Page 109 - Schaum's Outline of Differential Equations

92 wth-ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS [CHAP. 10

4
characteristic equation, so the characteristic equation must be (X - 4) = 0, or

The associated differential equation is

10.11. Find the general solution to a third-order linear homogeneous differential equation for y(x) with real
2
numbers as coefficients if two solutions are known to be e' * and sin 3x.
2x
If sin 3x is a solution, then so too is cos 3x. Together with e~ , we have three linearly independent solutions
to a third-order linear, homogeneous differential equation, and we can write the general solution as

10.12. Determine the differential equation described in Problem 10.11.
x
The characteristic equation of a third-order differential equation must have three roots. Because e^ and sin 3x
are solutions, we know that X = -2 and X = + 3 are roots of the corresponding characteristic equation, so this equation
must be

or

The associated differential equation is

10.13. Find the general solution to a sixth-order linear homogeneous differential equation for y(x) with real
2 7x
numbers as coefficients if one solution is known to be X e cos 5.x.
lx
lx
2 lx
If x e cos 5x is a solution, then so too are xe cos 5x and e cos 5x. Furthermore, because complex roots of
a characteristic equation come in conjugate pairs, every solution containing a cosine term is matched with another
2 lx
lx
solution containing a sine term. Consequently, x e sin 5x, xe lx sin 5x, and e sin 5x are also solutions. We now
have six linearly independent solutions to a sixth-order linear, homogeneous differential equation, so we can write
the general solution as
10.14. Redo Problem 10.13 if the differential equation has order 8.
An eighth-order linear differential equation possesses eight linearly independent solutions, and since we can
only identify six of them, as we did in Problem 10.13, we do not have enough information to solve the problem. We
can say that the solution to Problem 10.13 will be part of the solution to this problem.

2
10.15. Solve if one solution is xe *.
If xe 2x is a solution, then so too is e 2x which implies that (X- 2) 2 is a factor of the characteristic equation
. Now,

3x
so two other roots of the characteristic equation are X = +3, with corresponding solutions e^ x and e~ . Having
identified four linearly independent solutions to the given fourth-order linear differential equation, we can write the
general solution as