Question and Answers Forum

All Questions      Topic List

Differential Equation Questions

Previous in All Question      Next in All Question      

Previous in Differential Equation      Next in Differential Equation      

Question Number 193733 by Tawa11 last updated on 18/Jun/23

Find the ordinary differential equation  satisfy by:                       y  =  x^n (A  +  Blogx)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{ordinary}\:\mathrm{differential}\:\mathrm{equation} \\ $$$$\mathrm{satisfy}\:\mathrm{by}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{y}\:\:=\:\:\mathrm{x}^{\mathrm{n}} \left(\mathrm{A}\:\:+\:\:\mathrm{Blogx}\right) \\ $$

Answered by Rajpurohith last updated on 19/Jun/23

y ′=x^n ((B/x))+nx^(n−1) (A+Blog(x))  =Bx^(n−1) +((n/x))y  ⇒xy ′=Bx^n +ny....(1)  differentiate again  xy′′+y^′ =Bx^(n−1) +ny′  ⇒x^2 y′′+xy′=Bx^n +nxy′....(2)  subtract (1) from (2)  x^2 y′′+xy′−xy′=Bx^n +nxy′−Bx^n −ny  ⇒x^2 y′′−nxy′+ny=0  is the required ODE.

$${y}\:'={x}^{{n}} \left(\frac{{B}}{{x}}\right)+{nx}^{{n}−\mathrm{1}} \left({A}+{Blog}\left({x}\right)\right) \\ $$$$={Bx}^{{n}−\mathrm{1}} +\left(\frac{{n}}{{x}}\right){y} \\ $$$$\Rightarrow{xy}\:'={Bx}^{{n}} +{ny}....\left(\mathrm{1}\right) \\ $$$${differentiate}\:{again} \\ $$$${xy}''+{y}^{'} ={Bx}^{{n}−\mathrm{1}} +{ny}' \\ $$$$\Rightarrow{x}^{\mathrm{2}} {y}''+{xy}'={Bx}^{{n}} +{nxy}'....\left(\mathrm{2}\right) \\ $$$${subtract}\:\left(\mathrm{1}\right)\:{from}\:\left(\mathrm{2}\right) \\ $$$${x}^{\mathrm{2}} {y}''+{xy}'−{xy}'={Bx}^{{n}} +{nxy}'−{Bx}^{{n}} −{ny} \\ $$$$\Rightarrow{x}^{\mathrm{2}} {y}''−{nxy}'+{ny}=\mathrm{0} \\ $$$${is}\:{the}\:{required}\:{ODE}. \\ $$$$ \\ $$

Commented by Tawa11 last updated on 19/Jun/23

I appreciate sir.  God bless you.

$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com