Unformatted text preview: MEC702 Wk 5 Lect 2 Existence and Uniqueness of Solutions
Theorem. (Existence of a General Solution) If p(x) and q(x) are continuous on an open interval I ,
then
y 00 + p(x)y 0 + q(x)y = 0 has a general solution on I . Theorem. (A General Solution encompasses all solutions) If
y 00 + p(x)y 0 + q(x)y = 0 has continuous coecients p(x) and q(x) on some open interval I , then every solution y = V (x) of the
ODE on I is of the form
V (x) = c1 y1 + c2 y2 where y1 and y2 is any basis of solutions of the ODE on I and c1 , c2 are suitable constants. 1 NonHomogeneous Linear ODEs
We now advance from homogeneous to nonhomogeneous linear ODEs.
Consider the secondorder nonhomogeneous linear ODE
y 00 + p(x)y 0 + q(x)y = r(x) where r(x) 6= 0. Denition. (General Solution, Particular Solution) A general solution of the nonhomogeneous ODE
on an open interval I is a solution of the form y(x) = yh (x) + yp (x) ; where
yh (x) = c1 y1 + c2 y2 is a general solution of the homogeneous ODE on I and
yp (x) is any solution of nonhomogeneous ODE on I containing no arbitrary constants.
A particular solution of the nonhomogeneous ODE on I is a solution obtained from y(x) by assigning
specic values to the arbitrary constants c1 and c2 in yh . Theorem. (Relation of Solutions)
1. The sum
y(x) = yh (x) + yp (x) of a solution yp (x) of the nonhomogeneous linear ODE on some open interval I and the solution
yh (x) of the homogeneous linear ODE (with r(x) = 0) on I is a solution of the nonhomogeneous
linear ODE.
2. The dierence of two solutions of a nonhomogenous linear ODE on I is a solution of the homogeneous case on I. Theorem. (A General Solution of a Nonhomogeneous ODE Includes all Solutions) If the coecients
p(x), q(x) and the function r(x) in
y 00 + p(x)y 0 + q(x)y = r(x) are continuous on some open interval I , then every solution of the ODE is obtained by assigning suitble
values to c1 and c2 in a general solution
y(x) = yh (x) + yp (x) of the ODE on I . 2 Method of Undetermined Coecients
The theorems above suggests the following: To solve the nonhomogeneous linear ODE
y 00 + p(x)y 0 + q(x)y = r(x) or an IVP, we have to solve the homogeneous ODE and nd the any solution yp of the ODE so as to
obtain a general solutions.
To nd a solution yp , one method to use is the Method of Undetermined Coecients. This
method is suitable for linear ODEs with constant coecients a and b
y 00 + ay 0 + by = r(x) when r(x) is an exponential function, a power of x, a cosine or sine, or sums or products of such
functions. These functions have derivatives similar to r(x) itself. This gives the good idea. We choose
a form for yp similar to r(x), but with unknown coecients to be determined by substituting that yp
and its derivatives into the ODE. The table and the rules are given below. Choice Rules for the Method of Undetermined Coecients 1. Basic Rule. If r(x) in ODE is one of the functions in the rst column in the Table below,
choose yp in the same line and determine its undetermined coecients by substituting yp and its
derivatives into the ODE.
2. Modication Rule. If a term in your choice for yp happend to be a solution of the homogeneous
ODE corresponding to the nonhomogeneous ODE, multiply this term by x (or x2 if this solution
corresponds to a double root of the characteristic equation of the homogeneous ODE.) 3. Sum Rule. If r(x) is a sum of functions in the rst column of the Table below, choose for yp
the sum of the functions in the corresponding lines of the second column. Table for the Method of Method of Undetermined Coecients
Term in r(x)
γx ke
kx , (n = 0, 1, . . .)
k cos ωx
n k sin ωx
keαx cos ωx
keαx sin ωx Choice for yp (x) Ceγx
Kn x + Kn−1 x
+ · · · + K1 x + K0
n n−1 K cos ωx + M sin ωx
eαx (K cos ωx + M sin ωx) 3 Example. Application of the Basic Rule.
Solve the IVP y 00 + y = 0.001x2 , y 0 (0) = 1.5 . y(0) = 0, Solution.
Step 1. General Solution of the homogeneous ODE.
The ODE y 00 + y = 0 has the general solution yh = A cos x + B sin x. Step 2. Solution yp of the nonhomogeneous ODE. We rst try yp = Kx2 . Then yp00 = 2K . By substitution, y 00 + y = 0.001x2 = 2K + Kx2 = 0.001x2 . For this to hold for all x, we see that K = 0, a contradiction.
The second line in the Table suggests the choice
yp = K2 x2 + K1 x + K0 . Then for the ODE
yp00 + yp = 0.001x2 (2K2 ) + (K2 x2 + K1 x + K0 )
K2 x2 + (K1 x) + (2K2 + K0 ) = 0.001x2 = 0.001x2 Equating the coecient of x2 , x and x0 on both sides, we have K2 = 0.001, K1 = 0 and
2K2 + K0 = 0. Hence This gives
and K2 = 0.001 K1 = 0 K1 = −0.002. yp = 0.001x2 − 0.002
y = yh + yp = A cos x + B sin x + 0.001x2 − 0.002. Step 3. Solution of the IVP. Applying the initial condition, we have
y(0) = 0 = A − 0.002 =⇒ A = 0.002. By dierentiation and from the second initial condition,
y 0 = yh0 + yp0 = −A sin x + B cos x + 0.002x and y 0 (0) = B = 1.5. This gives the answer
y = 0.002 cos x + 1.5 sin x + 0.001x2 − 0.002. 4 Example. Application of the Modication Rule.
Solve the IVP y 00 + 3y 0 + 2.25y = −10e−1.5x , y 0 (0) = 0. y(0) = 1, Solution.
Step 1: General solution of the homogneneous ODE.
The characteristic equation of the homogeneous ODE is λ2 + 3λ + 2.25 = 0 =⇒ (λ + 1.5)2 = 0 =⇒ λ = −1.5. Hence, the homogeneous ODE has the general solution
yh = (c1 + c2 x)e−1.5x . Step 2: Solution yp of the homogeneous ODE. The function e−1.5x on the right would normally require the choice Ce−1.5x . But we see from yh
that this function is a solution of the homogeneous ODE, which corresponds to a double root of the
characteristic equation. Hence, according to the Modication Rule, we have to multiply our choice
function by x2 . That is, we choose
yp = Cx2 e−1.5x . Then
yp0 = C(2x − 1.5x2 )e−1.5x yp00 = C(2 − 3x − 3x + 2.25x2 )e−1.5x . We substitute these expressions into the given ODE and omit the factor e−1.5x . This yields
yp00 + 3yp0 + 2.25yp = −10e−1.5x C(2 − 6x + 2.25x2 ) + 3C(2x − 1.5x2 ) + 2.25Cx2 = −10. 0Cx2 + 0Cx + 2C = −10 C = −5. Hence the general solution of the given ODE is
y = yh + yp = (c1 + c2 x)e−1.5x − 5x2 e−1.5x . Step 3: Solution of the IVP:
Applying the initial condition y(0) = 1 = c1 . Dierentiating and applying the second initial condition
y0 = (c2 − 1.5c1 − 1.5c2 x)e−1.5x − 10xe−1.5x + 7.5x2 e−1.5x y 0 (0) = 0 = c2 − 1.5c1 gives c2 = 1.5. Then we obtain the full solution of the IVP
y = (1 + 1.5x)e−1.5x − 5x2 e−1.5x . 5 Example. Application of the Sum Rule.
Solve the IVP y 00 + 2y 0 + 0.75y = 2 cos x − 0.25 sin x + 0.09x, y 0 (0) = −0.43. y(0) = 2.78, Solution.
Step 1: General solution of the homogeneous ODE.
The characteristic equation of the homogeneous ODE is λ2 + 2λ + 0.75 = 0 which gives the general solution x y = c1 e− 2 + c2 e− 3x
2 . Step 2: Particular solution of the nonhomogeneous ODE.
We write yp = yp1 + yp2 and following the Table, we have
yp1 = K cos x + M sin x yp2 = K1 x + K0 . Dierentiation gives
0
yp1 = −K sin x + M cos x, 00
yp1 = −K cos x − M sin x, 0
yp2 = K1 , 00
yp2 = 0. For yp1 , substituting into the ODE and equating to r(x),
00
0
yp1
+ 2yp1
+ 0.75yp1 = 2 cos x − 0.25 sin x (−K cos x − M sin x) + 2(−K sin x + M cos x) + 0.75(K cos x + M sin x) = 2 cos x − 0.25 sin x (−K + 2M + 0.75K) cos x + (−M − 2K + 0.75M ) sin x = 2 cos x − 0.25 sin x −K + 2M + 0.75K = 2 −M − 2K + 0.75M = −0.25 For yp2 , substituting into the ODE and equating to r(x),
00
0
yp2
+ 2yp2
+ 0.75yp2 = 0.09x (0) + 2K1 + 075(K1 x + K0 ) = 0.09x 6 K0 = −0.32 K1 = 0.12. K = 0 M = 1. Hence the general solution of the nonhomogeneous ODE
x y = c1 e− 2 + c2 e− 3x
2 + sin x + 0.12x − 0.32. Step 3: Solution of the IVP.
Applying the initial conditions
y(0)
y 0 (0) 2.78 = c1 + c2 − 0.32
1
3
= −0.4 = − c1 − c2 + 1 + 0.12
2
2
= gives c1 = 3.1 and c2 = 0. The solution of the IVP is then
x y = 3.1e− 2 + sin x + 0.12x − 0.32. 7 Solution by Variation of Parameters
We continue our discussion of nonhomogeneous linear ODE
y 00 + p(x)y 0 + q(x)y = r(x). We have already used the Method of Undetermined Coecients to nd yp when r(x) is not too
complicated. However, since this method is restricted to functions r(x) whose derivatives are of a
form similar to r(x) itself (powers, exponential functions, etc), it is desirable to have a method valid
for general ODEs which we will now see. It is called the Method of Variation of Parameters.
Langrage's method gives a particular solution yp of the ODE on some open interval I in the form
ˆ
yp (x) = −y1 y2 r
dx + y2
W ˆ y1 r
dx
W where y1 , y2 form a basis of solutions of the corresponding homogeneous ODE
y 00 + p(x)y 0 + q(x)y = 0 on I and W is the Wronskian of y1 , y2 , y
W = 10
y1 y2 .
y20 Remark. Caution! The solution formula is obtained under the assumption that the ODE is written
in standard form, with y 00 as the rst term. If it starts with f (x)y 00 , divide rst by f (x). Also, the
integration in the formula may often cause diculties, and also the wronskian if y1 , y2 has variable
coecients. If you have a choice, use the previous method. It is simpler. Example. Solve the nonhomogeneous ODE
y 00 + y = sec x = 1
.
cos x Solution. A basis of solutions of the homogeneous ODE on any interval is y1 = cos x, y2 = sin x.
This gives the Wronskian
W (y1 , y2 ) = 1 Using the formula of the variation of parameters method, we have
ˆ yp (x) = ˆ
y2 r
y1 r
dx + y2
dx
W
W
ˆ
ˆ
− cos x
sin x sec xdx + sin x
cos x sec xdx
ˆ
ˆ
sin x
cos x
− cos x
dx + sin x
dx
cos x
cos x
ˆ
ˆ
− cos x
tan xdx + sin x
1dx = − cos x [ln  sec x] + sin x [x] . =
=
= −y1 Hence the general solution of the nonhomogeneous ODE is
y = yh + yp
= c1 cos x + c2 sin x − cos x · ln  sec x + x sin x. 8 ...
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 Winter '16
 Alveen Chand
 Derivative, Ode, Constant of integration, Complex number

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