Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

If $\left(D^2+16\right)y=\cos 4x$ Here the auxiliary equation is ${\mathrm{m}}^2+16=0$ $\Rightarrow \mathrm{m}=\pm 4$ $\therefore$ Complementary function $=(A\cos 4x+B\sin 4x)$ \& Particular Integral (P.I.) $=\frac{1}{{\mathrm{D}}^2+16}\cdot \cos 4\mathrm{x}$ But $\frac{1}{{\mathrm{D}}^2+{\mathrm{a}}^2}\cos \mathrm{ax}=\frac{\mathrm{x}}{2\mathrm{a}}\sin \mathrm{ax}$ $\therefore$ P.I. $=\frac{\mathrm{x}}{2\times 4}\cdot \sin 4\mathrm{x}=\frac{\mathrm{x}}{8}\sin 4\mathrm{x}$ $\therefore$ Solution $\mathrm{y}=$ Complementary function + Particular Integral $\Rightarrow \mathrm{y}=\mathrm{A}\cos 4\mathrm{x}+\mathrm{B}\sin 4\mathrm{x}+\frac{\mathrm{x}}{8}\sin 4\mathrm{x}$