# Second Order Differential Equations

Second Order Differential Equations

Second Order Differential Equations

Equivalent Forms; Initial Value Problems In general,

second order differential equations take the form

d2

y

= f

dx2 x, y, dy

dx

where f(x, y, z) is a continuous function of the variables x, y

and z in some region R ⊂ R 3 and is continuously differentiable

with respect to y and z throughout that region.

If we define z = dy

, then we can write the above equation

dx

equivalently as a system of two (coupled) first order equations:

dy

dx

= z;

dz

dx

,

= f(x, y, z).

Written in this form and drawing on experience with first order

equations, we would expect to have to give initial values

y(x0) = y0, z(x0) = z0,

in order to specify an initial value problem. Going back to the

second order equation, this means we need to give

y(x0) = y0, dy

dx (x0) = z0; equivalently : dy

dx (x0) = y1,

in order to specify an initial value problem for it.

General Solutions We say that an expression y(x, c1, c2) is a

general solution for the second order differential equation above

if, as a function of x, it satisfies the differential equation, and,

1

given initial values as above at a point x0, with (x0, y0, y1) in

the region R where f(x, y, z) has the required properties, we can

solve the equations

y(x0, c1, c2) = y0, dy

dx (x0, c1, c2) = y1

for c1 and c2 to satisfy that initial value problem.

Example 1 Consider the second order differential equation

d2y 2

=

dx2 x

dy

dx

− 2

x 2y

in the region x > 0, and let us try to find solutions of the form

y = x r . Substituting, we have

or

r(r − 1) x r−2 = 2r x r−2 − 2x r−2

x r−2 (r 2 − 3r + 2) = 0.

Since x r−2 is not identically zero, we need to solve the so-called

indicial equation

r 2 − 3r + 2 = 0 −→ r = 2 or r = 1.

So we conclude that y(x) = x and y(x) = x 2 are solutions,

which is easily checked out. Since the equation is linear with

respect to y (if y and ˆy are solutions, so is any linear combination

of the two) it then makes sense to try a general solution of the

form

y(x, c1, c2) = c1 x + c2 x 2 .

To satisfy initial conditions as specified above at x0 > 0 we

need

c1 x0 + c2 x0 2 = y0

c1 + 2 c2 x0 = y1

2

,

a system of two linear algebraic equations. The determinant of

the coefficients of c1 and c2 on the left hand side is 2 x0 2 − x0 2 =

x0 2 = 0, so we conclude we can always obtain a unique solution

of these equations. We conclude therefore that y(x, c1, c2) =

c1 x + c2 x 2 is the general solution of the second order differential

equation d2y dx2 = 2

x

dy

dx

− 2

x 2y for x > 0.

A General Method of Solution in the “Autonomous”

Case

If our second order differential equation does not depend explicitly

on x, which is technically referred to as the autonomous,

i.e., ”self-governing”, case, we can employ a transformation of

variables to reduce the second order equation to two first order

differential equations. Assuming the equation is

d2

y

= f

dx2 y, dy

dx

,

we let

dy

= z.

dx

Substituting this into the second derivative we have

d2y dz

=

dx2 dx

dz dy

=

dy dx

= z dz

dy

and thus the original second order equation becomes

z dz

dy

= f(y, z).

If we can solve this last equation to get z = z(y, c1), then the

equation used to define z becomes

dy

dx

= z(y, c1)

3

which, being in separable form

can be integrated to give

1

dy = dx

z(y, c1)

y 1

dη = x + c2.

z(η, c1)

Then, at least in principle, we can solve this to get y = y(x, c1, c2).

Success of the project ultimately depends on being able to get a

closed form solution for the transformed second order equation,

z dz = f(y, z), which has now become a first order equation in

dy

z and y.

Example 2 We consider the autonomous second order differential

equation

d2y dy

= y

dx2 dx .

Following the plan indicated above, we arrive at

z dz

dy

= y z or dz

dy

This is solved immediately to obtain

= y.

z = z(y, c1) = y2 + c1

.

2

Then we have

dy

dx = y2 + c1

.

2

There are three cases here, depending on whether c1 is 0, positive,

or negative. Taking just the case where c1 is positive, we

write c1 = c2 and we have

2

c2 + y2dy = dx.

4

Setting y = c w we have

2

c

and we integrate to obtain

1

1 + w2

dw = dx

2

c tan−1 (w) = x + c2.

Then w = tan( c

2 (x + c2)) and we have as the general solution

y = y(x, c, c2) = c tan( c

(x + c2)).

2

The cases corresponding to c1 = 0 and c1 < 0 are handled in

a similar way but different expressions are obtained. (Try those

cases as an exercise.)

Linear Second Order Differential Equations

It would be misleading to say that we can solve most second

order differential equations. In fact, the ones we can solve are

rather special. Prominent among these are the linear second

order differential equations

d2y dy

+ p(x)

dx2 dx

+ q(x) y = g(x),

where p(x), q(x) and g(x) are known, continuous (or piecewise

continuous) functions of x. Such a differential equation is called

homogeneous if g(x) ≡ 0 and inhomogeneous if g(x) ≡ 0, just

as in the first order case. For the present we consider only the

homogeneous case

d2y dy

+ p(x)

dx2 dx

+ q(x) y = 0.

A fundamental property of solutions of such a differential equation

is expressed in the following

5

Proposition If y1(x) and y2(x) are any two solutions of the

above linear homogeneous second order o.d.e., then for any constants

c1 and c2, the linear combination y(x) = c1 y1(x) +

c2 y2(x) is also a solution.

The proof is immediate when we use

d

c1 y1(x) + c2 y2(x)

dx

= c1

dy1 dy2

(x) + c2

dx dx

and the similar relationship for the second derivative of the linear

combination.

Constant Coefficient Linear Homogeneous Second Order

Differential Equations These are differential equations

of the form

d2y dy

+ p + q y = 0

2

dx

dx

where p and q are constants. These differential equations are

particularly easy to solve. Since in the linear homogeneous first

order constant coefficient case

dy

dx

+ p y = 0

we have the exponential solution y(x) = e −px it makes sense

to guess that the corresponding second order equation might

also have solutions of the form y(x) = e rx . Substituting this

proposed solution form into the differential equation we have

r 2 e rx + p r e rx + q e rx = e rx (r 2 + p r + q) = 0.

The exponential is never equal to zero so we must have

r 2 + p r + q = 0,

an algebraic equation which is called the characteristic equation

for the differential equation in question. Use of the quadratic

6

formula gives the two roots

r = −p ± √ p2 − 4 q

;

2

we will refer to the two roots thus obtained as r1 and r2. Then,

retracing our steps, we can see that with y(x) = e ri x , i = 1, 2,

we obtain

d2y dy

+ p

dx2 dx + q y = 0 = eri x

(ri 2 + p ri + q) = 0, i = 1, 2,

so that these are, indeed, solutions. Then, from the Proposition

we have solutions

y(x, c1, c2) = c1 e r1 x + c2 e r2 x .

It is natural to suspect that this might be the general solution.

To check on this we suppose that we want to solve

y(x0, c1, c2) = y0

d

dxy(x0, .

c1, c2) = y1

Substituting y(x, c1, c2) = c1 e r1 x + c2 e r2 x we have

c1 e r1 x0 + c2 e r2 x0 = y0

c1 r1 e r1 x0 + c2 r2 e r2 x0 = y1

The coefficients of the unknowns c1 and c2 form the array

This array has determinant

e r1 x0 e r2 x0

r1 e r1 x0 r2 e r2 x0

e (r1+r2)x0 (r2 − r1)

which is different from 0 if r1 = r2, i.e., if the roots of the

characteristic equation are distinct. Thus we have the form of

the general solution in this case. We postpone consideration of

7

.

.

the case r1 = r2, wherein r 2 + p r + q = (r − r1) 2 until a later

lecture.

Example 3 Find the solution of the initial value problem

d2y dy

− 7

dt2 dt + 12 y = 0, y(0) = 0, y′ (0) = 1.

Solution In this case the characteristic equation is

r 2 − 7 r + 12 = (r − 3)(r − 4) = 0,

yielding the roots r1 = 3, r2 = 4. Thus the general solution is

Its derivative is

At t = 0 we require

y(t, c1, c2) = c1 e 3t + c2 e 4t .

y ′ (t, c1, c2) = 3 c1 e 3t + 4 c2 e 4t .

y(0, c1, c2) = c1 e 3·0 + c2 e 4·0 = c1 + c2 = 0,

y ′ (0, c1, c2) = 3 c1 e 3·0 + 4 c2 e 4·0 = 3 c1 + 4 c2 = 1.

Multiplying the first equation by 3 and subtracting the result

from the second equation we obtain c2 = 1. Then the first

equation gives c1 = − 1 and the desired solution is

y(t) = − e 3t + e 4t .

Example 4 Find the general solution of

6 d2y dy

− 19

dt2 dt

Here the characteristic equation is

+ 15 y = 0.

6 r 2 − 19 r + 15 = 0.

8

The roots are not immediately obvious. We recall the quadratic

formula for finding the roots of a quadratic equation a r2 + b r +

c = 0:

r = −b ± √ b2 − 4 a c

.

2a

Applying this formula to our case we obtain the roots

r =

19 ±

(19) 2 − 4 · 6 · 15

2 · 6

which yields the roots

r1 = 20

12

Thus the general solution is

5

=

3 ; r2 = 18

12

= 19 ± √ 361 − 360

,

12

= 3

2 .

y(t, c1, c2) = c1 e 5t/3 + c2 e 3t/2 .

Example 5 Two blocks of material are placed in contact

with each other on a large stone table whose temperature is

maintained at 0 ◦ . The heat transfer coefficient between the two

blocks and between the blocks and the table is k > 0. Describe

the evolution of the temperature T1(t) in the first block.

Solution Applying the laws of heat conduction we obtained

a coupled system of two first order equations:

dT1

dt = − k (T1 − T2) − k (T1 − 0) = − 2 k T1 + k T2; (I)

dT2

dt = − k (T2 − T1) − k (T2 − 0) = − 2 k T2 + k T1. (II)

We differentiate the equation (I) with respect to t:

d2T1 dt1

= − 2 k

dt2 dt

dT2

+ k . (III)

dt

9

We substitute equations (I) and (II) into (III) and obtain

d 2 T1

dt 2 = − 2k ( − 2k T1 + k T2) + k ( − 2k T2 + k T1)

= 5 k 2 T1 − 4 k 2 T2. (IV)

To eliminate T2 in equation (IV) we multiply equation (I) by 4 k

to obtain

4k dT1

dt = − 8 k2 T1 + 4 k 2 T2. (V)

Then adding equations (IV) and (V) we obtain

or

d2T1 dT1

+ 4k

dt2 dt = − 3 k2 T1,

d 2 T1

dt2 dT1

+ 4k

dt + 3 k2 T1 = 0.

Here the characteristic equation is

with the roots

r 2 + 4k r + 3 k 2 = (r + k)(r + 3k) = 0

r1 = − k, r2 = − 3k.

Accordingly, th general solution is

T1(t, c1, c2) = c1 e −kt + c2 e −3kt .

Follow-up Question If k = 1, T1(0) = 10, T2(0) = 30,

compute T1(t).

Solution The initial condition for dT1

dt (0) is obtained with the

use of equation (I) above:

dT1

dt (0) = − 2k T1(0) + k T2(0) = − 20 + 30 = 10.

10

Thus, evaluating the given solution form and its derivative at

t = 0,

c1 e −0 + c2 e −3·0 = 10; − c1 e −0 − 3 c2 e −3·0 = 10.

From this we obtain c2 = − 10; c1 = 20 and the solution is

seen to be

T1(t) = 20 e −t − 10 e −3t ,

dT1

dt (t) = 30 e−3t − 20 e −2t .

Because the second block is relatively warm, the first block initially

warms up before ultimately cooling toward the temperature

of the table.

11

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