Linear Differential Equations

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Linear Differential Equations
In this lecture we discuss the methods of solving
first order linear differential equations and
reduction of order.
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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 a1(x) ? 0
on 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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Example 1. Solve
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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A first order equation that can be written in the
form
is known as Bernoullis equation .
It is clearly linear when n0 or 1.
For n gt 1, it can be reduced to a linear equation
by change of variable
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Divide equation () be yn yields
Taking z y1-n, we find that
and so () becomes
Since 1/(1-n) is just a constant,
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Example Solve the DE
This is a Bernoulli equation with n 3,
P(x) -5, and Q(x) -5x/2
We first divide by y3 to obtain
Next we make the substitution z y-2
Since dz/dx -2y-3 dy/dx, then above equation
reduces to
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or
which is linear. Hence
Substituting z y-2 gives the solution
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Reduction of Order
A general second order DE has the form
In this section we consider two special types of
second order equations that can be solved by
first order methods.
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Type A Dependent variable missing.
When y is not explicitly present, (1) can be
written as
Then (2) transforms into
If we can solve (3) for p, then (2) can be solved
for y.
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Problem (1 (g) p. 65) Solve
The variable y is missing
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which is linear,
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Problem (1b p. 65) Solve the DE
Ans.
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Type B Dependent variable missing.
When x is not explicitly present, then (1) can be
written as
Then (4) becomes
If we can solve (5) for p, then (4) can be solved
for y.
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Problem (1(e), P 65) Solve
Here x is not explicitly present
? (1) can be writen as,
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Problem (2b p. 65) Find the specified particular
solution of the DE
Ans.