Examples

Q. Solve $(D^4-5D^2+8D-4)y=e^2y$

Auxilary equation $(m^4-5m^2+8m-4)=0$

$$ \begin{align*} m=2,2,1\\ P.I.= \frac{1}{f(D)}(e^{2x}+e^{x})=\frac{1}{(D-2)^2(D-1)}(e^{2x}+e^{x})\\ \frac{1}{(D-2)^2(D-1)}e^{2x}+\frac{1}{(D-2)^2(D-1)}e^x\\ = e^{2x}\frac{1}{D^{2}}1+e^{x} \frac{1}{D} (1)\\ \\ e^{2x} \frac{x^{2}}{2}+ e^xx\\ \\ General\\ solution:\\ c_{1}e^x+(c_2+c_3x)e^{2x}+e^{2x} \frac{x^{2}}{2}+ e^xx \end{align*} $$

We use Particular solution of linear higher order ODE

2 #

$$ \begin{align*} (D^2-4)y=sin(2x)\\ m^2-4=0\\ m=2,-2\\ \\ P.I. &= \frac{1}{F(-a^{2})}sin(2x)=\frac{-1}{8}\sin(2x)\\ \\ General\\ solution:\\ c_{1}e^{2x}+c_{2}e^{-2x}+\frac{-1}{8}\sin(2x) \end{align*} $$

3 #

$$ \begin{align*} (3D^{2}+2D-8)y=5 \cos x\\ \\ 3x^{2}+2x-8=0\\ x=-2,\frac{4}{3}\\ \\ P.I. =5 \frac{1}{3D^{2}+2D-8} \cos x\\ 5 \frac{1}{3(-1)+2D-8} \cos x\\ 5 \frac{2D+11}{4D^{2}-121} \cos x\

• \frac{1}{25}(2\sin x + 11 \cos x)\\ \\ General\\ solution:\\ c_{1}e^{-2x}+c_{2}e^{\frac{4}{3}x}+ \frac{1}{25}(2\sin x - 11 \cos x) \end{align*} $$

4 #

$$ \begin{align*} (D^{2}+a^{2})y=\sin ax\\ m^2+a^2=0\\ m=-a,-ai\\ \\ P.I. = \frac{1}{D^{2}+a^{2}}\sin ax\\ = IM(\frac{1}{D^{2}+a^{2}} e^{iax})\\ =e^{iax} \frac{1}{(D+ia)^{2}+a^{2}} e^{0}\\ =e^{iax} \frac{1}{D^2+2aiD} e^{0}\\ =e^{iax} \frac{1}{2ai}\frac{1}{D} e^{0}\\ =e^{iax} \frac{1}{2ai} x\\ Im\\ part: \frac{-x}{2a} \cos ax\\ \\ General\\ solution:\\ \frac{-x}{2a} \cos ax + c_{1}e^{-ax}+c_{2}e^{-aix} \end{align*} $$

5 #

$$ \begin{align*} (D^4-2D^3+D^2)y=x^3\\ m^4-2m^3+m^2=0\\ m=0,0,1,1\\ y_{c}=c_1+c_{2}x+(c_3+c_{4}x)e^{x}\\ \\ P.I. y=\frac{1}{D^{2}(D^{2}-2D+1)}x^{3}\\ \frac{1}{D^{2}}(1-D)^{-2}x^{3}\\ \frac{1}{D^{2}}(1+2D+3D^2+4D^3+…)x^{3}\\ \frac{1}{D^{2}}(x^3+6x^2+18x+24+0..)\\ \frac{1}{D}(\frac{x^{4}}{4}+2x^{3}+9x^{2}+24x)\\ y_p=\frac{x^{5}}{20}+ \frac{x^{4}}{2} + 3x^{3}+ 12x^{2}\\ \\ General\\ solution:\\ c_1+c_{2}x+(c_3+c_{4}x)e^{x}+\frac{x^{5}}{20}+ \frac{x^{4}}{2} + 3x^{3}+ 12x^{2} \end{align*} $$

6 #

Find particular integral of: $$ \begin{align*} (D^{2}-1)y=x^{2}\cos x\\ \frac{1}{D^{2}-1}x^{2}\cos x\\ =Re{\frac{1}{D^{2}-1}x^{2}e^{ix}}\\ e^{ix}\frac{1}{(D+i)^{2}-1}x^{2}\\ e^{ix} \frac{1}{D^{2}+2iD-2} x^{2}\\ e^{ix} \frac{1}{-2(1 - \frac{D^2+2iD}{2})} x^2\\ \frac{-1}{2}e^{ix} (1+ \frac{D^2+2iD}{2} + \frac{(D^2+2iD)^2}{4}) x^2\\ \frac{-1}{2}e^{ix}\left(1+\frac{D^{2}}{2}+iD-D^{2}\right)x^2\\ \frac{-1}{2}e^{ix}(x^{2}-1+2ix) \end{align*} $$ find real part of that later