Initial Conditions and Separation of Variables
AP Calculus BC · Flashcards
AP CALCULUS BC · FLASHCARDS
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QUESTION
What is the particular solution for
d
y
d
x
=
−
3
y
\frac{dy}{dx} = -3y
d
x
d
y
=
−
3
y
with
y
(
2
)
=
1
y(2) = 1
y
(
2
)
=
1
?
QUESTION
What is the general solution for
d
y
d
x
=
−
y
\frac{dy}{dx} = -y
d
x
d
y
=
−
y
?
QUESTION
Convert
d
y
d
x
=
x
2
y
3
\frac{dy}{dx} = x^2y^3
d
x
d
y
=
x
2
y
3
into a separable form.
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ANSWER
d
y
y
3
=
x
2
d
x
\frac{dy}{y^3} = x^2 dx
y
3
d
y
=
x
2
d
x
. Move
y
3
y^3
y
3
to the left side and
x
2
x^2
x
2
to the right.
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