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Practice: Separable with Initial Condition
Related Topics
Wize University Calculus 2 Textbook > Differential Equations
Separable Differential Equations
6 Activities
Q:
\textbf{Q:}
Q:
Solve the differential equation
d
y
d
x
=
2
x
+
1
y
3
\dfrac{dy}{dx}=\dfrac{2x+1}{y^3}
d
x
d
y
=
y
3
2
x
+
1
given that
y
(
0
)
=
2
y\left(0\right)=2
y
(
0
)
=
2
Answer
I don't know
Check Submission
More Separable Differential Equations Questions:
Practice: Separable DE
Practice Question
Solve
d
y
d
x
=
4
e
5
x
y
\displaystyle \frac{dy}{dx}=\frac{4e^{5x}}{y}
d
x
d
y
=
y
4
e
5
x
Practice: Separable DE
Solve
d
y
d
x
=
4
e
5
x
y
\displaystyle \frac{dy}{dx}=\frac{4e^{5x}}{y}
d
x
d
y
=
y
4
e
5
x
Practice: Separable with Initial Condition
Solve the differential equation
d
y
d
x
=
2
x
+
1
y
3
\dfrac{dy}{dx}=\dfrac{2x+1}{y^3}
d
x
d
y
=
y
3
2
x
+
1
given that
y
(
0
)
=
2
y\left(0\right)=2
y
(
0
)
=
2
Practice: Separable DE
Practice Question
Solve
d
y
d
x
=
4
e
5
x
y
\displaystyle \frac{dy}{dx}=\frac{4e^{5x}}{y}
d
x
d
y
=
y
4
e
5
x