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task how to solve about the first order differential equation
Typology: Exercises
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Selesaikan persamaan diferensial order satu berikut :
y
'
=x
3
e
− y
, y
Penyelesaian:
N(y).y’=M(X)
y
'
=x
3
e
− y
e
y
y
'
=x
3
permisalan y
'
dy
dx
e
y
dy
dx
=x
3
e
y
dy=x
3
dx
∫
e
y
dy=
∫
x
3
dx
e
y
x
4
1
Untuk mencari nilai C 1
y=0; x=
e
0
1
1
1
1
1
e
y
x
4
ln e
y
=ln
(
x
4
)
y=ln
(
x
4
)
dy
dx
y
'
x
y = 2 cos ( 3 x )
p ( x )=
x
q ( x )= 2 cos ( 3 x)
I (x)=e
∫
p
( x
) dx
=e
∫
2
x
dx
=x
2
y=
I (x)
y=
x
2
y=
x
2
9 sin
3 x
x
2
− 2 sin
3 x
3 x
y=
18 x
2
sin
3 x
3 x
− 4 sin
3 x
27 x
2
2
cos3y+2y)dy = 0.
Penyelesaian:
2 xsin( 3 y )
dx
dx
(
3 x
2
cos( 3 y ) + 2 y )
dy
dx
Permisalan y’=dy/dx
2 xsin( 3 y ) +( 3 x
2
cos ( 3 y )+ 2 y ) y '= 0
Persamaan dalam bentuk exact: M(x,y)+N(x,y)y’=
x
( x , y ) =M ( x , y )= 2 xsin ( 3 y )
y
x , y
x , y
= 3 x
2
cos
3 y
x
y
. y
'
d Ψ ( x , y )
dx
Ψ ( x , y )=C
∂ M ( x , y)
∂ y
∂ N (x , y )
∂ x
: sama
∂ M ( x , y)
∂ y
= 6 xcos( 3 y )
∂ N ( x , y )
∂ x
= 6 xcos ( 3 y )
∫
(¿ 2 xsin ( 3 y ))dx=x
2
sin ( 3 y )¿
∫
(¿ 2 y)dy= y
2
x
2
sin
3 y
2
x
3
4
dx+ 2 y
3
dy= 0
Penyelesaian:
3
4
dx
dx
3 dy
dx
Substitusi dy/dx=y’
x
3
4
3
y '= 0
Karena Persamaan Differential Order 1 Bernoulli, maka ditulis seperti:
y ' + P
x
y=Q
x
y
n
y
'
x
y =
−x
3
y
− 3
x
x
x
−x
3
, n=− 3
v= y
1 −n
v= y
1 −(− 3 )
= y
4
y=v
1
4
dy
dx
v
− 3 / 4 dv
dx
dy
dx
x
y=
−x
3
y
− 3
v
− 3
4
dv
dx
x
v
1
4
−x
3
v
− 3
4
v
3
4
dv
dx
x
v=
−x
3
1 −n
v
'
v
'
x
v=
−x
3
v
'
−x
3
x
v
v
'
(
x
3
x
v
)
v
'
(
4 x
3
4 x
v
)
v
'
3
2
2
) dx – 2xy dy = 0 dengan y(2) = 6
Penyelesaian:
2
2
dx
dx
− 2 xy
dy
dx
y ' + P
x
y=Q
x
y
n
2
2
dy
dx
Substitusi dy/dx=y’
x
2
2
− 2 xy y
'
Karena Persamaan Differential Order 1 Bernoulli, maka ditulis seperti:
y ' + P ( x ) y=Q ( x ) y
n
y
'
2 x
y =
x
y
− 1
P ( x )=
2 x
, Q ( x ) =
x
, n=− 1
v= y
1 −n
v= y
1 −(− 1 )
= y
2
y=v
1
2
dy
dx
v
− 1 / 2
dv
dx
dy
dx
2 x
y=
x
y
− 1
v
− 1 / 2
dv
dx
2 x
v
1
2
x
v
− 1
2
v
1
2
dv
dx
2 x
v =
x
1 −n
v
'
v ' −
2 x
v =
x
v
'
x
2 x
v
v
'
(
x
2 x
v
)
v
'
= x+
x
v
v
'
x
v=x
dv
dx
x
v=x
I ( x )=e
∫
− 3
x
dx
x
4
e
∫
− 3
x
dx
v= ∫
( x ) e
∫
− 3
x
dx
dx
x
4
v= ∫
x
4
x dx=
−sgn ( x )
x
v=e
3
ln
(
−sgn ( x )
x
)
=−x
2
+x
3
Substitusi v=y
2
y
2
=−x
2
3
y=6, x=
2
2
3
y= √
−x
2
3
y=x √
√
− 1 + 2 x
