MA 1506 Mathematics II Tutorials Ngo Quoc Anh Question 1 Question 2 Question 3 Question 4 1/15 MA 1506 Mathematics II Tutorial 3 Second order differential equations Groups: B03 & B08 February 8, 2012 Ngo Quoc Anh Department of Mathematics National University of Singapore
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MA 1506Mathematics II
Tutorials
Ngo Quoc Anh
Question 1
Question 2
Question 3
Question 4
1/15
MA 1506 Mathematics II
Tutorial 3Second order differential equations
Groups: B03 & B08February 8, 2012
Ngo Quoc AnhDepartment of Mathematics
National University of Singapore
MA 1506Mathematics II
Tutorials
Ngo Quoc Anh
Question 1
Question 2
Question 3
Question 4
2/15
Question 1: General solution of homogeneous secondorder linear DE with constant coefficients y′′+ ay′+ by = 0
General solution can be found using the so-calledcharacteristic equation. Following is the method of solving.
Method of solving
By solving the characteristic equation λ2 + aλ+ b = 0, thereare three cases depending on the sign of a2 − 4b.
. If a2 − 4b > 0, then the characteristic equation has twodistinct real roots, say λ1 and λ2. The general solutionof the associated ODE is
yh = c1eλ1x + c2e
λ2x.
. If a2 − 4b = 0, then the characteristic equation has areal double root λ1 = λ2 = −a
Question 1: General solution of homogeneous secondorder linear DE with constant coefficients y′′+ ay′+ by = 0
Method of solving (cont’)
. If a2 − 4b < 0, then the characteristic equation has twocomplex roots, say λ1 and λ2, where
λ1,2 =−a±
√a2 − 4b
2= −a
2±√4b− a22︸ ︷︷ ︸w
i.
The general solution of the associated ODE is
yh =(c1 cos(wx) + c2 sin(wx)
)e−
a2x.
As you may see that general solution y of a 2nd order ODEbasically depends on two parameters (c1 and c2 as shownabove). (What happens to the 1st order ODEs?)
In order to specify these numbers, we need two initialconditions involving, for example, y(x0) and y′(x0).
Question 2: Finding yp: The method of undeterminedcoefficients
This is also known as the lucky guess method. In order tofind the particular integral, we need to guess its form, withsome coefficients left as variables to be solved for.
Below is a table of some typical functions r(x) and thesolution yp to guess for them.
Form of r(x) Form for ypkeax Ceax
kxn, n > 0 Knxn + · · ·+K1x+K0
k cos(ax) or k sin(ax) K cos(ax) +M sin(ax)keax cos(bx) or keax sin(bx) eax(K cos(bx) +M sin(bx))
More generally, with either(∑n
i=1 kixi)eax cos(bx) or(∑n
i=1 kixi)eax sin(bx), the function yp can be found using
yp = eax(P cos(bx) +Q sin(bx))
where P and Q are usually two polynomials of order n.
Question 3: Finding yp: The method of variationparameters
Suppose we have already solved the homogeneous equationy′′+ py′+ q = 0 and written the solution as yh = c1y1+ c2y2where y1 and y2 are linearly independent solutions.
In view of the method of variation parameters, we look foryp of the nonhomogeneous equation y′′ + py′ + q = r of theform
yp(x) = u1(x)y1(x) + u2(x)y2(x).
Method of solving
We first have
y′p = u′1y1 + u′2y2 + u1y′1 + u2y
′2.
Since we have two arbitrary functions u1 and u2, we imposetwo conditions in order to find u1 and u2. While onecondition is that yp verifies the ODE, we can choose theother condition so as to simplify our calculations
Recall that the ODE y′′ = F (y) can be solved using twosteps. In the first step, we write it as
d
dy
(1
2y′
2)
= F (y)
and then integrate both sides w.r.t. y to have
y′ = ±
√2
∫F (y)dy.
In the second step, since the preceding ODE is separable, wecan solve it to get y. It is important to note that generallyyou have to keep the constant C after calculating
∫F (y)dy.
Why? In addtion, the sign of y′ is also important.
For simplicity, we denote R = 150× 109. We now considerthe given ODE, see for a background,
Since r denotes the distance,r′ is clearly the speed of theEarth. Therefore, we initiallyhave r′(0) = 0. Besides, thereholds r(0) = R.
R
2R/3
Using these conditions, we find that C = − 1R . In order to
clarify the sign of r′, we observe that r′ < 0. Why? (This isbased on the fact that the speed is increasing and thedistance r is decreasing.) Our separable ODE is now just