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

Find the complementary function from the auxiliary equation, rewrite the RHS using $2\sin^2x=1-\cos2x$, then find the particular integral term by term.

1. Complementary function. Auxiliary equation: $36m^2-24m+13=0$.

   $m=\dfrac{24\pm\sqrt{24^2-4(36)(13)}}{2(36)}=\dfrac{24\pm\sqrt{576-1872}}{72}=\dfrac{24\pm\sqrt{-1296}}{72}=\dfrac{24\pm36i}{72}=\dfrac13\pm\dfrac{i}{2}$

   Complex roots $\dfrac13\pm\dfrac{i}{2}$ give

   $CF=e^{x/3}\left[C_1\cos\dfrac{x}{2}+C_2\sin\dfrac{x}{2}\right]$

2. Simplify the RHS. Using $2\sin^2x=1-\cos2x$:

   $2\sin^2x-e^{-x}+2=(1-\cos2x)-e^{-x}+2=3-\cos2x-e^{-x}$

3. PI for the constant term $3$. For $f(D)y=3$, $PI_1=\dfrac{3}{f(0)}=\dfrac{3}{13}$ (since $f(0)=13$).

4. PI for $-\cos2x$. Put $D^2=-2^2=-4$ in $f(D)=36D^2-24D+13$: $f(D)\to 36(-4)-24D+13=-131-24D$.

   $PI_2=\dfrac{-\cos2x}{-131-24D}=\dfrac{\cos2x}{131+24D}$

   Rationalise by multiplying by $\dfrac{131-24D}{131-24D}$, using $D^2\to-4$ in the denominator:

   $(131+24D)(131-24D)=131^2-576D^2=131^2-576(-4)=17161+2304=19465$

   Numerator: $(131-24D)\cos2x=131\cos2x-24\,D(\cos2x)=131\cos2x-24(-2\sin2x)=131\cos2x+48\sin2x$

   $PI_2=\dfrac{131\cos2x+48\sin2x}{19465}$

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