Page 97 - Schaum's Outline of Differential Equations

80 LINEAR DIFFERENTIAL EQUATIONS: THEORY OF SOLUTIONS [CHAR 8

8.26. Do the results of Problems 8.24 and 8.25 contradict Theorem 8.3?
No. Since the Wronskian of two linearly independent functions is identically zero, it follows from Theorem 8.3
3
that these two functions, x 3 and U !, are not both solutions of the same linear homogeneous differential equation of
the form L(y) = 0.

3
8.27. Two solutions of /' - (2lx)y r = 0 on [-1, 1 ] are y = x and y=\}?\. Does this result contradict the solution
to Problem 8.26?
3
3
No. Although W(x , U !) = 0 and both y = x 3 and y = I.* ! 3 are linearly independent solutions of the same linear
homogeneous differential equation y"—(2lx)y' = 0, this differential equation is not of the form \-(y) = 0. The
coefficient —2lx is discontinuous at x = 0.
8.28. The initial-value problem y' = 2^\y\; y(0) = 0 has the two solutions y = x\x\ and y = 0. Does this result
violate Theorem 8.1?
No. Here 0 = 2^/1 y , which depends on y; therefore, the differential equation is not linear and Theorem 8.1 does
I
not apply.

x >
8.29. Determine all solutions of the initial-value problem /' + e y + (x+ l)y = 0;y (1) = 0, /(I) = 0.
x
Here, b 2(x) = 1, b^x) = e , b 0(x) = x+ 1, andg(x) =0 satisfy the hypotheses of Theorems. 1; thus, the solution to
the initial-value problem is unique. By inspection, y = 0 is a solution. It follows that y = 0 is the only solution.
8.30. Show that the second-order operator L(y) is linear; that is

where c 1 and c 2 are arbitrary constants and y 1 and y 2 are arbitrary «-times differentiable functions.
In general,

Thus

8.31. Prove the principle of superposition for homogeneous linear differential equations; that is, if y l and
y 2 are two solutions of \-(y) = 0, then C 1y 1 + C 2y 2 is also a solution of \-(y) = 0 for any two constants c 1
and c 2.

Let y 1 and y 2 be two solutions of \-(y) = 0; that is, L(y 1) = 0 and \-(y 2) = 0. Using the results of Problem 8.30,
it follows that

Thus, c ly l + C 2y 2 is also a solution of \-(y) = 0.

8.32. Prove Theorem 8.4.
Since \-(y h) = 0 and \-(y p) = (f> (x), it follows from the linearity of L that

Thus, y is a solution.