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Solving Differential Equations by Separating Variables
mccp-dobson-1111
Introduction
Suppose we have the first order differential equation
dy
P (y)
= Q(x)
dx
where Q(x) and P (y) are functions involving x and y only respectively. For example
1
1 dy
x−3
dy
= 3
or
=
.
y2
dx
x
y 2 dx
x3
We can solve these differential equations using the technique of separating variables.
General Solution
By taking the original differential equation
dy
= Q(x)
dx
we can solve this by separating the equation into two parts. We move all of the equation involving
the y variable to one side and all of the equation involving the x variable to the other side, then we
dy
can integrate both sides. Although dx
is not a fraction, we can intuitively treat it like one to move
the ”dx” to the right hand side. So
Z
Z
dy
P (y)
= Q(x) ⇔ P (y) dy = Q(x) dx.
dx
P (y)
Example
Let us find the general solution of the differential equation
dy
1
y2
= 3.
dx
x
1
= 3 ⇔
y
dx
x
2 dy
⇔
Z
Z
2
y dy =
2
y dy =
Z
Z
1
dx
x3
x−3 dx
y3
−x−2
=
+ c where c is a constant
3
2
−3
⇔ y 3 = 2 + 3c
2x
r
3 −3
+ 3c
⇔y=
2x2
⇔
www.mathcentre.ac.uk
c
Katy
Dobson
University of Leeds
Alan Slomson
University of Leeds
Example
To find the general solution of the differential equation
mathcentre community project
y 2 (x − 3)
dy
=
dx
x3
encouraging academics to share maths support resources
we first need to move
the
y 2 to are
thereleased
left hand
side
of the Non-commerical
equation. Then
move
All mccp
resources
under an
Attribution
Sharewe
Alike
licencethe dx to the
right hand side of the equation and integrate both sides.
y 2 (x − 3)
1 dy
x−3
dy
=
⇔ 2
=
3
dx
x
y dx
x3
Z
Z
1
x−3
⇔
dy =
dx
2
y
x3
Z
Z
1
x
3
⇔
dy =
− 3 dx
2
3
y
x
x
Z
Z
1
3
⇔ y −2 dy =
− 3 dx
2
x
x
Z
Z
⇔ y −2 dy = x−2 − 3x−3 dx
3x−2
+c
⇔ −y = −x +
2
1
3
−1
=− + 2 +c
⇔
y
x 2x
1
1
3
⇔ = − 2 −c
y
x 2x
2x
3
2cx2
1
⇔ = 2− 2−
y
2x
2x
2x2
1
2x − 3 − 2cx2
⇔ =
y
2x2
2x2
⇔y=
2x − 3 − 2cx2
−1
where c is a constant
−1
Exercises
Find the general solution of
dy
= y(1 + ex )
dx
1.
2.
dy
x
=
dx
y
3.
dy
= 9x2 y
dx
4.
4 dy
1
=
y 3 dx
x
Answers
1.
x+ex +c
y=e
2.
√
y = ± x2 + 2c
3.
3x3 +c
y=e
4.
s
y=±
−2
ln |x| + c
Note that the ñ symbol like in Exercise
2 means that the differential equation has two sets of
√
solutions, y = x2 + 2c and y = − x2 + 2c.
www.mathcentre.ac.uk
c
Katy
Dobson
University of Leeds
Alan Slomson
University of Leeds