Answer
(a) This function is not a solution.
(b) This function is not a solution.
(c) This function is not a solution.
(d) This function is a solution.
Work Step by Step
(a) $y = sin~x$
$y' = cos~x$
$y'' = -sin~x$
We can evaluate the differential equation:
$y''+y = -sin~x+sin~x = 0 \neq sin~x$
This function is not a solution.
(b) $y = cos~x$
$y' = -sin~x$
$y'' = -cos~x$
We can evaluate the differential equation:
$y''+y = -cos~x+cos~x = 0 \neq sin~x$
This function is not a solution.
(c) $y = \frac{1}{2}x~sin~x$
$y' = \frac{1}{2}~sin~x+ \frac{1}{2}x~cos~x$
$y'' = \frac{1}{2}~cos~x + \frac{1}{2}~cos~x-\frac{1}{2}x~sin~x$
$y'' = cos~x-\frac{1}{2}x~sin~x$
We can evaluate the differential equation:
$y''+y = (cos~x-\frac{1}{2}x~sin~x)+(\frac{1}{2}x~sin~x) = cos~x \neq sin~x$
This function is not a solution.
(d) $y = -\frac{1}{2}x~cos~x$
$y' = -\frac{1}{2}~cos~x + \frac{1}{2}x~sin~x$
$y'' = \frac{1}{2}~sin~x + \frac{1}{2}~sin~x+\frac{1}{2}x~cos~x$
$y'' = sin~x+\frac{1}{2}x~cos~x$
We can evaluate the differential equation:
$y''+y = (sin~x+\frac{1}{2}x~cos~x)+(-\frac{1}{2}x~cos~x)= sin~x$
This function is a solution.