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) $y^{\prime \prime}-y=-2 \cos t, \quad y(0)=1, \quad y^{\prime}(0)=1$