B)

¿

Rewrite∈the formof a first order linear ODE : y ' ( x )+ p ( x ) · y=q (x )

ddx

p ( x )= −1x (−ln ( x )+1 )

,q (x )= 1−ln (x )+1

+ln (x)

−ln ( x )+1

−1x (−¿ ( x )+1 )

y+ ddx

( y )= 1−ln ( x )+1

+ln (x)

−ln ( x )+1

Find theintegration factor : μ ( x )=1−ln ( x )

Find theintegrating factor μ (x) , so that : μ(x )· p(x )=μ '(x )

p ( x )= 1−ln ( x )+1

ddx

(μ ( x ) )=μ ( x )( −1x (−ln ( x )+1 ) )

Divide both sidesby μ ( x )

ddx

(μ ( x ))

μ(x )=

μ (x )( 1

x (−ln ( x )+1 ) )μ(x )

ddx

(μ ( x ))

μ(x )=

−1x (1−ln ( x ) )

ddx

( ln (μ ( x ) ) )=ddx

(μ ( x ))

μ(x )

ddx

( ln (μ ( x ) ) )= −1x (1−ln ( x ))

Solveddx

( ln (μ ( x )) )= −1x (1−ln ( x ) )

: μ ( x )=¿

μ ( x )=(1−ln ( x ) )ⅇc1

Theentire differential equationwill bemultiplied by (1−ln ( x ) )ⅇc1

Therefore the constant part ,ⅇc1 , isredundant∧canbe ignored

μ ( x )=1−ln (x )

Put the equation∈the form (μ ( x ) · y )'=μ ( x ) · q ( x ) : ddx

( (1−ln (x ) ) y )=ln ( x )+1

Multiply by the integration factor , μ(x )∧rewrtie the equationas

(μ ( x ) · y ( x )) '=μ (x) · q(x )

−1x (1−ln ( x ))

y+ dydx

( y )− 1−ln ( x )+1

±ln (x)

−ln ( x )+1

Multiply both sidesby the integrating factor ,1−ln (x )

−(1−ln ( x ) ) 1x (−ln ( x )+1 )

y+(1−ln (x ) ) ddx

( y )=(1−ln ( x ) ) 1− ln ( x )+1

+ (1− ln ( x ) ) ln ( x )−ln ( x )+1

Refine :− yx

=ln ( x )+1

Apply the Product Rule : ( f · g )'=f ' · g+ f · g '

f=1− ln x , g= y :− yx= ddx

( (1− ln ( x ) ) y )

ddx

((1−ln ( x ) ) y )= ln (x )+1

Solveddx

( (1− ln ( x ) ) y )=ln ( x )+1 : y=x (ln ( x )+1 )−x+c1

1− ln (x )

ddx

((1−ln ( x ) ) y )= ln (x )+1

If f (x)=g (x) then ∫ f (x )dx=∫ g(x )dx ,up¿aconstant

∫ⅆxⅆ

((1−ln ( x ) ) y ) dx= ∫ 1n ( x )+1dx

Integrate eachside of the equation ∫ ln(x)+1dx=x ( ln(x)+1)−x+c1

∫ ln (x )+1dx

Apply integration By Parts : ∫ uv '=uv−∫ u ' v

u=ln ( x )+1 , u'=1x, v '=1 , v=x

¿ ( ln ( x )+1 ) x− ∫1xxdx

¿ x ( ln ( x )+1 )− ∫ 1dx

∫ 1dx=x

∫ 1dx

Integral of aconstant : ∫ f (a )dx=x · f (a )¿1 xsimplify :=x

¿ x (ln (x)+1)−x

agregamos unaconstante ala solucion¿ x ( ln ( x )+1 )−x+C1

∫ ( ddx ((1−ln ( x ) ) y))dx= (1−ln ( x ) ) y+c2

(1−ln ( x ) ) y+c2=x ( ln ( x )+1 )−x+c1

combinamos las constantes(1−ln ( x ) ) y=x ( ln ( x )+1 )−x+c1

isolate y : y=x ( ln ( x )+1 )−x+c1

1−ln (x)

divedeboth sidesby (1−ln (x ) )

(1− ln ( x ) ) y1−ln (x)

=x (ln ( x )+1 )−x+c1

1−ln (x)

y=x ( ln ( x )+1 )−x+c1

1−ln (x )

y=x ( ln ( x )+1 )−x+c1

1−ln (x )