5. Undetermined Coefficients. Here is an alternative way to solve the equation
2.2 Separable Differential Equations
▼ For example, if f(y) =y−y3/3, then d d x
y− y3
3
=(1−y2)d y d x.
Thus the second term in equation (7) is the derivative with respect toxofy−y3/3, and the first term is the derivative of−x3/3. Thus equation (7) can be written as
d d x
−x3 3
+ d
d x
y− y3 3
=0, or
d d x
−x3
3 +y− y3 3
=0.
Therefore, by integrating (and multiplying the result by 3), we obtain
−x3+3y−y3=c, (8) wherecis an arbitrary constant.
Equation (8) is an equation for the integral curves of equation (6). A direction field and several integral curves are shown in Figure 2.2.1. Any differentiable function y = φ(x) that satisfies equation (8) is a solution of equation (6). An equation of the integral curve passing through a particular point (x0,y0) can be found by substituting x0 and y0 forx and y, respectively, in equation (8) and determining the corresponding value ofc.
4
2
4 2
2 – 4
–
–2
–4 y
x
FIGURE 2.2.1 Direction field and integral curves ofy′=x2/(1−y2).
Essentially the same procedure can be followed for any separable equation. Returning to equation (4), letH1andH2be any antiderivatives ofM andN, respectively. Thus
H1′(x) =M(x), H2′(y) =N(y), (9) and equation (4) becomes
H1′(x)+H2′(y)d y
d x =0. (10)
Ifyis regarded as a function ofx, then according to the chain rule, H2′(y)d y
d x = d
d yH2(y) d y d x = d
d xH2(y). (11)
Consequently, we can write equation (10) as d
d x
H1(x)+H2(y)
=0. (12)
By integrating equation (12) with respect tox, we obtain
H1(x)+H2(y)=c, (13)
where c is an arbitrary constant. Any differentiable function y = φ(x) that satisfies equation (13) is a solution of equation (4); in other words, equation (13) defines the solution implicitly rather than explicitly. In practice, equation (13) is usually obtained from equation (5) by integrating the first term with respect to x and the second term with respect to y. The justification for this is the argument that we have just given.
The differential equation (4), together with an initial condition
y(x0) =y0, (14)
forms an initial value problem. To solve this initial value problem, we must determine the appropriate value for the constantcin equation (13). We do this by settingx=x0andy=y0 in equation (13) with the result that
c=H1(x0)+H2(y0). (15)
Substituting this value ofcin equation (13) and noting that H1(x)−H1(x0)=
x x0
M(s)ds, H2(y)−H2(y0) = y
y0
N(s)ds, we obtain
x x0
M(s)ds+ y
y0
N(s)ds=0. (16)
Equation (16) is an implicit representation of the solution of the differential equation (4) that also satisfies the initial condition (14). Bear in mind that to determine an explicit formula for the solution, you need to solve equation (16) foryas a function ofx. Unfortunately, it is often impossible to do this analytically; in such cases you can resort to numerical methods to find approximate values ofyfor given values ofx.
EXAMPLE 2
Solve the initial value problem d y
d x = 3x2+4x+2
2(y−1) , y(0)= −1, (17)
and determine the interval in which the solution exists.
Solution:
The differential equation can be written as
2(y−1)d y=(3x2+4x+2)d x.
Integrating the left-hand side with respect toyand the right-hand side with respect toxgives
y2−2y=x3+2x2+2x+c, (18)
wherecis an arbitrary constant. To determine the solution satisfying the prescribed initial condition, we substitutex=0 andy= −1 in equation (18), obtainingc=3. Hence the solution of the initial value problem is given implicitly by
y2−2y=x3+2x2+2x+3. (19)
To obtain the solution explicitly, we must solve equation (19) foryin terms ofx. That is a simple matter in this case, since equation (19) is quadratic iny, and we obtain
y=1± x3+2x2+2x+4. (20)
Equation (20) gives two solutions of the differential equation, only one of which, however, satisfies the given initial condition. This is the solution corresponding to the minus sign in equation (20), so we finally obtain
y=φ(x) =1− x3+2x2+2x+4 (21)
▼
▼ as the solution of the initial value problem (15). Note that if we choose the plus sign by mistake in equation (20), then we obtain the solution of the same differential equation that satisfies the initial conditiony(0) =3. Finally, to determine the interval in which the solution (21) is valid, we must find the interval in which the quantity under the radical is positive. The only real zero of this expression isx = −2, so the desired interval isx > −2. Some integral curves of the differential equation are shown in Figure 2.2.2. The green curve passes through the point (0,−1) and thus is the solution of the initial value problem (15). Observe that the boundary of the interval of validity of the solution (21) is determined by the point (−2, 1) at which the tangent line is vertical.
y
1 x 1
– –2
(– 2, 1)
2
– 2 –1 1 2 3
(0, –1)
FIGURE 2.2.2 Integral curves ofy′=(3x2+4x+2)/2(y−1); the solution satisfyingy(0)= −1 is shown in green and is valid forx> −2.
EXAMPLE 3
Solve the separable differential equation d y
d x = 4x−x3
4+y3 (22)
and draw graphs of several integral curves. Also find the solution passing through the point (0, 1) and determine its interval of validity.
Solution:
Rewriting equation (22) as
(4+y3)d y=(4x−x3)d x,
integrating each side, multiplying by 4, and rearranging the terms, we obtain
y4+16y+x4−8x2=c, (23)
wherecis an arbitrary constant. Any differentiable functiony=φ(x) that satisfies equation (23) is a solution of the differential equation (22). Graphs of equation (23) for several values ofcare shown in Figure 2.2.3.
To find the particular solution passing through (0, 1), we setx=0 andy=1 in equation (23) with the result thatc=17. Thus the solution in question is given implicitly by
y4+16y+x4−8x2=17. (24)
It is shown by the green curve in Figure 2.2.3. The interval of validity of this solution extends on either side of the initial point as long as the function remains differentiable. From the figure we see that the interval ends when we reach points where the tangent line is vertical. It follows from the differential equation (22) that these are points where 4+y3=0, ory=(−4)1/3∼= −1.5874. From equation (24) the corresponding values ofxarex∼= ±3.3488. These points are marked on the graph in Figure 2.2.3.
▼
▼
–3 –2 –1 1 2 y
–3 –2 –1 1 2 3
x
) 4 7 8 5 . 1 – , 8 8 4 3 . 3 ( )
4 7 8 5 . 1 – , 8 8 4 3 . 3 – (
FIGURE 2.2.3 Integral curves ofy′=(4x−x3)/(4+y3). The solution passing through (0, 1) is shown by the green curve.
Note 1:Sometimes a differential equation of the form (2):
d y
d x = f(x,y)
has a constant solutiony=y0. Such a solution is usually easy to find because if f(x,y0) =0 for some valuey0and for allx, then the constant functiony=y0is a solution of the differential equation (2). For example, the equation
d y
d x =(y−3) cosx
1+2y2 (25)
has the constant solutiony=3. Other solutions of this equation can be found by separating the variables and integrating.
Note 2:The investigation of a first-order nonlinear differential equation can sometimes be facilitated by regarding bothxandyas functions of a third variablet. Then
d y
d x = d y/dt
d x/dt. (26)
If the differential equation is
d y
d x = F(x,y)
G(x,y), (27)
then, by comparing numerators and denominators in equations (26) and (27), we obtain the system
d x
dt =G(x,y), d y
dt =F(x,y). (28)
At first sight it may seem unlikely that a problem will be simplified by replacing a single equation by a pair of equations, but in fact, the system (28) may well be more amenable to investigation than the single equation (27). Chapter 9 is devoted to nonlinear systems of the form (28).
Note 3:In Example 2 it was not difficult to solve explicitly for yas a function of x.
However, this situation is exceptional, and often it will be better to leave the solution in implicit form, as in Examples 1 and 3. Thus, in the problems below and in other sections where nonlinear equations appear, the words “solve the following differential equation” mean to find the solution explicitly if it is convenient to do so, but otherwise to find an equation defining the solution implicitly.
Problems
In each of Problems1through8, solve the given differential equation.
1. y′= x2 y
2. y′+y2sinx=0 3. y′=cos2(x) cos2(2y) 4. x y′=(1−y2)1/2 5. d y
d x = x−e−x y+ey 6. d y
d x = x2 1+y2 7. d y
d x = y x 8. d y
d x = −x y
In each of Problems9through16:
a. Find the solution of the given initial value problem in explicit form.
G b.Plot the graph of the solution.
c. Determine (at least approximately) the interval in which the solution is defined.
9. y′=(1−2x)y2, y(0)= −1/6 10. y′=(1−2x)/y, y(1)= −2 11. x d x+ye−xd y=0, y(0)=1 12. dr/dθ=r2/θ, r(1) =2 13. y′=x y3(1+x2)−1/2, y(0) =1 14. y′=2x/(1+2y), y(2)=0 15. y′=(3x2−ex)/(2y−5), y(0) =1 16. sin(2x)d x+cos(3y)d y=0, y(π/2)=π/3
Some of the results requested in Problems 17 through 22 can be obtained either by solving the given equations analytically or by plotting numerically generated approximations to the solutions. Try to form an opinion about the advantages and disadvantages of each approach.
G 17. Solve the initial value problem y′= 1+3x2
3y2−6y, y(0) =1 and determine the interval in which the solution is valid.
Hint: To find the interval of definition, look for points where the integral curve has a vertical tangent.
G 18. Solve the initial value problem y′= 3x2
3y2−4, y(1) =0 and determine the interval in which the solution is valid.
Hint: To find the interval of definition, look for points where the integral curve has a vertical tangent.
G 19. Solve the initial value problem y′=2y2+x y2, y(0) =1
and determine where the solution attains its minimum value.
G 20. Solve the initial value problem y′= 2−ex
3+2y, y(0) =0
and determine where the solution attains its maximum value.
G 21. Consider the initial value problem y′= t y(4−y)
3 , y(0)= y0.
a. Determine how the behavior of the solution ast increases depends on the initial valuey0.
b. Suppose thaty0=0.5. Find the timeTat which the solution first reaches the value 3.98.
G 22. Consider the initial value problem y′= t y(4−y)
1+t , y(0)=y0>0.
a. Determine how the solution behaves ast→ ∞.
b. Ify0=2, find the timeT at which the solution first reaches the value 3.99.
c. Find the range of initial values for which the solution lies in the interval 3.99< y< 4.01 by the timet=2.
23. Solve the equation d y
d x = ay+b cy+d, wherea,b,c, anddare constants.
24. Use separation of variables to solve the differential equation d Q
dt =r(a+b Q), Q(0)=Q0,
wherea, b, r, and Q0 are constants. Determine how the solution behaves ast→ ∞
Homogeneous Equations. If the right-hand side of the equation d y/d x = f(x,y) can be expressed as a function of the ratioy/x only, then the equation is said to be homogeneous.1Such equations can always be transformed into separable equations by a change of the dependent variable. Problem25illustrates how to solve first-order homogeneous equations.
...
1The word “homogeneous” has different meanings in different mathematical contexts. The homogeneous equations considered here have nothing to do with the homogeneous equations that will occur in Chapter 3 and elsewhere.
N 25. Consider the equation d y
d x = y−4x
x−y . (29)
a. Show that equation (29) can be rewritten as d y
d x = (y/x)−4
1−(y/x); (30)
thus equation (29) is homogeneous.
b. Introduce a new dependent variablev so thatv = y/x, or y=xv(x). Expressd y/d xin terms ofx,v, anddv/d x.
c. Replace yand d y/d x in equation (30) by the expressions from partbthat involvev anddv/d x. Show that the resulting differential equation is
v+xdv
d x = v−4 1−v, or
xdv
d x = v2−4
1−v. (31)
Observe that equation (31) is separable.
d. Solve equation (31), obtainingvimplicitly in terms ofx.
e. Find the solution of equation (29) by replacingvbyy/xin the solution in partd.
f. Draw a direction field and some integral curves for equation (29). Recall that the right-hand side of equation (29) actually depends only on the ratioy/x. This means that integral curves have the same slope at all points on any given straight line
through the origin, although the slope changes from one line to another. Therefore, the direction field and the integral curves are symmetric with respect to the origin. Is this symmetry property evident from your plot?
The method outlined in Problem 25 can be used for any homogeneous equation. That is, the substitutiony=xv(x) transforms a homogeneous equation into a separable equation. The latter equation can be solved by direct integration, and then replacing v by y/x gives the solution to the original equation. In each of Problems 26 through31:
a. Show that the given equation is homogeneous.
b. Solve the differential equation.
G c. Draw a direction field and some integral curves. Are they symmetric with respect to the origin?
26. d y
d x = x2+x y+y2 x2 27. d y
d x = x2+3y2 2x y 28. d y
d x = 4y−3x 2x−y 29. d y
d x = −4x+3y 2x+y 30. d y
d x = x2−3y2 2x y 31. d y
d x = 3y2−x2 2x y