Question 1.6.4: Consider the differential equation y″+y =0. Explain why the ...

In our upcoming study of differential equations, we will call the equation y″+y =0 a linear second-order homogeneous equation with constant coefficients. Equations of this form will be considered in chapter 3 and be the focus of chapter 4.

For now, we can intuitively understand why y = c_{1} cos t +c_{2} sin t is a solution to the equation. Note that in order to solve the equation y+y =0, we must find all functions y such that y″ =−y. From our experience in calculus,we know that

\frac {d}{dt}[sin t] = cos t  and \frac {d}{dt}[cost]=−sin t

Furthermore, if we consider second derivatives,

\frac{d^{2}}{d t^{2}}[\sin t]=\frac{d}{d t}[\cos t]=-\sin t \text { and } \frac{d^{2}}{d t^{2}}[\cos t]=\frac{d}{d t}[-\sin t]=-\cos t

Hence, the second derivative of each basic trigonometric function is the opposite of itself, which makes both y = cos t and y = sin t solutions to the equation y″ +y = 0.

Moreover, it is a straightforward exercise to show (using  properties of the derivative) that any scalar multiple (such as y = 3sin t ) of either function is also a solution to the differential equation, as is any combination of the form y = 2cos t +3sin t . More generally, this makes any function

y = c_{1} cos t +c_{2} sin t

a solution to the differential equation.