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Question Number 141716 by qaz last updated on 22/May/21

(x^2 lnx)y′′−xy′+y=0

(x2lnx)y″−xy′+y=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 ....⇑⇑⇑

yp1=ln(x)+1oryp1=xconsider:yp1=1+ln(x)because::(x2ln(x))[−1x2]−x(1x)+ln(x)+1=0yp2=v(x).yp1where:v(x)=∫1(1+ln(x))2e∫1xln(x)dxdx=∫1(1+ln(x))2e∫1xln(x)dx=∫eln(ln(x)(1+ln(x))2dx=∫ln(x)(1+ln(x))2dx=ln(x)=t∫tet(1+t)2dt=∫[1+t−1(1+t)2.et]dt=∫[11+t−1(1+t)2]etdt=et1+t⇒v(x)=x1+ln(x)yp2=v(x).yp1=x(1+ln(x))(1+ln(x))=x....yc=c1(1+ln(x))+c2x.....generalsolution....⇑⇑⇑

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