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## Differential Equations Cauchy Euler Notes

by: Jessica Notetaker

6

0

4

# Differential Equations Cauchy Euler Notes MA 3253

Marketplace > Mississippi State University > Mathematics (M) > MA 3253 > Differential Equations Cauchy Euler Notes
Jessica Notetaker
MSU
GPA 3.8

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These notes go over the Cauchy Euler differential equations and has 6 example problems showing the different methods to solves them as well as the different solutions possibe.
COURSE
Differential Equations
PROF.
Velinda Calvert
TYPE
Bundle
PAGES
4
WORDS
CONCEPTS
Math, Differential Equations, de, cauchy-euler, Euler, CAUCHY, Examples
KARMA
75 ?

## Popular in Mathematics (M)

This 4 page Bundle was uploaded by Jessica Notetaker on Tuesday March 29, 2016. The Bundle belongs to MA 3253 at Mississippi State University taught by Velinda Calvert in Spring 2016. Since its upload, it has received 6 views. For similar materials see Differential Equations in Mathematics (M) at Mississippi State University.

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Date Created: 03/29/16
Cauchy­Euler Equations n n−1 n d y n−1 d y 1 dy anx n+a n−1 x n−1+…+a x 1 +a 0=g(x) d x d x dx Where a ,0a 1 …, a nare constants. It can also be written as: 2 d y dy a x 2+bx +cy=g(x) d x dx Or 2 ax y +bxy'+c=g(x) y=x    y’=x   y”=m(m­1)x m­2 2 2 m−1 m m 2 m m m 2 a x (m −m )+bxm x +cx =g x =ax (m −m )+bx m+cx =x (am −am+bm+c) 1) Solve as homogeneous equation for m  and 1 , assu2ing x  does not equal zero. 2 am +m b−a +c=0 Case 1: m  and m  are real and distinct 1 m1 2 m2 yc  c 1 +c x2 Case 2: m =m  (repeated roots) 1 2  yc  c 1 +c x2ln|x| Case 3: m  1nd m  a2e complex numbers yc  c 1 cos(ln|x|)+c x2sin(ln|x|) 2) To solve for y  pse the variation of parameters or the undetermined coefficients, or use a  substitution to transform the Cauchy­Euler equation to an equation with constant  coefficients. Ex 1: Solve x y”+5xy’+3y=0 a=1 b=5 c= x (m +m(5­1)+3)=0 x  =/ 0 2 m +4m+3=0 (m+1)(m+3)=0 m 1­1 m =23 y=c 1 +c x2 ­3 2 Ex 2: Solve x y”­2xy’+3y=0 a=1 b=­2 c=3 m +m(­2­1)+3=0 m ­3m+3=0 3± √−12 3 √3 m= = ±i 2 2 2 3 3 y=c x cos √3 ln( )+c x cos √3ln( ) 1 ( 2 ) 2 (2 ) Ex 3: Solve x y”­3xy’+4y=0 a=1 b=­3 c=4 m( x 22+(­3­1)m+4)=0 2 m ­4m+4=0 (m­2) =0 m=2 y=c1x +c2x ln(x) 3 Ex 4: Solve x y”’+xy’­y=0 a=1 b=0 c=1 d=­1 x x (m)(m­1)(m­2)+x m­x =0 x (m­1)(m ­2m+1)=0 m=1 2 y=c1x+c2xln(x)+3 x(ln(x)) Ex 5: Solve 2x y”+5xy’+y=x ­x 1) 2x y”+5xy’+y=0 2m +m(5­2)+1=0      2m +3m+1=0     (2m+1)(m+1)=0  m =­1/2  m =­1 ­0.5 ­1 1 2 yc=c1x +c 2 2) Use variation of parameters method yp=u1x +u 2 ­1 y ’=­0.5x   y ’=­x2 1 2 −2 −1 x x −5 −5 −5 w= −3 =−x 2+ 1x 2= −1 x 2 |−1 x 2 −x−2| 2 2 2 −1 0−1 x 1 −1 −2 w1= 1−x −2= (−x +x ) | |2 −x 2 −2 x 0 −1 −3 w = −1 −3 1−x −1= (−x +x ) 2 2 | x2 | 2 2 2 1(−x +x −2) 3 1 5 3 2 2 2 −2 2 2 2 u1= ∫ −5 dx= ∫x +x dx= x + x −1 x 2 5 3 2 −x (¿¿2+x)dx= −1 x + x 2 3 2 −1 −3 1 2 2 2(−x +x ) u2= ∫ −5 dx= ¿∫ −1 2 x 2 Ex 6: Solve x y”­3xy’+5y=3+2 t Let x=e to transform the Cauchy­Euler equation into a DE with constant coefficients t t t ­t x=e  x’=e or dx/dt=e which means  dt/dx=e ­t  ­t y’= dy/dx= (dy/dt)*(dt/dx) = (dy/dt)*e = e (dy/dt) y” = (d y/dx ) = (d/dx)(dy/dx) = (d/dx)(e (dy/dt)) = (d/dt)( y= {{d} ^ {2} y} over {d {x} ^ {2}} = {d} over {dx} * {dy} over {dx} = {d} over {dx} ( {e} ^ {­t} {dy} over {dt} )= {d} over {dt} ( {e} ^ {­t} {dy} over {dt} ) {dx} over {dt} 2 2 −e −t dy+e −td y e =−e −2tdy +e−2td y ( dt dt2 ) dt dt 2 Now plug everything back in 2 e−t)2−e −2tdy +e−2td y −3e e −tdy +5y=3+2e t ( dt dt2) ( ) dt −dy d y2 dy d y dy + 2−3 +5y=3+2e = t 2−4 +5y dt d t dt dt dt Now solve homogeneous equation: 2 d y dy 2−4 +5y=0 d t dt 4± 1√−20 2 m= =2±i m ­4m+5=0 2 y = c e cos(t) + c e sin(t) c 1 2 Now use undetermined coefficients: y =A + Be t y’=Be t y”=Be t p t  t  t  t t  t Be – 4Be + 5A + 5Be = 3 + 2e 2Be + 5A = 3 + 2e 2Be=2e t B=1 5A=3 A=3/5 t 2t 2t t yp= 3/5 + e y=c e1cos(t) + c e s2n(t) + 3/5 + e and now replace e = xt 2 2 y=c 1 cos(ln(x)) + c x s2n(ln(x)) + 3/5 + x

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