Download Nonlinear Second Order Differential Equations

Pertemuan V
Nonlinear Second Order Differential Equations
In general, little is known about nonlinear second order differential equations
,
but two cases are worthy of discussion:
(1) Equations with the y missing
Let v = y'. Then the new equation satisfied by v is
This is a first order differential equation. Once v is found its integration
gives the function y.
Example 1: Find the solution of
Solution: Since y is missing, set v=y'. Then, we have
This is a first order linear differential equation. Its resolution gives
Since v(1) = 1, we get
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Consequently, we have
Since y' = v, we obtain the following equation after integration
The condition y(1) = 2 gives
Therefore, we have
Note that this solution is defined for x > 0.
(2) Equations with the x missing
Let v = y'. Since
we get
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This is again a first order differential equation. Once v is found then we
can get y through
which is a separable equation. Beware of the constants solutions.
Example 2: Find the general solution of the equation
Solution: Since the variable x is missing, set v=y'. The formulas above lead to
This a first order separable differential equation. Its resolution gives
Since
we get y' = 0 or
Since this is a separable first order differential equation, we get, after
resolution,
,
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where C and
are two constants. All the solutions of our initial
equation are
Note that we should pay special attention to the constant solutions when
solving any separable equation. This may be source of mistakes...
Homogeneous Linear Equations with Constant Coefficients
A second order homogeneous equation with constant coefficients is written as
where a, b and c are constant. This type of equation is very useful in many
applied problems (physics, electrical engineering, etc..). Let us summarize the
steps to follow in order to find the general solution:
(1) Write down the characteristic equation
This is a quadratic equation. Let r1 and r2 be its roots ( we have
;
(2) If r1 and r2 are distinct real numbers (this happens if b2 – 4 ac > 0 ), then
the general solution is
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(3) If r1 and r2 (which happens if b2 – 4ac = 0 ), then the general solution
is
(4) If r1 and r2 are complex numbers (which happens if b2 – 4 ac < 0 ), then
the general solution is
where
,
that is,
Example: Find the solution to the IVP
Solution: Let us follow the steps:
1 Characteristic equation and its roots
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Since 4-8 = -4<0, we have complex roots
. Therefore,
and
;
2 General solution
;
3 In order to find the particular solution we use the initial conditions to
determine and . First, we have
.
Since
, we get
From these two equations we get
,
which implies
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Summary
Equation: ay
Auxiliary equation: ak2 + bk + c = 0
Solutions to the Auxiliary
Equation
General Solution
Distinct real roots k1, k2
Equal real roots k1 = k2 (= k)
Complex roots k1 = A + iB and k2 =
A - iB
y = eA[Pcos(Bx) + Qsin(Bx)]
Example 1. Find the general solution of the following D.E.'s:
a. y” + 4y’ + 3y = 0
b. y” + 4y’ + 4y = 0
c. y” + 4y’ + 5y = 0
The two constants appearing in the complementary function are
usually determined by knowing two conditions associated with the
problem. If the conditions both apply at the same value of x, (usually
x = 0), the problem is called an initial value problem.
Example 2. Solve the initial value problem y + 4y + 3y = 0 given y = 1 and
y = 3 when x = 0.
Example: Find the general solution of
Solution: Let us follow these steps:
(1) Characteristic equation
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Its roots are the complex numbers
In the analytical form, these roots are
;
(2) Independent set of solutions
(2.1) The complex roots
and
generate the two solutions
;
(2.2) The complex roots
generate the two solutions
;
(3) The general solution is
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Soal – soal
1. y″ - 3y′ + 2 y = 0
2. y” + y’ – 30 y = 0
ans . y = C1 ex + C2 e2x
3. y”” – 4 y’” + 8 y” – 8 y’ + 4 y = 0
ans. Y = C1 e5x + C2e-6x
4. y”” + y” = 0 , y(0) = y’(0)
ans y = ex( ( C1 + C2x) cos x + (
= y”(0), y’”(0) = 1
C3 + C4) sinx )
ans
y = x – sin x
5 . y’” + 6y” + 12 y’ + 8y = 0
y(0) = y’(0) , y”(0) = 2
ans y = x2 e-2
Terima kasih
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