Answer
See explanations.
Work Step by Step
a. Step 1. We need to plug-in each given test solution into the original differential equation to see if they are the solutions.
Step 2. For $y=sin(x)$, we have $y'=cos(x), y''=-sin(x)$; thus $y''+y=-sin(x)+sin(x)=0$ which satisfies the equation.
Step 3. For $y=cos(x)$, we have $y'=-sin(x), y''=-cos(x)$, thus $y''+y=-cos(x)+cos(x)=0$ which satisfies the equation.
Step 4. For $y=a\ cos(x)+b\ sin(x)$, we have $y'=-a\ sin(x)+b\ cos(x), y''=-a\ cos(x)-b\ sin(x)$, thus
$y''+y=-a\ cos(x)-b\ sin(x)+a\ cos(x)+b\ sin(x)=0$ which satisfies the equation.
b. Step 1. For the equation $y''+4y=0$, we can modify the solutions as $y=sin(2x)$, $y=cos(2x)$, and $y=a\ cos(2x)+b\ sin(2x)$.
Step 2. For a generalized equation $y''+n^2y=0$, we can modify the solutions as $y=sin(nx)$, $y=cos(nx)$, and $y=a\ cos(nx)+b\ sin(nx)$.
