Tugas Soal Dan Pembahasan Persamaan Differensial Eksak

Irang maulana

Soal dan pembahasan persamaan differensial eksak

1. ( x + 2y ) dx + ( 4y + 2x ) dy = 0



F(x,y)

=∫ (

)

= ∫(

( ) )

( )

= 2y2 + 2xy + Q(x) = 2y + Q’(x) x + 2y = 2y + Q’(x) Q’(x)

=x

Q(x)

=∫ = x2

F(x,y) = x2 + 2xy + 2y2 

F(x,y)

=∫

(

)

= ∫(

( ) )

= x2 + 2xy + Q(y) = 2x + Q’(y) 4y + 2x = 2x + Q’(y) Q’(y)

= 4y

Q(y)

=∫ = 2y2

F(x,y) = x2 + 2xy + 2y2

( )

Irang maulana

2. ( x2 + y ) dx + ( 2y2 + x ) dy = 0



F(x,y)

(

=∫

)

= ∫(

( ) )

( )

= x3 + xy + Q(y) = x + Q’(y) 2y2 + x = x + Q’(y) Q’(y)

= 2y2

Q(y)

=∫ =

y3

F(x,y) = x3 + xy + y3 

F(x,y)

=∫ ( = ∫(

)

( ) )

= y3 + xy + Q(x) = y + Q’(x) x2 + y = y + Q’(x) Q’(x)

= x2

Q(x)

=∫ = x2

F(x,y) = x3 + xy + y3

( )

Irang maulana

3. ( 2x + 2y ) dx + ( y3 + 2x ) dy = 0



F(x,y)

(

=∫

)

= ∫(

( ) )

( )

= x2 + 2xy + Q(y) = 2x + Q’(y) y3 + 2x = 2x + Q’(y) Q’(y)

= y3

Q(y)

=∫ =

y4

F(x,y) = x2 + 2xy + y4 

F(x,y)

=∫ (

)

= ∫(

( ) )

= y4+ 2xy + Q(x) = 2y + Q’(x) 2x + 2y = 2y + Q’(x) Q’(x)

= 2x

Q(x)

=∫ = x2

F(x,y) = x2 + 2xy + y4

( )

Irang maulana

Soal dan pembahasan Faktor integrasi 1. Y dx + ( y + 2x ) dy = 0 -

=1

-

=2 F(y)

=(

)

=(

)

=U

=

∫

= =y U(x,y)M(x,y)dx + U(x,y)N(x,y)dy = 0 y (y) dx + y(y +2x)dy = 0 y2 dx + (y2 + 2xy) dy = 0 (eksak)

Irang maulana

2. 2y dx + x dy = 0 -

=2

-

=1 F(y)

=( =(

) )

= U

=

∫

= =x U(x,y)M(x,y)dx + U(x,y)N(x,y)dy = 0 x (2y) dx + x(x)dy = 0 2xy dx + x2 dy = 0 (eksak)