1000 Solved Problems in Modern Physics

(Romina) #1

1.3 Solutions 57


1.3.8 OrdinaryDifferentialEquations......................


1.55

dy
dx

=

x^3 +y^3
3 xy^2
The equation is homogenous becausef(λx,λy)= f(x,y). Use the trans-
formation
y=Ux,dy=Udx+xdU
Udx+xdu
dx

=

x^3 +U^3 x^3
3 x.U^2 x^2

=

1 +U^3

3 U^2

3 U^3 dx+ 3 xU^2 du=(1+U^3 )dx
or (2U^3 −1)dx+ 3 xU^2 du= 0
Dividing byx(2U^2 −1),
dx
x

+

3 U^2 du
2 U^3 − 1

= 0

Integrating, lnx+^12 ln(2U^3 −1)=C

2lnx+ln

(

2 y^3
x^3

− 1

)

=C

or 2y^3 −x^3 =Cx

1.56

d^3 y
dx^3

− 3

d^2 y
dx^2

+ 4 y= 0

The auxiliary equation is
D^3 − 3 D^2 + 4 = 0
Solving, the roots are− 1 , 2 ,2.
The root−1 gives the solutione−x.
The double root 2 gives two solutionse^2 x,xe^2 x.
The general solution is
y=C 1 e−x+C 2 e^2 x+C 3 xe^2 x

1.57

d^4 y
dx^4

− 4

d^3 y
dx^3

+ 10

d^2 y
dx^2

− 12

dy
dx

+5y= 0

The auxiliary equation is
D^4 − 4 D^3 + 10 D^2 − 12 D+ 5 = 0
Solving, the roots are 1, 1 , 1 ± 2 i
The pair of imaginary roots 1± 2 igives the two solutionsexcos 2x and
exsin 2x.
The double root gives the two solutionsex,xex.
The general solution is
Y=C 1 ex+C 2 xex+C 3 excos 2x+C 4 exsin 2x
or,y=(C 1 +C 2 x+C 3 cos 2x+C 4 sin 2x)ex.
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