Question Number 123207 by benjo_mathlover last updated on 24/Nov/20
$$\:\:{x}^{\mathrm{2}} \:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:−\mathrm{3}{x}\:\frac{{dy}}{{dx}}\:+\:\mathrm{4}{y}\:=\:\ell{n}^{\mathrm{2}} \left({x}\right)−\ell{n}\left({x}\right) \\ $$
Commented by mnjuly1970 last updated on 24/Nov/20
$$\:\:\:\:\:{cauchy}\:{differential}\:{equation}. \\ $$
Commented by mohammad17 last updated on 24/Nov/20
 Yp=(1/4){u^2 −u+2u−1+(3/2)} Yp=(1/4)u^2 +(1/4)u+(1/8) put:u=lnx ∴Yh=(C_1 +C_2 lnx)x^2 ∴Yp=(1/4){ln^2 x+lnx+(1/2)} Y=Yh+Yp=(C_1 +C_2 lnx)x^2 +(1/4){ln^2 x+lnx+(1/2)} by:⟨m.o⟩](https://www.tinkutara.com/question/Q123236.png)
$${put}:{x}={e}^{{u}} \Rightarrow{u}={lnx} \\ $$$$ \\ $$$${let}:{xD}={D}_{\mathrm{1}} \Rightarrow{x}^{\mathrm{2}} {D}^{\mathrm{2}} ={D}_{\mathrm{1}} ^{\mathrm{2}} −{D}_{\mathrm{1}} \\ $$$$ \\ $$$$\Rightarrow\left({D}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{4}{D}_{\mathrm{1}} +\mathrm{4}\right){y}={u}^{\mathrm{2}} −{u} \\ $$$$ \\ $$$$\left({m}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{0}\Rightarrow{m}_{\mathrm{1}} ={m}_{\mathrm{2}} =\mathrm{2} \\ $$$$ \\ $$$${Yh}=\left({c}_{\mathrm{1}} +{c}_{\mathrm{2}} {x}\right){e}^{\mathrm{2}{u}} \\ $$$$ \\ $$$${Yp}=\frac{\mathrm{1}}{\left({D}_{\mathrm{1}} −\mathrm{2}\right)^{\mathrm{2}} }\left({u}^{\mathrm{2}} −{u}\right) \\ $$$$ \\ $$$${Yp}=\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{1}}{\left(\mathrm{1}−\frac{{D}_{\mathrm{1}} }{\mathrm{2}}\right)^{\mathrm{2}} }\left({u}^{\mathrm{2}} −{u}\right) \\ $$$$ \\ $$$${Yp}=\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{1}+{D}_{\mathrm{1}} +\mathrm{3}\frac{{D}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{4}}+\mathrm{4}\frac{{D}_{\mathrm{1}} ^{\mathrm{3}} }{\mathrm{8}}+….\right]\left({u}^{\mathrm{2}} −{u}\right) \\ $$$$ \\ $$$${Yp}=\frac{\mathrm{1}}{\mathrm{4}}\left\{{u}^{\mathrm{2}} −{u}+\mathrm{2}{u}−\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}}\right\} \\ $$$$ \\ $$$${Yp}=\frac{\mathrm{1}}{\mathrm{4}}{u}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}{u}+\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$ \\ $$$${put}:{u}={lnx} \\ $$$$ \\ $$$$\therefore{Yh}=\left({C}_{\mathrm{1}} +{C}_{\mathrm{2}} {lnx}\right){x}^{\mathrm{2}} \\ $$$$ \\ $$$$\therefore{Yp}=\frac{\mathrm{1}}{\mathrm{4}}\left\{{ln}^{\mathrm{2}} {x}+{lnx}+\frac{\mathrm{1}}{\mathrm{2}}\right\} \\ $$$$ \\ $$$${Y}={Yh}+{Yp}=\left({C}_{\mathrm{1}} +{C}_{\mathrm{2}} {lnx}\right){x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}\left\{{ln}^{\mathrm{2}} {x}+{lnx}+\frac{\mathrm{1}}{\mathrm{2}}\right\} \\ $$$$ \\ $$$${by}:\langle{m}.{o}\rangle \\ $$
Answered by Lordose last updated on 24/Nov/20
$$\mathrm{I}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{but}\:\mathrm{can}'\mathrm{t}\:\mathrm{type} \\ $$