DIFFERENTIAL EQUATIONS
10.5 MORE GENERAL SECOND-ORDER EQUATIONS
= M g L
[L2 4 − h20
] .
Alternatively, we can calculate the kinetic energy from the difference in potential energy between the initial and final positions of the chain. Taking the pulley as the zero point for potential energy, the chain initially has its center of mass at (L2+ 4h20)/4L and finally at L/2; The difference of these quantities is
L 2 −
(L 4 +h20
L )
=L 4 −h20
L . Multiplying by M g, we confirm our earlier result.
10.5 MORE GENERAL SECOND-ORDER EQUATIONS
Exercises
10.5.1. For the ODE y′′+ yy′ = 0, find a general solution containing two independent constants and, in addition, a singular solution that contains only one constant. Does your symbolic computing system find the singular solution?
Hint. Consider what happens to your solution procedure if one of the constants of integration is zero.
168 CHAPTER 10. ORDINARY DIFFERENTIAL EQUATIONS Solution:
Following the procedure at Eq. (10.74), set y′ = u and y′′= u(du/dy), leading to
udu
dy + yu = 0, satisfied if u = 0 or du dy =−y .
From u = 0 (i.e., y′= 0), we get y = C; from the other equation we have u = y′=−y2
2 +C′
2 , equivalent to 2 dy
C′− y2 = dx , which (assuming C′ to be nonzero) integrates to
x = 1 C′ ln
(C′+ y C′− y
) + C′′. Solving for y, this is equivalent to
y = C′tanh
(C′x + C′′
2 )
.
However, if C′ = 0, the integration of y′=−y2/2 leads to y = 2/(x + C′′′).
Because this is a nonlinear equation all the solutions we have found should be checked against the original ODE. They pass that test; noting that the tanh solution reduces to y = C′ in the limit C′′=∞, a complete solution is included when we write
y = C′tanh
(C′x + C′′
2 )
or y = 2 x + C .
The singular solution is not found by the current versions of either maple or mathematica.
Find general solutions for the following ODEs.
10.5.2. 2yy′′− y′2= 0 .
Solution:
Set y′ = u, y′′= u(du/dy), so the ODE becomes 2yu(du/dy)− u2= 0.
One solution is u = 0, i.e., y = C; other solutions can be obtained from 2y(du/dy)− u = 0. Solving that equation:
2y du = u dy −→ du u = dy
2y −→ ln u = ln y
2 + C or u = Cy1/2. Setting u = dy/dx and integrating again,
dy
y1/2 = C dx −→ 2y1/2= Cx + C′. Absorbing the “2” into the constants, we have y = (Cx + C′)2.
10.5. MORE GENERAL SECOND-ORDER EQUATIONS 169 10.5.3. xy′′− y′− y′2= 0 . In addition, this ODE has the singular solution y = C.
Solution:
Set y′ = u, y′′= u′. Then xu′− u − u2= 0. This ODE is separable:
x du = (u + u2)dx or du
u(u + 1) =dx x . Decompose 1/u(u + 1) into partial fractions and integrate:
ln u− ln(u + 1) = ln x + ln C −→ u
u + 1 = Cx . If C̸= 0 one can solve for u as follows and integrate again:
u = y′= Cx
1− Cx −→ y =
∫ Cx
1− Cxdx , so y =−ln(1− Cx)
C − x + C′.
However, if C = 0 we get the simple result u = 0, corresponding to y = C′. 10.5.4. xy′′+ y′= 0 .
Solution:
Set y′ = u and y′′ = u′, so the ODE becomes xu′+ u = 0. This equation is separable:
du
u =−dx
x −→ u = C x. Integrating again,
dy = C dx
x , −→ y = C ln x + C′. 10.5.5. x2y′′− xy′− 3y = x4.
Solution:
This is an inhomogeneous Euler equation. The corresponding homogeneous equation has solutions y = xp, with p(p− 1) − p − 3 = 0, so p = 3 or p = −1, so the homogeneous equation has general solution
y0(x) = C1x3+ C2x−1.
The inhomogeneous equation has a particular solution Ax4; substitution into the ODE shows that A = 1/5. The general solution is then y0(x) + x4/5.
10.5.6. x2y′′− 3xy′+ 4y = x .
Solution:
This is an inhomogeneous Euler equation. The corresponding homogeneous equation has solutions y = xp, with p(p− 1) − 3p + 4 = 0, which has a double root at p = 2. The homogeneous equation therefore has general solution
y0(x) = C1x2+ C2x2ln x .
170 CHAPTER 10. ORDINARY DIFFERENTIAL EQUATIONS A particular solution to the inhomogeneous equation is y = x, so this problem has general solution
y0(x) = x + C1x2+ C2x2ln x .
10.5.7. x2y′′− 3xy′+ 4y = x2.
Hint. Try solutions of the form xplnqx.
Solution:
This is an inhomogeneous Euler equation with the same corresponding homo-geneous equation as Exercise 10.5.6, with the general solution given as y0(x) of that Exercise. Because the inhomogeneous term has the same form as one of the terms of y0, a particular solution of the inhomogeneous equation must have the form Ax2ln2x. Substituting this form, we find A = 1/2, so the ODE has general solution y0(x) + (x2ln2x)/2.
10.5.8. The ODE satisfied by a quartic oscillator of mass m is of the form md2x
dt2 =−kx3.
Find a first integral of this ODE that can be interpreted as the law of conservation of energy for the oscillator.
Solution:
This problem is of the type illustrated in Eq. (10.87). Multiplying both sides of the ODE by dx/dt and integrating with respect to t, we get, giving dx/dt the name v,
mv2
2 =−kx4
4 + C , equivalent to mv2 2 +kx4
4 = C .
The two left-hand-side terms can be interpreted respectively as the kinetic and potential energy of the oscillator, while C is its total energy.
10.5.9. Consider the surface of the Earth to be a sphere of radius R. An object of mass m, located a distance r > R from the Earth’s center will experience a gravitational force due to the Earth in amount
F =mgR2 r2 .
Neglecting air resistance and other extraterrestrial masses, find the minimum upward velocity (escape velocity) an object must have when launched from the Earth’s surface in order for it to escape from the earth’s gravitational field.
Solution:
Assuming the motion of the object to be vertically upward, it will move subject to the equation
md2r
dt2 =−F = −mgR2 r2 .
Obtaining a first integral as illustrated in Eq. (10.87), and writing v for dr/dt, we have
mv2(r)
2 =−
∫ mgR2
r dr =mgR2 r + C .
10.6. GENERAL PROCESSES FOR LINEAR EQUATIONS 171 We now choose C to have the value corresponding to v = 0 when r =∞, namely C = 0. Then, setting r = R, we find
mv2(R)
2 = mgR , corresponding to v(R) =√ 2gR .
10.5.10. A particle of mass m initially at rest at x = 0 is subject, starting at time t = 0 to a force F = (a− x)
[ 1 + 2 ln
(a− x a
)]
.
By a method that does not require a complete solution of the equation of motion mx′′= F show that the particle will move only from x = 0 to x = a.
Solution:
Obtain a first integral as illustrated in Eq. (10.87), writing the integral of F in a form that ensures that x′ = 0 at x = 0,
m(x′)2
2 =
∫ x 0
F dx =
∫ x 0
(a− x) [
1 + 2 ln (a− x
a )]
dx . Evaluating the definite integral, we have
m(x′)2
2 =−(a − x)2ln (a− x
a )
.
The formula for F shows that F > 0 at x = 0, so the particle will initially move to positive values of x. Since the formula for (x′)2 remains nonzero for all x between x = 0 and x = a, x′ must stay positive over the interior of that interval. But both (x′)2and F become zero at x = a, so the particle will come to a permanent stop at x = a.