGURE9.1 RC circuit.
9.2.3 The decay of a population by catastrophic two-body collisions is described by dN
dt = −kN2.
This is a first-order, nonlinear differential equation. Derive the solution N (t )=N0
1+ t
τ0
−1
,
whereτ0=(kN0)−1. This implies an infinite population att= −τ0.
9.2.4 The rate of a particular chemical reactionA+B→Cis proportional to the concentra- tions of the reactantsAandB:
dC(t ) dt =α
A(0)−C(t ) B(0)−C(t ) . (a) FindC(t )forA(0)=B(0).
(b) FindC(t )forA(0)=B(0).
The initial condition is thatC(0)=0.
9.2.5 A boat, coasting through the water, experiences a resisting force proportional tovn,v being the boat’s instantaneous velocity. Newton’s second law leads to
mdv
dt = −kvn.
Withv(t=0)=v0,x(t=0)=0, integrate to findv as a function of time andv as a function of distance.
9.2.6 In the first-order differential equationdy/dx=f (x, y)the functionf (x, y)is a func- tion of the ratioy/x:
dy
dx=g(y/x).
Show that the substitution ofu=y/xleads to a separable equation inuandx.
9.2.7 The differential equation
P (x, y) dx+Q(x, y) dy=0 is exact. Construct a solution
ϕ(x, y)= x
x0
P (x, y) dx+ y
y0
Q(x0, y) dy=constant.
9.2.8 The differential equation
P (x, y) dx+Q(x, y) dy=0 is exact. If
ϕ(x, y)= x
x0
P (x, y) dx+ y
y0
Q(x0, y) dy, show that
∂ϕ
∂x =P (x, y), ∂ϕ
∂y =Q(x, y).
Henceϕ(x, y)=constant is a solution of the original differential equation.
9.2.9 Prove that Eq. (9.26) is exact in the sense of Eq. (9.21), provided that α(x)satisfies Eq. (9.28).
9.2.10 A certain differential equation has the form
f (x) dx+g(x)h(y) dy=0,
with none of the functionsf (x),g(x),h(y)identically zero. Show that a necessary and sufficient condition for this equation to be exact is thatg(x)=constant.
9.2.11 Show that
y(x)=exp
− x
p(t ) dt x
exp s
p(t ) dt
q(s) ds+C
is a solution of
dy
dx +p(x)y(x)=q(x)
by differentiating the expression fory(x)and substituting into the differential equation.
9.2.12 The motion of a body falling in a resisting medium may be described by mdv
dt =mg−bv
when the retarding force is proportional to the velocity,v. Find the velocity. Evaluate the constant of integration by demanding thatv(0)=0.
9.2.13 Radioactive nuclei decay according to the law dN
dt = −λN,
N being the concentration of a given nuclide andλ, the particular decay constant. In a radioactive series ofndifferent nuclides, starting withN1,
dN1
dt = −λ1N1, dN2
dt =λ1N1−λ2N2, and so on.
FindN2(t )for the conditionsN1(0)=N0andN2(0)=0.
9.2.14 The rate of evaporation from a particular spherical drop of liquid (constant density) is proportional to its surface area. Assuming this to be the sole mechanism of mass loss, find the radius of the drop as a function of time.
9.2.15 In the linear homogeneous differential equation dv dt = −av
the variables are separable. When the variables are separated, the equation is exact.
Solve this differential equation subject tov(0)=v0by the following three methods:
(a) Separating variables and integrating.
(b) Treating the separated variable equation as exact.
(c) Using the result for a linear homogeneous differential equation.
ANS.v(t )=v0e−at. 9.2.16 Bernoulli’s equation,
dy
dx +f (x)y=g(x)yn,
is nonlinear forn=0 or 1. Show that the substitutionu=y1−n reduces Bernoulli’s equation to a linear equation. (See Section 18.4.)
ANS.du
dx +(1−n)f (x)u=(1−n)g(x).
9.2.17 Solve the linear, first-order equation, Eq. (9.25), by assumingy(x)=u(x)v(x), where v(x)is a solution of the corresponding homogeneous equation[q(x)=0]. This is the method of variation of parameters due to Lagrange. We apply it to second-order equa- tions in Exercise 9.6.25.
9.2.18 (a) Solve Example 9.2.1 for an initial velocityvi=60 mi/h, when the parachute opens.
Findv(t ).(b) For a skydiver in free fall use the friction coefficientb=0.25 kg/m and massm=70 kg. What is the limiting velocity in this case?