Chapter 2 - The Derivative - 2.5 Derivatives of Trigonometric Functions - Exercises Set 2.5 - Page 151: 27

$$\eqalign{ & \left( {\text{a}} \right)y = x\sin x{\text{ is a solution of }}y' + y = 2\cos x \cr & \left( {\text{b}} \right)y = x\sin x{\text{ is a solution of }}{y^{\left( 4 \right)}} + y'' = - 2\cos x \cr} $$

$$\eqalign{ & \left( {\text{a}} \right){\text{Let }}y = x\sin x \cr & {\text{Compute the derivatives }}y'{\text{ and }}y'' \cr & \,y' = \frac{d}{{dx}}\left[ {x\sin x} \right] \cr & \,y' = x\left( {\cos x} \right) + \sin x\left( 1 \right) \cr & and \cr & \,y'' = x\cos x + \sin x \cr & \,y'' = \frac{d}{{dx}}\left[ {x\cos x} \right] + \frac{d}{{dx}}\left[ {\sin x} \right] \cr & \,y'' = x\left( { - \sin x} \right) + \cos x + \cos x \cr & \,y'' = - x\sin x + 2\cos x \cr & \cr & {\text{Substitute }}y'{\text{ and }}y{\text{ into }}y' + y = 2\cos x \cr & \left( { - x\sin x + 2\cos x} \right) + x\sin x = 2\cos x \cr & {\text{Simplify}} \cr & - x\sin x + 2\cos x + x\sin x = 2\cos x \cr & 2\cos x = 2\cos x \cr & {\text{Then }}y = x\sin x{\text{ is a solution of }}y' + y = 2\cos x \cr & \cr & \left( {\text{b}} \right){\text{Let }}y = x\sin x \cr & \,y'' = - x\sin x + 2\cos x \cr & {\text{Compute }}y'''{\text{ and }}{y^{\left( 4 \right)}} \cr & \,y''' = - x\left( {\cos x} \right) - \sin x - 2\sin x \cr & \,y''' = - x\cos x - 3\sin x \cr & and \cr & \,{y^{\left( 4 \right)}} = \frac{d}{{dx}}\left[ { - x\cos x - 3\sin x} \right] \cr & \,{y^{\left( 4 \right)}} = - x\left( { - \sin x} \right) - \cos x - 3\cos x \cr & \,{y^{\left( 4 \right)}} = x\sin x - 4\cos x \cr & \cr & {\text{Substitute }}y''{\text{ and }}{y^{\left( 4 \right)}}{\text{ into }}{y^{\left( 4 \right)}} + y'' = - 2\cos x \cr & \left( {x\sin x - 4\cos x} \right) + \left( { - x\sin x + 2\cos x} \right) = - 2\cos x \cr & {\text{Simplify}} \cr & x\sin x - 4\cos x - x\sin x + 2\cos x = - 2\cos x \cr & - 2\cos x = - 2\cos x \cr & {\text{Then }}y = x\sin x{\text{ is a solution of }}y' + y = 2\cos x \cr} $$