Differential Equations Quiz
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FIRST-ORDER DIFFERENTIAL EQUATIONS
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A first-order differential equation is given by
d
y
d
x
=
−
2
x
y
3
+
sec
2
(
x
)
3
x
2
y
2
\frac{dy}{dx} = -\frac{2xy^3 + \sec^2(x)}{3x^2y^2}
d
x
d
y
=
−
3
x
2
y
2
2
x
y
3
+
s
e
c
2
(
x
)
. What is the general solution?
A
x
2
y
3
+
tan
(
x
)
=
C
x^2y^3 + \tan(x) = C
x
2
y
3
+
tan
(
x
)
=
C
B
x
2
y
3
−
tan
(
x
)
=
C
x^2y^3 - \tan(x) = C
x
2
y
3
−
tan
(
x
)
=
C
C
y
3
+
tan
(
x
)
x
2
=
C
y^3 + \frac{\tan(x)}{x^2} = C
y
3
+
x
2
t
a
n
(
x
)
=
C
D
The equation is not exact and cannot be solved this way.
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