FORM SIX MATHEMATICS STUDY NOTES DIFFERENTIAL EQUATION
UNAWEZA JIPATIA NOTES ZETU KWA KUCHANGIA KIASI KIDOGO KABISA:PIGA SIMU:0787237719
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A differential equation is a relationship between the rates of different variables (i.e an independent variables x, a dependent variable, y, and one or more differential coefficient of y with respect to x).
Eg + y sin x = 0
xy + y + = 0
+ – 4y = 0
The order of a differential equation
The order of a differential equation is given by the highest derivative involved in the equation.
– y^{2} = 0 is an eqn of the 1^{st} order
xy – y^{2} sin x = 0 is an eqn of the 2^{nd} order
xy – y + = 0 is an eqn of the 3^{rd} order
The degree of a differential equation
Eg + y sin x = 0 the degree is 1b
Linear differential equation (L.D.E)
A linear differential equation should look in the form
+ + + ……… + pn y = Q
Where P_{1}, P_{2}, P_{3} …Pn, Q are functions of x, or constant n = 1, 2, 3, 4…….
Examples
i) + + 6y = 0
ii) + + =
iii) = 4
Examples of non L.D.E
+ + 6y = 0
+ + 6y = 0
Note
A D. E will be linear if
 Variable y and its derivatives occur at the first degree only
 No product of y and its derivatives
 No transcendental function of y or x
Exercise
Show which of the following differential equations is the L.D.E or N.L.D.E
i) = y……………linear
ii) (x^{2 }– 1) = y……linear
iii) + y^{2} + 4 = 0…………..non – linear
iv) = ……………non – linear
v) = 1 + xy………linear
vi) (x^{2} – y^{2}) = 2xy……Non – linear
Formulation of a differential
Differential equation may be formed when arbitrary constant are eliminated from a given function.
Examples 1
Y = A sin x + B cos x, where A and B are two arbitrary constants
Solution
= A cos x – B sin x
= A sin x – B cos x
his is identical to the original eqn with opposite signs
= 1(A cos x + B sin x)
= y
Form a DE whose solution is the form y = Ae^{2x} + Be^{3X }Where A and B are constant.
Example 2
Form a different equation from the function y = x +
Solution
y = x + = x + Ax^{1}
= 1 – Ax^{2 }= 1 –
From the given equation
Y = x +
yx = x +
x(y – x) = A
= 1 x (y – x) x^{2}
= 1 –
= x^{2 }– x (y – x)
= x – (y – x)
^{ }= 2x – y
Example 3
Form the DE for y = Ax^{2} + Bx
Y = Ax^{2 }+ Bx…… (i)
= 2Ax + B
– 2Ax = B……. (ii)
= 2A
Substituting both (ii) and (iii) into (i) gives
– 2x + 2y = 0
Solution to a DE
This involves finding the function for which the equation is true (i.e manipulating the eqn so as to eliminate all the differential coefficients and have a relationship between x and y)
E.g. verify that the function
i) Y = 3e^{2x} is a soln of DE = – 2y = 0 for all x
ii) y (x) = sin x – cos x + 1 is a soln of eqn + y = 1 for all value of x
Solution
Substitute equation iv into iii.
= 2y
– 2y = 0
Direct integration
Direct integration is used to solve equation which is arranged in the form = f (x)
Examples
1. Solves = 3x^{2 }– 6x + 5
Solution
= 3x^{2} – 6x + 5
Then =
= x^{3 }– 3x^{2} + 5x + c
Y = x^{3} – 3x^{2 }+ 5x + c
2. Solve = 5x^{3} + 4
Solution
Rearranging in the form = f (x)
= 5x^{2 }+
Then = dx
Y = + 4ln x + c
Y = + 4ln x + c
The solution is a called general solution since it consist a constant C (i.e. unknown constant)
3. Find the solution of the eqn e^{x} = 4 given that y = 3
When x = 0
Soln: rearrange the given equation
= = 4e^{x} + 7
This is called a particular soln since it contains a non – unknown variable.
First order D.E
Separating the variables
This is a method used to solve the D.E when is in the form = f (x, y)
The variables y on the right handside prevents solving by direct integration
Example
1. Solve =
Solution =
= (y + 1) = 2x
Integrating both sides with respect to x
2. Solve =
Solution
=
=
dy = dx
3. By separating variables solve the differential equation (xy + x) dx =(x^{2}y^{2} + y^{2 }+ x^{2} + 1) dy
Solution
Given Equation;
(xy + x) dx = (x^{2}y^{2} + y^{2} + x^{2} + 1) dy
X (y + 1) dx = (y^{2 }(x^{2 }+ 1) + (x^{2 }+1)) dy
X (y + 1) dx = (x^{2 }+ 1) (y^{2} + 1) dy
dx = dy
Integrating both sides
dx = dy
R.H.S == y1+
dx =) dy
½ ln (x^{2} + 1) = – y + 2 ln (y + 1) + C
First order homogenous D.E
In first order H.D.E all the terms are of the same dimension
Consider the table about dimensions
Term  X  x^{2}  1/x  x^{n}  3  y  y^{n}  x^{2/}y  x^{2}y 
Dimension  1  2  1  n  O  1  N  1  3 
dy/dx  dy^{2}/dx^{2}  xdy/dx  yx  
0  1  2 
O shows there is no effect on the term
Which of the following equation are first order homogeneous
a) x^{2 }= y^{2}
b) xy = x^{2 }+ y^{2}
c) x^{2} = 1 + xy
d) (x^{2} =
e) (x^{2} – y^{2}) = 2xy
f) (1 + y^{2}) = x
Solution of 1^{st} order Homogeneous differential equation
1^{st} order homogeneous d.e can be written in the form = Q
Where both P and Q are function of and have same dimension.
Suppose p and Q hence the dimension
Then divide by x^{n} and use the substitution
Y = vx = v =
i.e
Example 1.
Solve =
Solution
Since all terms are of degree 2, i.e the equation is homogeneous
Let y = vx
= v +
And
=
= ()
=
V + =
= – v
=
=
=
= dx
= lnx + c
But v =
^{2} = ln x + c
= lnx + c
Y^{2 = }2x^{2} (lnx + c)
Example 2
Solve xy = x^{2 }+ y^{2}
Solution
All terms are of the order 2
xy = x^{2} + y^{2}……
Let y = vx
= v +
From the equation…
= +
But y = vx
= = +
= x^{2}
=
V + =
=
=
=
=
= lnx + A
But v =
^{2} = lnx + A
Y^{2} = 2x^{2} (lnx + A)
Example 3
Solve the following
(x^{2} + xy) = xy – y^{2}
Solution
Let y = vx
= v +
And
=
=
v + =
= v
=
=
=
[( dv + dv)] =
½ ( + ln v) = lnx + c
= ln (x^{2} V) + 2c
But v =
= ln (x^{2}) + 2c
– 2c = ln (x^{2}.)
Let ln A = 2c
+ ln A = ln xy
Xy = Ae^{x/y}
EXERCISE
Solve the differential equation
1. x^{2 }= y (x + y)
2. (x^{2} + y^{2}) = xy
First order exact differential equation
We know that
Now consider
+ y = e^{x}
Integrating both side w.r.t.x
= xy = e^{2} + c
Y = (e^{x} + c)
Y = (e^{x} + C)
2. Find the general solution of the following exact differential equation.
i)e^{x}y + e^{x }= 2
ii) cos x – ysin x = x^{2}
Solution
i) Given
(e^{x}y) = 2
Integrating both sides with respect to x
(e^{x}y) dx =
ii) Given
cosx – y sin x = x^{2}
(y cos x) = x^{2}
Integrating both side with respect to x
= (e^{x}y) dx =
Y cos x = + c
3ycosx = x3 + 3C
3ycosx = x3 + 4
3. + lny = x + 1
Soln
Given + lny = x + 1
Integrating both sides w.r.t x
(x ln y) dx =
Xln y = ½ ^{x2} + x + C
2xlny + x^{2} + 2x + 2c
2x lny + x^{2} + x + A
Integrating factors
Consider first order differential equation of the form
+ py =Q
Where p and Q are functions of x
Multiplying by integrating factor F both sides will make an exact equation
i.e F + Fpy + FQ…….i
= +
+ …..ii
Comparing (i) (ii)
+ Fpy = +
^{ }Fpy =
= Fp
dF = by separating of variables
F = is the required integrating factor
Examples
1. Solve + y = x^{3}
^{ }Solution
+ y = x^{3}
+ = x^{2}
Compare with + py = Q
P =
Then = dx
= ln x
If F = e^{ln x}
F = e^{ln x}
(xy) = x^{2} x
(xy) =
Xy = x^{4} + C
2. Solve
(x + 1) + y + (x + 1)^{2}
^{ }Soln
(x + 1) + y + (x + 1)^{2}
= + = x + 1
= + , y = x + 1
P =
F = e^{sp}
= e^{ln (x + 1)}
= x + 1
Y (x + 1) =
=
= x^{3 }+ x^{2 }+ x + c
= + c
Y = +
3. Solve (1 – x^{2}) – xy = 1
Solution – y =
P =
= dx
= ½ ln (1 – )
F = e ^{ln }(1^{ – ½}
^{ } = (1 – x^{2}) ^{½ }
Y = dx
= dx
Y =
4. tanx + y = e^{x} tan x
Solution
+ , y = e^{x}
P = = cot x
=
= ln
F = e^{ln }
=
Y = dx
Integrating by parts the R.H.S
= = e^{x} – dx
= e^{x} sin x – (e^{x} cos x + dx
= e^{x} – e^{x} – dx
2 sin x dx = e^{ x} ( –)
= ()
Y sin x = () + c
Bernoulli’s equation
This is a first order D.E of the form
+ p (x) y = Q (x) y ^{n}
p (x) and Q (x) are functions of x or constant
Steps
+ p (x) y = Q (x) y^{n}……i
Divide both sides by y^{n} gives
Y^{n } + p (x) y^{1 – n} = Q (x)…… (ii)
Let z = y^{1n}
(1 – n) y^{n }
Multiplying (ii) by (1 – n) both sides gives
(1 – n) y^{n } + (1 – n) p (x) y (1 – n) = (1 – n) Q (x)
+ (1 – n) p (x) y ^{(1 – n)} = (1 – n) Q (x)
+ p_{1} (x) y ^{(1 – n)} = Q_{1} (x)
p_{1} (x) and Q_{1} (x) are functions of x or constant
But z = y ^{(1 n)}
= + p_{1} (x) z = Q_{1} (x)……..ii
iii) is linear the use of integrating factor can be used
Example 1
Solve + = xy^{2}
Solution
Dividing both sides by gives
Let Z =
Multiply (ii) by 1 gives
But Z =
=
=
=
Then Z.F=
= 1
=
But Z =
Example 2
Solve
Solution
1^{st} Expressing the equation in the form
–
–
Let Z =
But Z =
=
=
Then
But Z =
Example 3
Solve
Solution
Expressing the equation in the form
Then dividing by
let Z =
Multiply (ii) by 2 gives
But Z =
Then,
But Z =
Exercise
Solve the following first order D.es
1.
2.
Second order Differential Equations
Second order differential equation is of the form of;
Where a, b, c are constant coefficients and is a given function of x
If the equation is homogeneous otherwise it is a nonhomogeneous
Which of the following are linear H.D.Es
1.
2.
3.
4.
5.
Characteristic (Auxiliary) Equation for H.DE
Consider a linear nonhomogeneous 2^{nd} order D.E
Let y = u and y=v be two solution of the equation
Where u and v a functions of x
Adding (i) & (ii) gives
Then,
and
Becomes
If y = u and y = v are the solutions of the equation
Suppose a = 0
i.e.
(separable)
(where is constant)
Take m for –k
– is the solution of dy +ky
Also will be the solution of the equation
if it satisfies the equation
If
The values will be and
If are two solution
Note that:
If the Auxiliary equation has
(i) Two real roots
(ii) Equal roots
The solution is
(iii) 2 Complex roots to the auxiliary equation
The solution is
Examples
Solve the following 2^{nd} order Des
1.
Solution
Auxiliary equation
2.
Solution
The auxiliary equation is
3.
Solution
Auxiliary equation is
In this case P=2 and q =
Exercise
Solve the following
1.
Non homogeneous 2^{nd} order D.E
Consider the equation
If

The general solution of (i) is given by
Note that
is called complementary function matas R.H.S Zero
Y = f (x) is called particular intergral makes R.H.S 0
i.e if assume
Examples
1. Solve
Solution
C.F solve L.H.S = 0
Assume
Substituting in the given equation gives
2. Solve
Solution
(i) C.F:
(ii)
Substituting into the given equation
Collecting like terms
Comparing L.H.S to R.H.s
General solution
3. Solve
Given that
Solution
C.F:
I: Assume
+2
Substitute into (i)
But y = 1,
1= A + B + 1
A + B = 0
A2B+3
………………………….(i)
B = 3
Substituting into (ii) gives
A + 3 = 0
A = 3
4. Solve
Solution
C.F:
Let y =
5
2
2C=1
Note that: if P.I is contained in the C.F multiply the assumed P.I by x and go on
Example
Solve
Solution
C.F:
P.I: Assume y =
Since is already contained
+B
Exercise
Solve the following
(i)
2^{nd} order equations which are reducible to 1^{st} order
Consider 2^{nd} order equation which can not be written in the form
i.e
Such equation will be solved by the substitution of: –
Where A and B are constants
Example
Solve
Solution
Let
dx
Example
Solve
Solution
Let p =
Example
Solution
Let
In p= ln Ay where c= ln A
P = Ay
i.e.
Note that:
– If P.I is contained in C.F, and C.F is of real roots Assume if
– If C.F is a distinct root, but one root is the same as that on P.I, assume if
– If , assume independently followed by then add to obtained a P.I
Applications of differential Equations
1. i. Population growth
The rate of population growth id differently proportional to the number of inhabitants present at specific time.
If N is the number of inhabitants at specific time.
When
i.e.
growth equation
If the population of a certain place doubles in 50 years. After how many years will the population treble and the assumption that the rate of increase is proportional to the number of inhabitant in the place.
Solution
Let N be the number of inhabitants at time t.
When
………………………..(i)
When t = 50 years, N=2
=
K =
Substituting into (i)
ln N = kt +ln N_{o}
Example
The population of Kenya is known to increase to a rate proportional to the number of people living there at any given time. In 1986 the population was 1.1 times that of 1984 and 1987 the population was 18,000,000. What was the population in 1984?
Solution
Let N be the number of people at time, t