Concavity of the Solution Curve In the discussion of direction fields in
Section 1.3, you saw how the differential equation defines the slope of the
solution curve at a point in the ty-plane. In particular, given the initial
value problem \(y^{\prime}=f(t, y), y\left(t_{0}\right)=\) \(y_{0}\), the slope of
the solution curve at initial condition point \(\left(t_{0}, y_{0}\right)\) is
\(y^{\prime}\left(t_{0}\right)=f\left(t_{0}, y_{0}\right)\). In like manner, a
second order equation provides direct information about the concavity of the
solution curve. Given the initial value problem \(y^{\prime \prime}=f\left(t,
y, y^{\prime}\right), y\left(t_{0}\right)=\) \(y_{0},
y^{\prime}\left(t_{0}\right)=y_{0}^{\prime}\), it follows that the concavity of
the solution curve at the initial condition point \(\left(t_{0}, y_{0}\right)\)
is \(y^{\prime \prime}\left(t_{0}\right)=f\left(t_{0}, y_{0},
y_{0}^{\prime}\right)\). (What is the slope of the solution curve at that
point?)
Consider the four graphs shown. Each graph displays a portion of the solution
of one of the four initial value problems given. Match each graph with the
appropriate initial value problem.
(a) \(y^{\prime \prime}+y=2-\sin t, \quad y(0)=1, \quad y^{\prime}(0)=-1\)
(b) \(y^{\prime \prime}+y=-2 t, \quad y(0)=1, \quad y^{\prime}(0)=-1\)
(c) \(y^{\prime \prime}-y=t^{2}, \quad y(0)=1, \quad y^{\prime}(0)=1\)
(d) \(y^{\prime \prime}-y=-2 \cos t, \quad y(0)=1, \quad y^{\prime}(0)=1\)
