Chapter 8 - Linear Differential Equations of Order n - 8.10 Chapter Review - Additional Problems - Page 576: 38

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We are given $y''+4y=7\cos^2x$ We have: $F(x)=7\cos^2 x=\frac{7}{2}+\frac{7}{2}\cos 2x$ Obtain: $\frac{7}{2}\cos 2x(D^2+4)=0$ and $\frac{7}{2}(D)=0$ So, $D^2+4$ is the annihilator of $\frac{7}{2}\cos 2x$ and $D$ is the annihilator of $\frac{7}{2}$ Therefore, $D(D^2+4)(D^2+4)=0$ The trial solution is $y_p(x)=A_0+A_1x\sin x+A_2x\cos x$