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Question #300868
dy/dx+ycotx=5e
cosx
Expert's answer
Integrating factor
μ
(
x
)
=
e
∫
cot
x
d
x
=
e
ln
(
sin
x
)
=
sin
x
\mu(x)=e^{\int\cot xdx}=e^{\ln (\sin x)}=\sin x
μ
(
x
)
=
e
∫
c
o
t
x
d
x
=
e
l
n
(
s
i
n
x
)
=
sin
x
∫
cot
x
d
x
=
∫
cos
x
sin
x
d
x
=
ln
(
∣
sin
x
∣
)
+
C
\int \cot xdx=\int\dfrac{\cos x}{\sin x}dx=\ln(|\sin x|)+C
∫
cot
x
d
x
=
∫
sin
x
cos
x
d
x
=
ln
(
∣
sin
x
∣
)
+
C
y
′
sin
x
+
y
cos
x
=
5
e
cos
x
sin
x
y'\sin x+y\cos x=5e^{\cos x}\sin x
y
′
sin
x
+
y
cos
x
=
5
e
c
o
s
x
sin
x
d
(
y
sin
x
)
=
5
e
cos
x
sin
x
d
x
d(y\sin x)=5e^{\cos x}\sin xdx
d
(
y
sin
x
)
=
5
e
c
o
s
x
sin
x
d
x
Integrate
∫
d
(
y
sin
x
)
=
∫
5
e
cos
x
sin
x
d
x
\int d(y\sin x)=\int 5e^{\cos x}\sin xdx
∫
d
(
y
sin
x
)
=
∫
5
e
c
o
s
x
sin
x
d
x
y
sin
x
=
−
5
e
cos
x
+
C
y\sin x=-5e^{\cos x}+C
y
sin
x
=
−
5
e
c
o
s
x
+
C
y
=
−
5
e
cos
x
csc
x
+
C
csc
x
y=-5e^{\cos x}\csc x+C\csc x
y
=
−
5
e
c
o
s
x
csc
x
+
C
csc
x
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on Dec 2023
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