Question 1.11.7: Consider the differential equation y″+y = 0 Show that y1 =...

This example is very similar to example 1.6.4. Because of its importance, we discuss the current problem in full detail here as well.

For any equation, a solution is an object that makes the equation true. In the above differential equation, y represents a function. The equation asks “for which functions y is the sum of y and its second derivative equal to zero?”

Observe first that if we let y_{1} = sin t , then \acute{y_{1}}= cos t, so y_{1}^{″}=−sin t , and therefore y_{1}^{″}+y_{1}=−sin t +sin t = 0. In other words, y_{1} is a solution to the differential equation. Similarly, for y_{2} = cos t , \acute{y_{2}}=−sin t and y_{2}^{″}=−cos t, so that y_{2}^{″}+y_{2} =−cos t +cos t = 0. Thus, y_{2} is also a solution to the differential equation.

Now, consider any function y of the form y =c_{1}y_{1}+c_{2}y_{2}. That is, let y be any linear combination of the two solutions we have already found. We then have

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

so that, using standard properties of the derivative (properties which are linear in nature), it follows that

y\acute{y}= c_{1} cos t −c_{2} sin t

and

y″ =−c_{1} sin t −c_{2} cos t

We, therefore see that

y″ +y = (−c_{1} sin t −c_{2} cos t )+(c_{1} sin t +c_{2} cos t )
=−c_{1} sin t +c_{1} sin t −c_{2} cos t +c_{2} cos t
= 0

so that y is indeed also a solution of y″ +y = 0.