Differential Equations MATH C241

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FIRST ORDER DIFFERENTIAL EQUATIONS
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In this lecture we discuss various methods of
solving first order differential equations. These
include

• Variables separable
• Homogeneous equations
• Exact equations
• Equations that can be made exact by multiplying
  by an integrating factor

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Non-linear
Linear
Integrating Factor
Integrating Factor
Separable
Homogeneous
Exact
Transform to Exact
Transform to separable
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Variables Separable
The first-order differential equation
(1)
is called separable provided that f(x,y) can be
written as the product of a function of x and a
function of y.
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Suppose we can write the above equation as
We then say we have separated the variables.
By taking h(y) to the LHS, the equation becomes
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Integrating, we get the solution as
where c is an arbitrary constant.
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Example 1. Consider the DE
Separating the variables, we get
Integrating we get the solution as
or
an arbitrary constant.
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Example 2. Consider the DE
Separating the variables, we get
Integrating we get the solution as
or
an arbitrary constant.
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Homogeneous equations
Definition A function f(x, y) is said to be
homogeneous of degree n in x, y if
for all t, x, y
Examples
is homogeneous of degree
2.
is homogeneous of degree
0.
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A first order DE
is called homogeneous if
are homogeneous functions of x and y of the same
degree.
This DE can be written in the form
where
is
clearly homogeneous of degree 0.
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The substitution y z x converts the given
equation into variables separable form and
hence can be solved. (Note that z is also a (new)
variable,) We illustrate by means of examples.
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Example 3. Solve the DE
That is
Let y z x. Hence we get
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or
Separating the variables, we get
Integrating we get
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We express the LHS integral by partial fractions.
We get
or
an arbitrary constant.
Noting z y/x, the solution is
c an arbitrary constant
or
c an arbitrary constant
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Working Rule to solve a HDE
1. Put the given equation in the form
2. Check M and N are Homogeneous function of the
same degree.
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4. Differentiate y z x to get
5. Put this value of dy/dx into (1) and solve the
equation for z by separating the variables.
6. Replace z by y/x and simplify.
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Example 4. Solve the DE
Let y z x. Hence we get
or
Separating the variables, we get
Integrating we get
cosec z cot z c x
where
and c an arbitrary constant.
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Non-homogeneous equations
We shall now see that some equations can be
brought to homogeneous form by appropriate
substitution.
Example 5 Solve the DE
That is
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We shall now put x uh, y vk where h, k
are constants ( to be chosen).
Hence the given DE becomes
We now choose h, k such that
Hence
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Hence the DE becomes
which is homogeneous in u and v.
Let v z u. Hence we get
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or
Separating the variables, we get
Integrating we get
i.e.
or
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Example 6 Solve the DE
That is
Now the previous method does not work as the lines
are parallel. We now put u 3x 2y.
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The given DE becomes
or
Separating the variables, we get
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Integrating, we get
i.e.
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EXACT DIFFERENTIAL EQUATIONS
A first order DE
is called an exact DE if there exits a function
f(x, y) such that
Here df is the total differential of f(x, y)
and equals
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Hence the given DE becomes df 0
Integrating, we get the solution as
f(x, y) c, c an arbitrary constant
Thus the solution curves of the given DE are the
level curves of the function f(x, y) .
Example 8 The DE
is exact as it is d (xy) 0
Hence the solution is x y c
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Example 7 The DE
is exact as it is d (x2 y2) 0
Hence the solution is x2 y2 c
Example 9 The DE
is exact as it is
Hence the solution is
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Test for exactness
Suppose
is exact. Hence there exists a function f(x, y)
such that
Hence
Assuming all the 2nd order mixed derivatives of
f(x, y) are continuous, we get
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Thus a necessary condition for exactness is
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We saw a necessary condition for exactness is
We now show that the above condition is also
sufficient for M dx N dy 0 to be exact. That
is, if
then there exists a function f(x, y) such that
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Integrating
partially w.r.t. x, we get
. ()
where g(y) is a function of y alone
We know that for this f(x, y),
Differentiating () partially w.r.t. y, we get
N gives
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()
or
We now show that the R.H.S. of () is
independent of x and thus g(y) (and so f(x, y))
can be found by integrating () w.r.t. y.
0
Q.E.D.
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Note (1) The solution of the exact DE d f 0 is
f(x, y) c.
Note (2) When the given DE is exact, the
solution f(x, y) c is found as we did in the
previous theorem. That is, we integrate M
partially w.r.t. x to get
We now differentiate this partially w.r.t. y and
equating to N, find g? (y) and hence g(y).
The following examples will help you in
understanding this.
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Example 8 Test whether the following DE is
exact. If exact, solve it.
Here
Hence exact.
Now
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Differentiating partially w.r.t. y, we get
Hence
Integrating, we get
(Note that we have NOT put the arb constant )
Hence
Thus the solution of the given d.e. is
c an arb const.
or
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Example 9 Test whether the following DE is
exact. If exact, solve it.
Here
Hence exact.
Now
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Differentiating partially w.r.t. y, we get
Hence
Integrating, we get
(Note that we have NOT put the arb constant )
Hence
Thus the solution of the given d.e. is
or
c an arb const.
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In the above problems, we found f(x, y) by
integrating M partially w.r.t. x and then
equated
We can reverse the roles of x and y. That is we
can find f(x, y) by integrating N partially
w.r.t. y and then equate
The following problem illustrates this.
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Example 10 Test whether the following DE is
exact. If exact, solve it.
Here
Hence exact.
Now
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Differentiating partially w.r.t. x, we get
gives
Integrating, we get
(Note that we have NOT put the arb constant )
Hence
Thus the solution of the given d.e. is
or
c an arb const.
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Integrating Factors
The DE
is NOT exact
but becomes exact when
as it becomes
multiplied by
i.e.
is an Integrating Factor of the given DE
We say
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Definition If on multiplying by ?(x, y), the DE
becomes an exact DE, we say that ?(x, y) is an
Integrating Factor of the above DE
are all integrating factors of
the non-exact DE
We give some methods of finding integrating
factors of an non-exact DE
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Problem Under what conditions will the DE
have an integrating factor that is a function of
x alone ?
Solution. Suppose ? h(x) is an I.F.
Multiplying by h(x) the above d.e. becomes
Since () is an exact DE, we have
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i.e.
or
or
or
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Hence if
is a function of x alone, then
is an integrating factor of the given DE
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Rule 2 Consider the DE
, a function of y alone,
If
then
is an integrating factor of the given DE
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Problem Under what conditions will the DE
have an integrating factor that is a function of
the product z x y ?
Solution. Suppose ? h(z) is an I.F.
Multiplying by h(z) the above d.e. becomes
Since () is an exact DE, we have
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i.e.
or
or
or
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Hence if
is a function of z x y alone, then
is an integrating factor of the given DE
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Example 11 Find an I.F. for the following DE and
hence solve it.
Here
Hence the given DE is not exact.
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Now
a function of x alone. Hence
is an integrating factor of the given DE
Multiplying by x2, the given DE becomes
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which is of the form
Note that now
Integrating, we easily see that the solution is
c an arbitrary constant.
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Example 12 Find an I.F. for the following DE and
hence solve it.
Here
Hence the given DE is not exact.
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Now
a function of y alone. Hence
is an integrating factor of the given DE
Multiplying by sin y, the given DE becomes
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which is of the form
Note that now
Integrating, we easily see that the solution is
c an arbitrary constant.
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Example 13 Find an I.F. for the following DE and
hence solve it.
Here
Hence the given DE is not exact.
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Now
a function of z x y alone. Hence
is an integrating factor of the given DE
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Multiplying by
the given DE becomes
which is of the form
Integrating, we easily see that the solution is
c an arbitrary constant.
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Problem Under what conditions will the DE
have an integrating factor that is a function of
the sum z x y ?
Solution. Suppose ? h(z) is an I.F.
Multiplying by h(z) the above DE becomes
Since () is an exact DE, we have
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i.e.
or
or
or
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Hence if
is a function of z x y alone, then
is an integrating factor of the given DE
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Linear Equations

• A linear first order equation is an equation that
  can be expressed in the form

where a1(x), a0(x), and b(x) depend only on the
independent variable x, not on y.
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We assume that the function a1(x), a0(x), and
b(x) are continuous on an interval and that
a1(x) ? 0on that interval. Then, on dividing by
a1(x), we can rewrite equation (1) in the
standard form
where P(x), Q(x) are continuous functions on the
interval.
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Lets express equation (2) in the differential
form
If we test this equation for exactness, we find
Consequently, equation(3) is exact only when P(x)
0. It turns out that an integrating factor ?,
which depends only on x, can easily obtained the
general solution of (3).
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Multiply (3) by a function ?(x) and try to
determine ?(x) so that the resulting equation
is exact.
We see that (4) is exact if ? satisfies the DE
Which is our desired IF
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In (2), we multiply by ?(x) defined in (6) to
obtain
We know from (5)
and so (7) can be written in the form
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Integrating (8) w.r.t. x gives
and solving for y yields
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Working Rule to solve a LDE
1. Write the equation in the standard form
2. Calculate the IF ?(x) by the formula
3. Multiply the equation in standard form by
?(x) and recalling that the LHS is just
obtain
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4. Integrate the last equation and solve for y by
dividing by ?(x).
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Ex 1. Solve
Solution - Dividing by x cos x, throughout, we
get
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yields
Multiply by
Integrate both side we get
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Problem (2g p. 62) Find the general solution of
the equation
Ans.
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The usual notation implies that x is
independent variable y the dependent variable.
Sometimes it is helpful to replace x by y and y
by x work on the resulting equation. When
diff equation is of the form
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Q. 4 (b) Solve
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