Question

Solve the differential equation $(36D^2-24D+13)y=2{\sin }^{2}x-e^{-x}+2$

Answer 1

The characteristic equation is $36p^2-24p+13=0$
$\Rightarrow p=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$=\frac{24\pm \sqrt{{24}^2-4\left(36\right)\left(13\right)}}{72}$
$(a=36,b=-24,c=13)$
$\frac{24\pm \sqrt{-1296}}{72}=\frac{24\pm 26i}{72}=\frac{12(2\pm 3i)}{72}$
$=\frac{2\pm 3i}{6}=\frac{2}{6}+\frac{3i}{6}=\frac{1}{3}+\frac{i}{2}$

Complementary functions is $e^{\frac{1}{3}x}[A\cos \frac{1}{2}x+B\sin \frac{x}{2}]....\left(1\right)$
$PI=2{\sin }^{2}x-e^{-x}+2=1-\cos 2x-e^{-}x+2=3-\cos 2x-e^{-x}[2.{\sin }^{2}x=1-\cos 2x]$
$P{I}_1=\frac{3.e^{ax}}{36D-24D+13}=\frac{3.e^{ax}}{0-0+13}=\frac{3}{13}{e}^{ax}=\frac{3}{13}....\left(2\right)$
$P{I}_2=\frac{-e^{-x}}{36D^2-24D+13}=\frac{-e^{-x}}{36(-1{)}^2-24(-1)+13}....\left(3\right)$
$=\frac{-e^{-x}}{36+24+13}=\frac{-e^{-x}}{73}$
$P{I}_3=\frac{-\cos 2x}{36D^2-24D+13}=\frac{-\cos 2x}{36(-4)-24D+13}$
$=\frac{-cos2x}{-144-24D+13}=\frac{-\cos 2x}{-24D-131}$
$=\frac{\cos 2x}{24D+131}\times \frac{24D-131}{24D-131}=\frac{(24D+131)\cos 2x}{576D^2-17161}$
$=\frac{(24D+131)\cos 2x}{576(-4)-17161}=\frac{(24D-131)\cos 2x}{-2304-17161}$
$=\frac{(24D-131)\cos 2x}{-19465}=\frac{24\left(2\right)(-sin2x)-131\cos 2x}{-19465}$
$=\frac{-48\sin 2x-131\cos 2x}{-19465}$
$=\frac{-48\sin 2x+131\cos 2x}{-19465}$
$\therefore$ Solution is $y=CF+P{I}_1+P{I}_2+P{I}_3$
$y=e^{\frac{x}{3}}[A\cos \frac{x}{2}+B\sin \frac{x}{2}]+\frac{3}{13}-\frac{e^{-x}}{73}+\frac{48\sin 2x+131\cos 2x}{19465}$