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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)
