Differential Equations Part 1

dy

dx

+ (

2x+1

x

) y = e−2x
I.F. = e∫(

2x+1

x

)dx = e∫(2+

1

x

)dx = e2x+lnx = e2x . x
So , y(xe2x) = ∫e−2x.xe2x + C
⇒ xyex = ∫ x dx + C
⇒ 2xye2x = x2 + 2C
It passes through (1,

1

2

e−2) we get C = 0
y =

xe−2x

2

⇒

dy

dx

=

1

2

e−2x (- 2x + 1)
⇒ f (x) is decreasing in (

1

2

,1)
y (loge2) =

(loge2)e−2(loge2)

2

=

1

2

loge 2