Question

(cosx.tany+cos(x+y))DX+(sinxsecsecy+cos(x+y))dy=0 find exact differential equation​

Answer 1

cos()tan()+cos(+))+(sin()sec2()+cos(+))=0
(
cos
⁡
(
x
)
tan
⁡
(
y
)
+
cos
⁡
(
x
+
y
)
)
d
x
+
(
sin
⁡
(
x
)
sec
2
⁡
(
y
)
+
cos
⁡
(
x
+
y
)
)
d
y
=
0


The question claims this is an exact differential equation. Verify it meets the exactness condition first.

(,)+(,)=0,∂∂(,)=∂∂(,)⇒(,)=
M
(
x
,
y
)
d
x
+
N
(
x
,
y
)
d
y
=
0
,
∂
M
∂
y
(
x
,
y
)
=
∂
N
∂
x
(
x
,
y
)
⇒
F
(
x
,
y
)
=
C


(,)=cos()tan()+cos(+)
M
(
x
,
y
)
=
cos
⁡
(
x
)
tan
⁡
(
y
)
+
cos
⁡
(
x
+
y
)


(,)=sin()sec2()+cos(+)
N
(
x
,
y
)
=
sin
⁡
(
x
)
sec
2
⁡
(
y
)
+
cos
⁡
(
x
+
y
)


∂∂=∂∂=cos()sec2()−sin(+)
∂
M
∂
y
=
∂
N
∂
x
=
cos
⁡
(
x
)
sec
2
⁡
(
y
)
−
sin
⁡
(
x
+
y
)


The exactness condition has been met. Now, solve ∂∂(,)=(,)
∂
F
∂
x
(
x
,
y
)
=
M
(
x
,
y
)
to get the function (,)
F
(
x
,
y
)
.

∂∂=(,)=cos()tan()+cos(+)
∂
F
∂
x
=
M
(
x
,
y
)
=
cos
⁡
(
x
)
tan
⁡
(
y
)
+
cos
⁡
(
x
+
y
)


(,)=(−1)sin()tan()+(−1)sin(+)+()
F
(
x
,
y
)
=
(
−
1
)
sin
⁡
(
x
)
tan
⁡
(
y
)
+
(
−
1
)
sin
⁡
(
x
+
y
)
+
g
(
y
)


The constant of integration is really an unknown function of ,()
y
,
g
(
y
)
. Compute ∂∂(,)=(,)
∂
F
∂
y
(
x
,
y
)
=
N
(
x
,
y
)
to determine ()
g
(
y
)
.

∂∂=(−1)sin()sec2()+(−1)cos(+)+′()
∂
F
∂
y
=
(
−
1
)
sin
⁡
(
x
)
sec
2
⁡
(
y
)
+
(
−
1
)
cos
⁡
(
x
+
y
)
+
g
′
(
y
)


(−1)sin()sec2()+(−1)cos(+)+′()=sin()sec2()+cos(+)⇒′()=0
(
−
1
)
sin
⁡
(
x
)
sec
2
⁡
(
y
)
+
(
−
1
)
cos
⁡
(
x
+
y
)
+
g
′
(
y
)
=
sin
⁡
(
x
)
sec
2
⁡
(
y
)
+
cos
⁡
(
x
+
y
)
⇒
g
′
(
y
)
=
0


′()=0⇒()=
g
′
(
y
)
=
0
⇒
g
(
y
)
=
A
a constant

This ()
g
(
y
)
is not always a constant! Yet, it must always be function of
y
alone.

(,)=(−1)sin()tan()+(−1)sin(+)+=
F
(
x
,
y
)
=
(
−
1
)
sin
⁡
(
x
)
tan
⁡
(
y
)
+
(
−
1
)
sin
⁡
(
x
+
y
)
+
A
=
B


sin()tan()+sin(+)=,=−
sin
⁡
(
x
)
tan
⁡
(
y
)
+
sin
⁡
(
x
+
y
)
=
C
,
C
=
A
−
B


Answer

sin()tan()+sin(+)=
sin
⁡
(
x
)
tan
⁡
(
y
)
+
sin
⁡
(
x
+
y
)
=
C

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