AP Calculus BC Quiz
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DIFFERENTIAL EQUATIONS
Question 1
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Does
y
=
1
x
2
y=\dfrac{1}{x^2}
y
=
x
2
1
satisfy
d
y
d
x
=
−
2
x
y
\dfrac{dy}{dx}=-\dfrac{2}{x}y
d
x
d
y
=
−
x
2
y
on
x
≠
0
x\ne 0
x
=
0
?
A
Yes, because
y
′
=
−
2
x
3
y'=-\dfrac{2}{x^3}
y
′
=
−
x
3
2
and
−
2
x
y
=
−
2
x
⋅
1
x
2
=
−
2
x
3
-\dfrac{2}{x}y=-\dfrac{2}{x}\cdot\dfrac{1}{x^2}=-\dfrac{2}{x^3}
−
x
2
y
=
−
x
2
⋅
x
2
1
=
−
x
3
2
.
B
No, because
y
′
=
2
x
3
y'=\dfrac{2}{x^3}
y
′
=
x
3
2
.
C
No, because
−
2
x
y
=
−
2
x
2
-\dfrac{2}{x}y=-\dfrac{2}{x^2}
−
x
2
y
=
−
x
2
2
.
D
Yes, because
y
′
=
−
2
x
2
y'=-\dfrac{2}{x^2}
y
′
=
−
x
2
2
and RHS is
−
2
x
2
-\dfrac{2}{x^2}
−
x
2
2
.
E
No, because
y
′
=
−
1
x
3
y'=-\dfrac{1}{x^3}
y
′
=
−
x
3
1
.
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