1000 Solved Problems in Modern Physics

(Romina) #1

64 1 Mathematical Physics


1.67 (a)y′−


2 y
x

=

1

x^3

(1)

Lety=px,y′=p+xp′
Then (1) becomes
xp′−p= 1 /x^3
Nowddx

(p
x

)

=xpx− 2 p

∴xp′−p=x^2

d
dx

(p
x

)

=

1

x^3
d
dx

(p
x

)

=

1

x^5

or d

(p
x

)

=

dx
x^5
Integrating
p
x

=−

1

4 x^4

+C

or

y
x^2

=−

1

4 x^4

+C

y=−

1

4 x^2

+Cx^2
It is inhomogeneous, first order.
(b)y′′+ 5 y′+ 4 y= 0
D^2 + 5 D+ 4 = 0
(D+4)(D+1)= 0
D=− 4 ,− 1
y=Ae−^4 x+Be−x
It is inhomogeneous, second order.

1.68 (a)

dy
dx

+y=e−x
Compare with the standard equation
dy
dx

+py=Q

P=1;Q=e−x

yexp

(∫

pdx

)

=

[∫

Qexp

(∫

pdx

)]

dx+C

yexp

(∫

1dx

)

=

[∫

e−xexp

(∫

1dx

)]

dx+C

yex=x+C
y=xe−x+Ce−x

(b)d

(^2) y
dx^2



  • 4 y=2 cos(2x)(1)
    The complimentary function is obtained fromy′′+ 4 y= 0
    y=U=C 1 sin 2x+C 2 cos 2x
    Differentiate (1) twice

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