Lesson 1.
4
In this lesson, the learners shall solve separable differential equations using antidifferentiation.
Before learning how to solve separable differential equations using antidifferentiation, we look
into the meanings of terms that we are to meet for the very first time here.
Differential equation (DE) is an equation containing a function and its derivatives.
Example 1:
dy
=3 x 2 +5 x+ 2 , here we see a derivative and a function in one equation
dx
Example 2:
d2 y
dy
+ 3 y =0 , this includes some derivatives and a function
2
dx
dx
2
The derivatives
dy d y d y d y
,
,
,
can be written in another way as
dx dx 2 dx 3 dx 4
y ' , y ' ' , y '' ' , y (4)
Example 3:
(4 )
(3)
''
'
3 y + 9 y +18 y +3 y 7 y=0 , derivatives written in another form
The order or degree of the differential equation is the order or degree of the highest derivative
present in the equation.
A linear differential equation of order n is a differential equation written in the general form:
dn y
d n1 y
dy
an ( x ) n +a n1 ( x ) n1 ++ a1 ( x ) +a0 ( x ) y =f ( x)
dx
dx
dx
where an (x ) 0 , because this function dictates the order of the differential equation.
Example 4:
In Example 1 above, n=1. We say that the differential equation is of order one or
first order differential equation
Example 5:
In Example 2, n=2.
The differential equation is order two or second order DE.
Example 6:
In Example 3, n=4 . The DE is referred to as fourth order.
A differential equation usually represents the relationship between a continuously varying
quantity and its rate of change. This is a relationship usually investigated in scientific works.
Thus we can see the practicality of using differential equations in biology, chemistry, physics and
other natural sciences, along with areas such as engineering and economics.
A strange attractor generated
equation (Figure 1):
from
differential
This is an illustration of heat transfer in a
particular material. This is created by
solving the heat equation represented by
a differential equation. In this material,
heat is being generated internally and
being cooled at the extremities,
providing a steady state of temperature
distribution. (Figure 2)
More complicated differential equations can be used, for instance, to model the relationship
between predators and prey. As predators increases, the prey decreases as more get eaten. As a
consequence of this, the predators will start to die out because they will have less to eat, which
allows more prey to survive. The interactions between the two populations are connected by
differential equations.
Other common practical uses of differential equations include, but are not limited to modelling
cancer growth or the spread of disease in the field of medicine; describing the movement of
electricity, usually in physics and engineering; modelling chemical reactions in chemistry;
finding optimum investment strategies in economics; describe the motion of waves, pendulums
or chaotic systems in physics. With such versatility to model real world phenomena, being able
to understand differential equations is an important skill for mathematicians.
A solution to a differential equation is a function that satisfies the DE, unlike in algebra where
solutions are just numbers.
There are many ways to determine the solution of a differential equation, if such exists, however
for this lesson, we look into a particular technique applicable to separable differential equations.
This technique is called separation of variables and is one of the most common ways of solving
differential equations.
Separable differential equations are differential equations wherein we could algebraically
separate the independent and dependent variables, usually x and y respectively, into two
functions equated to each other and then integrating each function using techniques we have
already learned from the previous lessons. Not all differential equations are separable.
Example 7:
Determine the solution to the following DE:
dy
=6 xy
dx
This involves three easy steps. First step is to separate variables algebraically. A
function involving one variable on the left side of the equation and a function involving
the other variable on the right side:
dy
=6 xdx
y
Notice that on the left side, the function now contains only variable y and on the right,
only variable x.
Second step is to integrate left side function and right side function using techniques we
have already encountered in previous lessons. For the left side function:
dy
y =ln y +c 1
And for the right side function:
6 xdx=3 x 2 +c 2
Putting this together we get ln y +c 1=3 x + c 2 . Simply, we can write this as
2
ln y=3 x +c , because the sum or difference of constants is still a constant.
And finally, the third step is to simplify in such a way that we have y on one side of the
equation. To do this on this particular problem, we apply the exponential function on
both sides since it is the inverse of a logarithmic function. This action will leave y on one
side.
ln y=3 x 2 +c
2
e ln y =e 3 x +c
3x
y=e e
y=c e
3x
applying the exponential function on both sides
using rules on exponentials
c
since e is just another constant, we can write it simply
as c
Example 8:
Find the solution of
First step:
Second step:
Third step:
dy 3 cos 2 t
=
dt
y
ydy=3 cos 2 t dt
ydy = 2y
algebraically separating variables
+ c1
for the left side.
3
3 cos 2 t dt= sin 2 t+c 2
2
for the right side. Remember the chain rule.
y2 3
= sin 2 t+ c
2 2
combining and simplifying
y= 3 sin 2t +c
solving for y and further simplifying
Example 9:
Show how to solve the differential equation
First step:
Second step:
2
dy ( 1+2 x )
=
dx
y
x
dy
=ln y + c1
y
( 1+ 2 x 2 )
dy
=y
dx
x
separating variables by algebra
integrating the left side function
( 1+2 x 2 )
x
dx=
dx
+ 2 x dx=ln x + x 2 +c 2
x
ln y=ln x +x +c3
Third step:
ln y
=e
integrating right side
combining the integral of both sides
ln x +x +c 3
property of natural logarithm and
exponential functions
1
x
=cx e
y
y=
1
cx e x
property of exponential functions
solving for y and simplifying
Example 10:
Solve
dw t 2 w4 w
=
dt
t+ 2
2
Step 1:
dw w(t 4) w(t2)(t +2)
=
=
dt
t+ 2
t+2
dw
=(t2) dt
w
Step 2:
simplifying through algebra
variables w and t separated
dw
=ln w+c 1
w
antiderivative of left side function
( t2 ) dt= 2t 2 t +c 2
antiderivative of right side function
ln w=
t
2t +c
2
combining and simplifying
Step 3:
w=
t
2
ce
e 2t
applying exponential function on both sides
By now, we are familiar with the three easy steps used in finding solutions to separable
differential equations. We are now ready to consider initial value problems (IVP) which brings
us closer to how differential equations are used in real world situations. Initial value problems
are differential equations along with an appropriate number of initial conditions. Usually, the
number of initial conditions depends on the order of the differential equation. For this lesson, we
are limiting our discussions to first order linear differential equations and these usually only have
one initial condition. You will probably meet higher orders in a separate differential equation
course.
Example 11:
4 x 2 y ' ' +12 x y ' +6 y=0
2
3
y (5 )= , y ' ( 5 ) =
5
64
The above IVP has a second order DE with two initial conditions
Example 12:
2t y ' + 4 y=6
y (1)=7
The IVP above is of order one. It comes with one initial condition
Example 13:
Determine the solution to the following IVP :
dy
=6 xy
dx
y (1)=2
To do this, we solve the corresponding DE. This is what we did in Example 7 to find the
3x
solution y=c e
. In an IVP, our goal is to solve for the constant c in the general
2
solution to come up with a particular solution. The initial condition above, y (1)=2 ,
means that we will be solving for c using the condition that when x=1 , then y=2 .
x=1
Simply substitute
for c:
and
y=2
into the general solution
y=c e 3 x
then solve
2=c e 3(1)
2
=c
e3
c=0.09957413674
Thus the particular solution is
y=0.099 e 3 x
Example 14:
Solve for the particular solution to
dy 3 cos 2 t
=
dt
y
when
y (0)=1
From Example 8, we found out that the general solution to this DE is y= 3 sin 2t +c .
Now, using the initial condition which implies that y=1 when t=0 , we solve for
c.
y= 3 sin 2t +c
the general solution
substituting y=1 and t=0
1= 3 sin 0+c
c=1
solving for c
y= 3 sin 2t +1
Finally, the particular solution is
Example 15:
Find the particular solution to
ds
=sin t +cos t
dt
when s ( ) =1
The general solution is s=sin t cos t +c . Verify this as your exercise. To look for the
particular solution, we need to solve for c when t= and s=1 .
s=sin t cos t +c
the general solution
1=sin cos + c
substituting t=
c=0
and s=1
solving for c
Thus the particular solution is s=sin t cos t
Exercises: Determine if the following DE are separable, if so, solve for their general solutions
dy
y 2
1. dw =3 e w
cos x
ysin y
2.
y'=
3.
dy
2 2
=9 t y
dt
4.
y ' =2 x e y
5.
x y =2( y4 )
'
Exercises: Solve for the particular solution given the following initial value problems:
'
y (0)=1
6. y =10x ,
7.
dy
=9 t 24 t +5 ,
dt
y (1)=0
8.
ds
=32 t2 ,
dt
y (1/2)=4
2 wy + 4 y =3 ,
dy t
10. dt = y ,
9.
y (1)=4
y (2)=1
http://www.sosmath.com/diffeq/basicdef/basicdef.html
http://www.maths.surrey.ac.uk/explore/vithyaspages/differential.html
http://www.saylor.org/site/wp-content/uploads/2012/09/MA102-5.5.4-Equations-and-InitialValue-Problems.pdf
https://en.wikipedia.org/wiki/Differential_equation
https://ibmathsresources.com/2014/02/28/differential-equations-in-real-life/
http://www.analyzemath.com/calculus/Differential_Equations/separable.html
