Question-136815 Tinku Tara June 3, 2023 Differential Equation 0 Comments FacebookTweetPin Question Number 136815 by rs4089 last updated on 26/Mar/21 Answered by Dwaipayan Shikari last updated on 26/Mar/21 y=eλxx2λ2+xλ+(x2−14)=0⇒λ=−x±x2−4x2(x2−14)2x2=−12x±2x2−4x42x2=−1±2−2x22xy=Λe−1+i2x2−22+Φe−1−i2x2−22=κe−12cos(2x2−22)+τe−12sin(2x2−22) Answered by Olaf last updated on 26/Mar/21 x2y″+xy′+(x2−14)y=0(1)Letu=yxy=ux−1/2y′=u′x−1/2−12ux−3/2y″=u″x−1/2−12u′x−3/2−12u′x−3/2+34ux−5/2y″=u″x−1/2−u′x−3/2+34ux−5/2(1):u″x3/2−u′x1/2+34ux−1/2+u′x1/2−12ux−1/2+(x3/2−14x−1/2)u=0u″x3/2+x3/2u=0u″+u=0u=aeix+be−ix=Acosx+iBsinxy=1x(Acosx+iBsinx) Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-71273Next Next post: Determine-S-S-a-2-ar-a-r-ar-2-a-2r-ar-n-1-a-n-1-r- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.