(A) From Answer 25, we know that the d.e. for slope field II is y′ = y.
The general solution is y = cex. For a solution curve to pass through point
(0, −1), we have −1 = ce^0 and c = −1.
(C) Euler’s method for y′ = x, starting at (1, 5), with ∆x = 0.1, yields
(B) We want to compare the true value of y(1.2) to the estimated value of
5.21 obtained using Euler’s method in Solution 31. Solving the d.e.
yields , and initial condition y(1) = 5 means that , or C =
4.5. Hence . The error is 5.22 − 5.21 = 0.01.
(A) Slopes depend only on the value of y, and the slope field suggests
that y′ = 0 whenever y = 0 or y = −2.
(D) The slope field suggests that the solution function increases (or
decreases) without bound as x increases, but approaches y = 1 as a
horizontal asymptote as x decreases.
(D) The slope segments are not parallel in either the x or y direction, so
the d.e. must include both x and y in the definition; this excludes (B) and
(E). Next, (A) would result in all positive slopes. This is not the case, so
(A) is eliminated. Finally, (C) will have a zero slope at the point (1, 1),
and (D) will have a zero slope at the point (1, −1), thus, (D) will create
this slope field.
