MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture14_D2_prt.pdf · Suppose emx is...
Transcript of MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture14_D2_prt.pdf · Suppose emx is...
MA 108 - Ordinary Differential Equations
Santanu Dey
Department of Mathematics,Indian Institute of Technology Bombay,
Powai, Mumbai [email protected]
March 27, 2014
Santanu Dey Lectures 14
Outline of the lecture
Second order vs nth order
Method of variation of parameters
Method of undetermined coefficients
Santanu Dey Lectures 14
Abel’s formula
We have seen that the Wronskian of any two solutions y1(x), y2(x)of y ′′ + p(x)y ′ + q(x)y = 0 is given by
W (y1, y2)(x) = W (y1, y2)(x0)e−
∫ xx0
p(t)dt,
for any x0, x ∈ I . Note that the formula depends only on p(x) andnot on q(x). What should be the n-th order analogue?
Theorem
If y1, y2, . . . , yn are solutions of
y (n) + p1(x)y (n−1) + . . .+ pn(x)y = 0,
then,
W (y1, . . . , yn)(x) = W (y1, . . . , yn)(x0)e−
∫ xx0
p1(t)dt .
Santanu Dey Lectures 14
Proof
Proceeding as in the proof for second order case, we need to show
W ′ = −p1(x)W .
Notice that the derivative of∣∣∣∣ y1 y2y ′1 y ′2
∣∣∣∣is ∣∣∣∣ y ′1 y ′2
y ′1 y ′2
∣∣∣∣+
∣∣∣∣ y1 y2y ′′1 y ′′2
∣∣∣∣ ,which is ∣∣∣∣ y1 y2
y ′′1 y ′′2
∣∣∣∣ .In this, we had substituted y ′′i = −p1(x)y ′i − p2(x)yi to concludethe claim.
Santanu Dey Lectures 14
Proof
The derivative of W (y1, y2, . . . , yn) is the sum of n determinantswith derivative being taken in the first row in the first one, in thesecond row in the second one, etc. Except the last one, all vanishbecause two rows are identical. In the last one, make thesubstitution
y(n)i = −p1(x)y
(n−1)i − p2(x)y
(n−2)i − . . .− pn(x)yi .
Expand this into sum of n determinants. Once again, all but onevanish. The non-vanishing one gives
−p1(x) ·W (y1, y2, . . . , yn).
Thus, for solutions of linear homogeneous DE’s, the Wronskian iseither never zero or identically zero on I !
Santanu Dey Lectures 14
Basis of solutions & General solution (2nd order) - RECALL
If p(x) and q(x) are continuous on an open interval I , then
y ′′ + p(x)y ′ + q(x)y = 0
has a basis of solutions y1, y2 on I .
y(x) = C1y1(x) + C2y2(x)
is the general solution of the DE.That is, every solution y = Y (x) of the DE has the form
Y (x) = C1y1(x) + C2y2(x),
where C1 and C2 are arbitrary constants.
Santanu Dey Lectures 14
Basis of solutions & General solution (nth order)
If p1(x), . . . , pn(x) are continuous on an open interval I , then
y (n) + p1(x)y (n−1) + . . .+ pn(x)y = 0
has a basis of solutions y1, . . . , yn on I .
y(x) = C1y1(x) + C2y2(x) + · · ·+ Cnyn(x)
is the general solution of the DE.That is, every solution y = Y (x) of the DE has the form
Y (x) = C1y1(x) + · · ·+ Cnyn(x),
where C1, · · · ,Cn are arbitrary constants. (Prove this ! )
Santanu Dey Lectures 14
Non-homogeneous 2nd order Linear ODE’s - RECALL
Consider the non-homogeneous DE
y ′′ + p(x)y ′ + q(x)y = r(x)
where p(x), q(x), r(x) are continuous functions on an interval I .Let yp(x) be any solution of
y ′′ + p(x)y ′ + q(x)y = r(x)
and y1(x), y2(x) be a basis of the solution space of the correspondinghomogeneous DE.Then the set of solutions of the non-homogeneous DE is
{c1y1(x) + c2y2(x) + yp(x) | c1, c2 ∈ R}.
Summary:
In order to find the general solution of a non-homogeneous DE, we need to
get the general solution of the corresponding homogeneous DE.
get one particular solution of the non-homogeneous DE
Santanu Dey Lectures 14
Non-homogeneous nth order Linear ODE’s
Consider the non-homogeneous DE
y (n) + p1(x)y (n−1) + . . .+ pn(x)y = r(x)
where p1(x), · · · , pn(x), r(x) are continuous functions on an interval I .Let yp(x) be any solution of
y (n) + p1(x)y (n−1) + . . .+ pn(x)y = r(x)
and y1(x), · · · , yn(x) be a basis of the solution space of thecorresponding homogeneous DE.Then the set of solutions of the non-homogeneous DE is
{c1y1(x) + · · ·+ cnyn(x) + yp(x) | c1, · · · , cn ∈ R}.
Summary:
In order to find the general solution of a non-homogeneous DE, we need to
get the general solution of the corresponding homogeneous DE.
get one particular solution of the non-homogeneous DE
Santanu Dey Lectures 14
General Solution of Homogeneous Equations withConstant Coefficients
Considery (n) + p1y
(n−1) + . . .+ pny = 0
where p1, · · · , pn are in R; that is, an nth order homogeneous linearODE with constant coefficients.Suppose emx is a solution of this equation. Then,
mnemx + p1mn−1emx + · · ·+ pne
mx = 0,
and this implies
mn + p1mn−1 + · · ·+ pn = 0.
This is called the characteristic equation or auxiliary equation ofthe linear homogeneous ODE with constant coefficients.
Santanu Dey Lectures 14
This polynomial of degree n has n zeros say m1, · · · ,mn, some ofwhich may be equal and hence the characteristic polynomial canbe written in the form
(m −m1)(m −m2) · · · (m −mn).
Depending on the the nature of these zeros (real & unequal, real &equal, complex ), we write down the general solution of thehomogeneous DE in a similar way as we did for 2nd order equations.
Santanu Dey Lectures 14
Example 1
Find a basis of solutions and general solution of the DE:
y (3) − 7y ′ + 6y = 0.
1 The auxiliary equation is
m3 − 7m + 6 = 0.
2 The roots are 1, 2,−3.
3 Hence a basis for solutions is {ex , e2x , e−3x} Why?.
4 Thus, the general solution is of the form
c1ex + c2e
2x + c3e−3x ,
where c1, c2, c3 ∈ R.
Santanu Dey Lectures 14
Example 2
Find the general solution of the DE:
L(y) = (D3 − D2 − 8D + 12)(y) = 0.
1 The auxiliary equation is
m3 −m2 − 8m + 12 = (m − 2)2(m + 3) = 0.
2 The roots are 2, 2,−3.
3 Hence a basis for solutions is {e2x , xe2x , e−3x}. Why?
4 Thus, the general solution is of the form
c1e2x + c2xe
2x + c3e−3x ,
where c1, c2, c3 ∈ R.
Santanu Dey Lectures 14
Examples 3 & 4
Find the general solution of the DE:
L(y) = (D6 + 2D5 − 2D3 − D2)(y) = 0.
The general solution is
c1 + c2x + c3ex + c4e
−x + c5xe−x + c6x
2e−x ,
with ci ∈ R.Find the general solution of
y (5) − 9y (4) + 34y (3) − 66y (2) + 65y ′ − 25y = 0.
The characteristic polynomial is
(m − 1)(m2 − 4m + 5)2.
The roots are1, 2± ı, 2± ı.
Hence, the general solution is
y = c1ex + c2e
2x cos x + c3xe2x cos x + c4e
2x sin x + c5xe2x sin x ,
where ci ∈ R.Santanu Dey Lectures 14
Variation of Parameters- finding yp
Recall the case n = 2: We replaces c1, c2 in the general solution ofthe associated homogeneous equation by functions v1(x), v2(x), sothat
y(x) = v1(x)y1(x) + v2(x)y2(x)
is a solution of
y ′′ + p(x)y ′ + q(x)y = r(x).
Now,y ′ = v1y
′1 + v2y
′2 + v ′1y1 + v ′2y2.
We also demandedv ′1y1 + v ′2y2 = 0.
Santanu Dey Lectures 14
Variation of Parameters
Then we substituted y , y ′, y ′′ in the given non-homogeneous ODE,and rearranged terms, to obtain:
v ′1y′1 + v ′2y
′2 = r(x).
Together with,v ′1y1 + v ′2y2 = 0,
this yielded :
v ′1 =
∣∣∣∣ 0 y2r(x) y ′2
∣∣∣∣W (y1, y2)
, v ′2 =
∣∣∣∣ y1 0y ′1 r(x)
∣∣∣∣W (y1, y2)
,
and hence
v1 = −∫
y2r(x)
W (y1, y2)dx , v2 =
∫y1r(x)
W (y1, y2)dx .
Thus,
y = v1y1 + v2y2 = y2
∫y1r(x)
W (y1, y2)dx − y1
∫y2r(x)
W (y1, y2)dx .
Santanu Dey Lectures 14
Variation of Parameters
For general n, we proceed exactly the same way.
c1y1 + c2y2 → c1y1 + c2y2 + . . .+ cnyn
y = v1y1 + v2y2 → y = v1y1 + v2y2 + . . .+ vnyn
y ′ = v1y′1 + v2y
′2 + v ′1y1 + v ′2y2
→
y ′ = v1y′1 + . . .+ vny
′n + v ′1y1 + . . .+ v ′nyn
Santanu Dey Lectures 14
Variation of Parameters
Demandv ′1y1 + v ′2y2 = 0
→
Demandv ′1y1 + . . .+ v ′nyn = 0.
Takey ′′ = v1y
′′1 + . . .+ vny
′′n + v ′1y
′1 + . . .+ v ′ny
′n
and demandv ′1y′1 + . . .+ v ′ny
′n = 0.
. . . . . .
Santanu Dey Lectures 14
Variation of Parameters
Take
y (n−1) = v1y(n−1)1 + . . .+ vny
(n−1)n + v ′1y
(n−2)1 + . . .+ v ′ny
(n−2)n
and demandv ′1y
(n−2)1 + . . .+ v ′ny
(n−2)n = 0.
Santanu Dey Lectures 14
Variation of Parameters
Substituting y , y ′, y ′′, → Substituting y , y ′, . . . , y (n),
rearranging → rearranging
and using L(y1) = L(y2) = 0 → and using L(y1) = . . . = L(yn) = 0
get v ′1y′1 + v ′2y
′2 = r(x) → get
v ′1y(n−1)1 + . . .+ v ′ny
(n−1)n = r(x).
Santanu Dey Lectures 14
Variation of Parameters
Thus, [y1 y2y ′1 y ′2
] [v ′1v ′2
]=
[0
r(x)
]takes the shape
y1 y2 · yny ′1 y ′2 · y ′n· · · ·
y(n−1)1 y
(n−1)2 · y
(n−1)n
v ′1v ′2·v ′n
=
00·
r(x)
.Use Cramer’s rule to solve for
v ′1, v′2, . . . , v
′n,
and thus getv1, v2, . . . , vn,
and formy = v1y1 + v2y2 + . . .+ vnyn.
Santanu Dey Lectures 14
Example 5
Solvey (3) − y (2) − y (1) + y = r(x).
Characteristic polynomial for the homogeneous equation is
m3 −m2 −m + 1 = (m − 1)2(m + 1).
Hence, a basis of solutions is
{ex , xex , e−x}.
We need to calculate W (x). Use Abel’s formula:
W (x) = W (0)e−∫ x0 p1(t)dt = W (0) · ex .
Santanu Dey Lectures 14
Now,
W (x) =
∣∣∣∣∣∣ex xex e−x
ex ex + xex −e−xex 2ex + xex e−x
∣∣∣∣∣∣ .Thus,
W (0) =
∣∣∣∣∣∣1 0 11 1 −11 2 1
∣∣∣∣∣∣ = 4.
Hence,W (x) = 4ex .
Santanu Dey Lectures 14
Variation of Parameters
Let
W1(x) =
∣∣∣∣∣∣0 xex e−x
0 ex + xex −e−xr(x) 2ex + xex e−x
∣∣∣∣∣∣ = −r(x)(2x + 1).
Similarly,W2(x) = 2r(x), W3(x) = r(x)e2x .
Therefore,
y(x) = ex∫ x
0
−r(t)(2t + 1)
4etdt+xex
∫ x
0
2r(t)
4etdt+e−x
∫ x
0
r(t)e2t
4etdt.
Santanu Dey Lectures 14
Undetermined Coefficients - Example 6
Find a particular solution of
y (4) + 2y ′′ + y = 3 sin x − 5 cos x .
What’s the general solution of the homogeneous equation?The auxiliary equation is
m4 + 2m2 + 1 = (m2 + 1)2,
and therefore a basis of Ker L is
{cos x , sin x , x cos x , x sin x}.So the candidate is
y = x2(a cos x + b sin x).
Substituting, we get:
−8a cos x − 8b sin x = 3 sin x − 5 cos x .
Thus,
y(x) = x2(5
8cos x − 3
8sin x).
Santanu Dey Lectures 14
Example 7
Find the candidate solution for
y (4) − y (3) − y ′′ + y ′ = x2 + 4 + x sin x .
Note that the auxiliary/ characteristic equation is
m4 −m3 −m2 + m = m(m − 1)2(m + 1).
Thus a basis of Ker L is {1, ex , xex , e−x}. What’s the candidatefor solution?x2 + 4 on the right suggests ax2 + bx + c , but we should modifythis to
x(ax2 + bx + c),
since a constant is a solution of the homogeneous equation.x sin x suggests
(αx + β) cos x + (γx + δ) sin x .
Do we need to modify this? No. So our final candidate is
x(ax2 + bx + c) + (αx + β) cos x + (γx + δ) sin x .
Santanu Dey Lectures 14
Example 8
Find the candidate solution for
y (4) − 2y ′′ + y = x2ex + e2x .
Note that the AE is
m4 − 2m2 + 1 = (m − 1)2(m + 1)2.
So a basis of Ker L is {ex , xex , e−x , xe−x}.e2x on the right suggests
ae2x ,
and no modification required.x2ex initially suggests (bx2 + cx + d)ex , but this should bemodified to
x2(bx2 + cx + d)ex .
So final candidate would be
ae2x + x2(bx2 + cx + d)ex .
Santanu Dey Lectures 14
Review Question 1
For the non-homogeneous equation y ′′ − 5y ′ + 6y = x cos x , theform of yp using the method of undetermined coefficients is————-.(a) A1x cos x + B1x sin x(b) (A0 + A1x) cos x + (B0 + B1x) sin x(c) x((A0 + A1x) cos x + (B0 + B1x) sin x)(d) None of the above.
Solution : 4 (b)
Santanu Dey Lectures 14