1000 Solved Problems in Modern Physics

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.