Question Number 141716 by qaz last updated on 22/May/21
$$\left({x}^{\mathrm{2}} {lnx}\right){y}''−{xy}'+{y}=\mathrm{0} \\ $$
Answered by mnjuly1970 last updated on 23/May/21
![y_p_1 =ln(x)+1 or y_p_1 =x consider : y_p_1 =1+ln(x) because :: (x^2 ln(x))[((−1)/x^2 )]−x((1/x))+ln(x)+1=0 y_p_2 =v(x).y_p_1 where : v(x)=∫(1/((1+ln(x))^2 ))e^(∫(1/(xln(x)))dx) dx =∫(1/((1+ln(x))^2 ))e^(∫((1/x)/(ln(x)))dx) =∫ (e^(ln(ln(x)) /((1+ln(x))^2 ))dx =∫((ln(x))/((1+ln(x))^2 ))dx=^(ln(x)=t) ∫((te^t )/((1+t)^2 ))dt =∫[((1+t−1)/((1+t)^2 )).e^t ]dt=∫[(1/(1+t))−(1/((1+t)^2 ))]e^t dt =(e^t /(1+t))⇒ v(x)=(x/(1+ln(x))) y_p_2 =v(x).y_p_1 =(x/((1+ln(x))))(1+ln(x))=x .... y_c =c_1 (1+ln(x))+c_2 x..... general solution ....⇑⇑⇑](https://www.tinkutara.com/question/Q141751.png)
$$\:\:{y}_{{p}_{\mathrm{1}} } ={ln}\left({x}\right)+\mathrm{1}\:\:\:\:\:\:\:\:{or}\:\:\:\:{y}_{{p}_{\mathrm{1}} } ={x} \\ $$$$\:\:\:\:{consider}\::\:\:\:{y}_{{p}_{\mathrm{1}} } =\mathrm{1}+{ln}\left({x}\right) \\ $$$$\:\:{because}\:::\:\left({x}^{\mathrm{2}} {ln}\left({x}\right)\right)\left[\frac{−\mathrm{1}}{{x}^{\mathrm{2}} }\right]−{x}\left(\frac{\mathrm{1}}{{x}}\right)+{ln}\left({x}\right)+\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:{y}_{{p}_{\mathrm{2}} } ={v}\left({x}\right).{y}_{{p}_{\mathrm{1}} } \:\:{where}\:: \\ $$$$\:\:\:\:{v}\left({x}\right)=\int\frac{\mathrm{1}}{\left(\mathrm{1}+{ln}\left({x}\right)\right)^{\mathrm{2}} }{e}^{\int\frac{\mathrm{1}}{{xln}\left({x}\right)}{dx}} {dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\int\frac{\mathrm{1}}{\left(\mathrm{1}+{ln}\left({x}\right)\right)^{\mathrm{2}} }{e}^{\int\frac{\frac{\mathrm{1}}{{x}}}{{ln}\left({x}\right)}{dx}} =\int\:\frac{{e}^{{ln}\left({ln}\left({x}\right)\right.} }{\left(\mathrm{1}+{ln}\left({x}\right)\right)^{\mathrm{2}} }{dx} \\ $$$$\:\:\:=\int\frac{{ln}\left({x}\right)}{\left(\mathrm{1}+{ln}\left({x}\right)\right)^{\mathrm{2}} }{dx}\overset{{ln}\left({x}\right)={t}} {=}\int\frac{{te}^{{t}} }{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }{dt} \\ $$$$\:\:\:=\int\left[\frac{\mathrm{1}+{t}−\mathrm{1}}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }.{e}^{{t}} \right]{dt}=\int\left[\frac{\mathrm{1}}{\mathrm{1}+{t}}−\frac{\mathrm{1}}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }\right]{e}^{{t}} {dt} \\ $$$$\:\:\:=\frac{{e}^{{t}} }{\mathrm{1}+{t}}\Rightarrow\:{v}\left({x}\right)=\frac{{x}}{\mathrm{1}+{ln}\left({x}\right)} \\ $$$$\:\:\:\:\:{y}_{{p}_{\mathrm{2}} } ={v}\left({x}\right).{y}_{{p}_{\mathrm{1}} } =\frac{{x}}{\left(\mathrm{1}+{ln}\left({x}\right)\right)}\left(\mathrm{1}+{ln}\left({x}\right)\right)={x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….\:\:{y}_{{c}} ={c}_{\mathrm{1}} \left(\mathrm{1}+{ln}\left({x}\right)\right)+{c}_{\mathrm{2}} {x}….. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{general}\:{solution}\:….\Uparrow\Uparrow\Uparrow \\ $$$$ \\ $$