Calculus 2 Quiz
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DIFFERENTIAL EQUATIONS
Question 1
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The general solution to the differential equation
d
y
d
x
=
1
+
y
2
1
−
x
2
\frac{dy}{dx} = \frac{1+y^2}{\sqrt{1-x^2}}
d
x
d
y
=
1
−
x
2
1
+
y
2
is which of the following?
A
y
=
tan
(
arcsin
(
x
)
)
+
C
y = \tan(\arcsin(x)) + C
y
=
tan
(
arcsin
(
x
))
+
C
B
y
=
A
e
2
arcsin
(
x
)
−
1
y = \sqrt{A e^{2\arcsin(x)} - 1}
y
=
A
e
2
a
r
c
s
i
n
(
x
)
−
1
C
y
=
tan
(
arcsin
(
x
)
+
C
)
y = \tan(\arcsin(x) + C)
y
=
tan
(
arcsin
(
x
)
+
C
)
D
y
=
tan
(
ln
∣
x
+
x
2
−
1
∣
+
C
)
y = \tan(\ln|x+\sqrt{x^2-1}| + C)
y
=
tan
(
ln
∣
x
+
x
2
−
1
∣
+
C
)
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A
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C
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