3 Higher-order linear equations

3 Higher-order linear equations

Here we consider the general n-th order linear ODE, namely,

P_n(x)y⁽ⁿ⁾+Pn−1(x)y⁽ⁿ⁻¹⁾+· · ·+P₁(x)y⁰ +P₀(x)y=g(x). (3.1) Before looking at the solution of such equations, we need to look at the concept of linear dependence.

3.1 Linear independence of solutions: Wronskian

Defn: The functions y₁, . . . , y_m are said to be linearly dependent on the in- terval (α, β)if there exist c_i not all such that

m

X

i=1

c_iy_i(x)≡0, α < x < β. (3.2) They are said to be linearly independent if they are not linearly dependent.

The importance of this concept becomes clearer when we consider the homo- geneous equation

P_n(x)y⁽ⁿ⁾+Pn−1(x)y⁽ⁿ⁻¹⁾+· · ·+P₁(x)y⁰ +P₀(x)y= 0. (3.3) (It is convenient to consider the homogeneous equation rst as it turns out that the solution of the nonhomogeneous equation depends on the solution of the homogeneous equation.)

Theorem: Suppose the functions P0, P1, . . . , Pn in Eq. 3.3 are continuous on the interval (α, β)with P_n(x) never zero on this interval. Ify₁, . . . , y_n are n linearly independent solutions of the homogeneous equation Eq. 3.3 on the interval (α, β)then any solution of Eq. 3.3 may be written as

y=c₁y₁+· · ·+c_ny_n, (3.4) where c₁, . . . , c_n are arbitrary constants. Moreover, a set of n linearly inde- pendent solutions always exists.

Proof. Omitted.

Three consequences of this theorem are:

(a) Each solution of the homogeneous equation Eq. 3.3 is a linear combi- nation of n linearly independent solutions;

(b) these n linearly independent solutions are not unique;

(c) the homogeneous equation is guaranteed to have a solution.

A test for the linear independence of n solutions is to calculate a certain determinant.

Defn: If y1, . . . , yn are n solutions of Eq. 3.3, then we dene

W[y₁, . . . , y_n] =

y1(x) y2(x) · · · yn(x) y₁⁰(x) y⁰₂(x) · · · y_n⁰(x)

... ... ...

y₁⁽ⁿ⁻¹⁾(x) y⁽ⁿ⁻¹⁾₂ (x) · · · yn⁽ⁿ⁻¹⁾(x)

. (3.5)

The functionW(x) is known as the Wronskian of the n solutions y₁, . . . , y_n. The n solutions are linearly independent on the interval (α, β) if and only if their Wronskian is nonzero on this interval.

It may be shown that either the Wronskian is nonzero for all x ∈ (α, β) or identically zero. Thus we may check the value of the Wronskian at any convenient point x₀ ∈ (α, β). If the Wronskian is nonzero at x₀, then it is nonzero for allx∈(α, β).

Example 15

For the equation y⁰⁰ +y = 0, it is easily veried that y1(x) = sin(x) and y₂(x) = cos(x) are two solutions. Then

W[y₁, y₂] =

sin(x) cos(x) cos(x) −sin(x)

=−1, (3.6)

so the two solutions are linearly independent. Thus the general solution of the equation is

y(x) =c₁sin(x) +c₂cos(x). (3.7) If we have initial conditions y(0) = 0 and y⁰(0) = 1, then we obtain c₂ = 0 and c₁ = 1. ##

Now we go back to the nonhomogeneous equation. If y_a and y_p are two solutions of the nonhomogeneous equation, then their dierencez =y_a−y_p satises the homogeneous equation. Analogously to Theorem 2.3, we then conclude that the general solution of the nonhomogeneous equation is given by

y=yp+c1y1+· · ·+cnyn, (3.8) where yp is any solution of the nonhomogeneous equation (known as a par- ticular solution).

Exercises

1. By nding constants c₁, c₂, c₃, and c₄ which are not all zero such that c₁x+c₂e^x +c₃xe^x+c₄(2−3x)e^x = 0, show that the functions x, e^x, xe^x, and (2−3x)e^x are linearly dependent.

2. Use the Wronskian to show that the functions e^x, e^2x, and e^3x are linearly independent.

3. Use the Wronskian to show that the functions e^x, cos(x), and sin(x) are linearly independent.

4. Verify that each of the giveny₁ and y₂ below are solutions of the given homogeneous dierential equation and use the Wronskian to verify that y1 and y2 are linearly independent.

(a) y₁(x) = e^xcos(x), y₂(x) = e^xsin(x), y⁰⁰−2y⁰+ 2y= 0. (b) y1(x) = x², y2(x) = 1/x, x²y⁰⁰−2y= 0.

(c) y₁(x) = 1,y₂(x) =√

x, yy⁰⁰+ (y⁰)² = 0. (d) y₁(x) = e^x, y₂(x) = xe^x, y⁰⁰−2y⁰+y= 0.

3.2 Constant coecients case

3.2.1 Homogeneous DEs

Though homogeneous higher-order linear equations always have a general solution (under appropriate conditions), nding these solutions may be dif- cult and it is usually necessary to use numerical methods. However, in the situation where there are constant coecients, the procedure is quite straightforward.

It hinges on assuming that the solutions are of the form y(x) = e^rx. Example 16

Consider the DE Classify:

y⁰⁰−5y⁰+ 6y= 0. (3.9)

Substituting y=e^rx into this yields r²−5r+ 6

e^rx = 0

⇒ r²−5r+ 6 = 0 = (r−3)(r−2)

So we have 2 solns

y₁(x) =e^3x and y₂(x) = e^2x

The general solution is then an arbitrary linear combination of these two solutions, namely

y(x) =Ay1(x) +By2(x)

whereA,B are (arbitrary) constants.^‡ ##

General Case.

The most general (linear, const coef.) homogeneous DE can be written a_ny⁽ⁿ⁾+an−1y⁽ⁿ⁻¹⁾+· · ·+a₁y⁰+a₀y= 0, (3.10) wherea₀, a₁, . . . , a_n are constants.

Since this is ann^th-order linear DE, it will have n linearly independent solu- tions.

The solution procedure is as follows

1. Assume the solutions are of the form y=e^rx.

2. Substitute into the homog. DE. Since theny^(j)(x) =r^je^rx, this gives a polynomial in r, which must be zero.

3. Find the roots r₁, r₂, . . . , r_n of this polynomial.

4. The general solution is y(x) =A₁e^r1x+A₂e^r2x+. . .+A_ne^rnx. [Unless some of the roots are double, etc].

What are the A1, A₂, etc?

The polynomial which arises, namely

a_nrⁿ+an−1rⁿ⁻¹+· · ·+a₁r+a_n= 0. (3.11) is called the characteristic equation for the dierential equation. If the n zeros of the characteristic polynomial are r₁, . . . , r_n, then n solutions are given by

e^r1x, e^r2x, . . . , e^rnx. (3.12) Actually, we have to be careful as the details of the solution depend on the nature of the roots ri (real and distinct; complex conjugates; repeated).

Specically,

‡Determined by two initial conditions, if given.

(i) If ther_i are all real and distinct, then it may be shown that the Wron- skian W[e^r1x, . . . , e^rnx] is nonzero so that the general solution of the homogeneous equation is

y(x) = A₁e^r1x+· · ·+A_ne^rnx. (3.13) (ii) Suppose we have a zero r_j of the form λ+iµ where i = √

−1. Since complex zeros occur in conjugate pairs, there must exist another zero of the formλ−iµ. It can be shown that two linearly independent solutions corresponding to these two complex zeros are given bye^λxcos(µx)and e^λxsin(µx)

(iii) Suppose we have a zero r_j which occurs k times. Then it may be shown that k linearly independent solutions corresponding to this zero are given by

e^rjx, xe^rjx, . . . , x^k−1e^rjx. (3.14) Example 17

Consider the general second-order linear ODE with constant coecients a2y⁰⁰+a1y⁰+a0y= 0. (3.15) Then the characteristic polynomial is given by a₂r²+a₁r+a₀, with rootsr₁ and r₂, say. We then have the following conclusions:

(i) If r₁ and r₂ are real and not equal,

then the general solution is y(x) = Ae^r1x+Be^r2x. (ii) If r₁ =λ+iµ and r₂ =λ−iµ,

then the general solution is y(x) = Ae^λxcos(µx) +Be^λxsin(µx). (iii) If r₁ =r₂,

then the general solution is y(x) = Ae^r1x+Bxe^r1x= (A+Bx)e^r1x.

##

Proof about form of soln for complex conjugate roots.

Suppose we have a 2nd-order ODE with r₁ = r^∗₂ = λ+iµ, as in (ii) above.

Then our theory tells us that the general solution is λ, µ∈R

y = Ae^r1x+Be^r2x (3.16)

= Ae^(λ+iµ)x+Be^(λ−iµ)x (3.17)

= e^λx

Ae^iµx+Be^−iµx

(3.18)

= e^λx[A(cosµx+isinµx) +B(cosµx−isinµx)] (3.19)

= e^λx[(A+B) cosµx+i(A−B) sinµx] (3.20)

≡ e^λx[Ccosµx+Dsinµx], (3.21)

whereC, D must be REAL constants since we want y(x)to be real.

But waithow can that be so ifA, B are also real constants??

Answer: It can't be so! In fact, A, B are not real-valued constants but complex-valued ones. The essential requirement is that y(x) is real (not complex), and if some intermediate constants need to be complex-valued, that's ne.^§

3.2.2 Nonhomogeneous DEs

We now turn to the nonhomogeneous equation

a₀y⁽ⁿ⁾+a₁y⁽ⁿ⁻¹⁾+· · ·+an−1y⁰+a_ny=g(x), (3.22) wherea₀, a₁, . . . , a_n are constants.

We shall consider two methods, the method of undetermined coecients and the method of variation of parameters. The rst method is easier to apply, but in practice is only useful when g is a polynomial, exponential function, trigonometric function or combinations of these. In this method we try to nd a particular solution of the same form as g.^¶

For example, suppose g is a polynomial of degreem, that is,

g(x) = b₀x^m+b₁x^m−1+· · ·+bm−1x+b_m. (3.23) Then one should try a particular solution which is also a polynomial of degree m, namely

y_p(x) = A₀x^m+A₁x^m−1+· · ·+A_m−1x+A_m. (3.24) Ifg(x) =Ce^αxsin(βx) org(x) =Ce^αxcos(βx), then one should try a partic- ular solution of the form

yp(x) =Ae^αxcos(βx) +Be^αxsin(βx). (3.25) (The case g(x) = Ce^αx is covered by taking β = 0.) The most general case may be found in the textbook.

Example 18

Consider the initial value problem given by

y⁰⁰−3y⁰+ 2y = 2x², y(0) =y⁰(0) = 0. (3.26)

§In fact, sinceiexplicitly appears in Eqns. 3.173.20 theAandB MUST be complex to allow the RHS to be real-valued.

¶This is in fact the same method we used for 1st-order linear equations when guessing a particular solution. See page 14.

The characteristic polynomial is r² −3r+ 2 which has zeros 1 and 2. Thus the general solution of the homogeneous equation is

y(x) = c₁e^x+c₂e^2x. (3.27) The right-hand side g(x) is a quadratic so the particular solution to try is

y_p(x) =A₀x²+A₁x+A₂. (3.28) Substitution into the equation yields

2A₀−3(2A₀x+A₁) + 2(A₀x²+A₁x+A₂) = 2x². (3.29) Thus

2A₀−3A₁+ 2A₂ = 0, −6A₀+ 2A₁ = 0, 2A₀ = 2. (3.30) Upon solving, we nd

A₀ = 1, A₁ = 3, A₂ = ⁷₂. (3.31) Thus the solution of the nonhomogeneous equation is

y(x) = c1e^x+c2e^2x+x²+ 3x+⁷₂. (3.32) Imposing the two initial conditions shows that the unique solution to the initial value problem is

y(x) = −4e^x+¹₂e^2x+x²+ 3x+⁷₂. (3.33)

##

However, there is a special proviso to the technique given above. If the above recommendations yields a particular solution which contains a solution of the homogeneous equation, then one needs to multiply such a particular solution by x^s for some integer s ≥ 1. The value of s should be just large enough so that the particular solution obtained does not contain a solution of the homogeneous equation.

Example 19

Consider the dierential equation given by

y⁰⁰−4y⁰+ 4y =e^2x. (3.34) For this equation, the homogeneous solution is c₁e^2x +c₂xe^2x. Following the above recommendation, one would try yp(x) = Ae^2x. However, Ae^2x is solution of the homogeneous equation. Taking s= 1 is not enough as Axe^2x

is also a solution of the homogeneous equation. Here we need to take s = 2 and hence use y_p(x) = Ax²e^2x. ##

Example 20

A dierential equation describing damped forced simple harmonic motion (for example, the motion of a mass on a spring subject to an applied external periodic force and a resistive force) is given by

my¨+γy˙+ky =asin(ωt), (3.35) where t is time, y(t) is the distance from the equilibrium point, m is the mass, γ ≥0is the damping coecient, and k > 0is the spring constant.

(3.36) The characteristic polynomial is mr²+γr+k with zeros

r₁, r₂ = −γ±p

γ²−4mk

2m . (3.37)

Let us consider the homogeneous equation rst (the situation when there is no forcing term). We see that the solution of the homogeneous equation depends on the sign ofγ²−4mk. We have

y(t) =







Ae^r1t+Be^r2t, γ²−4mk >0;

Ae^−γt/(2m)+Bte^−γt/(2m), γ²−4mk = 0;

e^−γt/(2m)[Asin(µt) +Bcos(µt)], γ²−4mk <0;

(3.38)

where

µ=

p4mk−γ²

2m >0. (3.39)

It may be veried that in all three cases y(t)→0ast→ ∞. In the rst case no oscillatory behaviour occurs and the system is said to be overdamped. The transition in behaviour occurs when γ = 2√

mk and this value is known as critical damping. The most important case is the third case when γ is small compared to mk (this is the situation in which the damping is small and is called weak damping or underdamping). Note that if know the values of A and B from initial conditions and we letβ = tan⁻¹(A/B), then

sin(β) = A

√A²+B² and cos(β) = B

√A²+B². (3.40) Since cos(µt −β) = sin(β) sin(µt) + cos(β) cos(µt), it follows that we can write

y(t) = Re^−γt/(2m)cos(µt−β), (3.41) where R=√

A²+B². As indicated above, this solution of the homogeneous equation exhibits oscillatory behaviour. Although the motion is not periodic, the parameter µ determines the frequency with which the mass oscillates.

Consequently, µis known as the quasi-frequency.

We now consider the third case in more detail. In particular, we consider the nonhomogeneous equation. First, let us look at the situation when γ = 0 (that is, no damping) and µ = ω. Then µ² = k/m and the solution of the homogeneous equation is

y(t) =c₁sin(ωt) +c₂cos(ωt). (3.42) Then we look for a particular solution of the form

up(t) = Atcos(ωt) +Btsin(ωt). (3.43) (The particular solution we try out is not u_p(t) = Acos(ωt) + Bsin(ωt) because this u_p is a solution of the homogeneous equation.) By substitution into the equation, one may verify that A=−a/(2mω) andB = 0. Thus the solution is

u(t) =Rcos(ωt−β)− a

2mωtcos(ωt). (3.44) This solution is unbounded as t → ∞ and this phenomenon is known as resonance. In practice the spring would probably break.

We now look at the case when γ >0. Then we look for a particular solution of the form

u_p(t) =Acos(ωt) +Bsin(ωt). (3.45)

Hence

asin(ωt) = −mω²(Acos(ωt) +Bsin(ωt)) (3.46) +γω(−Asin(ωt) +Bcos(ωt)) (3.47) +k(Acos(ωt) +Bsin(ωt)). (3.48) Thus we require

−mω²A+γωB+kA= 0 and −mω²B−γωA+kB =a. (3.49) Upon solution we nd

u_p(t) = a

(k−mω²)²+ (γω)²[−γωcos(ωt) + (k−mω²) sin(ωt)]. (3.50) In the undamped case we saw thatu_p was unbounded. This will not happen in the present case. However, for some values of ω, the maximum may be so large that it causes the system to be destroyed. This is called practical resonance and may arise when γ is small and ω² is close to k/m. For those interested, the textbook contains a lot more detail about such mechanical vibrating systems. ##

3.3 Method of Variation of Parameters

We now look at the method of variation of parameters. This method may be used for general linear ODEs and is not restricted to just those with constant coecients. For simplicity, we shall consider only the n = 2 case. Suppose we have solved the homogeneous equation so that we have a general solution y(x) =c1y1(x) +c2y2(x). (3.51) In the method of variation of parameters we seek a particular solution of the form

yp(x) = w1(x)y1(x) +w2(x)y2(x), (3.52) wherew₁andw₂ are functions to be determined. Since we have two functions to determine, we need to specify two conditions. One obvious condition is that y_p satisfy the dierential equation. The other condition is chosen so as to `make life easy'. In particular, we impose the condition

w⁰₁y₁+w⁰₂y₂ = 0. (3.53) The reason for this choice is that since

y_p⁰ =w₁y₁⁰ +w₂y₂⁰ +w⁰₁y₁+w⁰₂y₂, (3.54)

we do not get any second derivatives of w₁ and w₂ iny_p⁰⁰ if this condition is imposed. With two equations, we can then solve for w⁰₁ and w₂⁰ and hence nd w₁ and w₂.

Example 21

We use the method of variation of parameters to solve the equation given in Example 18. Then we seek a particular solution of the form

y_p(x) = w₁(x)e^x+w₂(x)e^2x. (3.55) Imposing the condition

w⁰₁(x)e^x+w⁰₂(x)e^2x = 0 (3.56) means that

y_p⁰(x) = w₁(x)e^x+ 2w₂(x)e^2x (3.57) and

y⁰⁰_p(x) =w₁(x)e^x+ 4w₂(x)e^2x+w⁰₁(x)e^x+ 2w⁰₂(x)e^2x. (3.58) We require y_p to satisfy the dierential equation. Substitution into the dif- ferential equation and some algebra yields

w₁⁰(x)e^x+w₂⁰(x)2e^2x = 2x². (3.59) Solving the pair of simultaneous equations in w⁰₁ and w₂⁰ given by Eq. 3.56 and 3.59 yields

w₂⁰(x) = 2x²e^−2x and w⁰₁(x) =−2x²e^−x. (3.60) Integration by using a table of integrals yields

w₁(x) = −2e^−x(−x²−2x−2) +c₁, (3.61) w2(x) = 2e^−2x(−x²/2−x/2−1/4) +c2. (3.62) Thus

y_p(x) = 2x²+ 4x+ 4 +c₁e^x−x²−x−1/2 +c₂e^2x (3.63)

= x²+ 3x+ 7/2 +c₁e^x+c₂e^2x. (3.64) If we discard the c₁e^x and c₂e^2x terms which are solutions of the homoge- neous equation, we end up with the same particular solution as before. The technique for general n may be found in the textbook. ##

Exercises

1. Solve the following homogeneous ODEs with given initial conditions.

(a) y⁰⁰−2y⁰−5y= 0, y(0) = 0, y⁰(0) = 3.

(b) y⁰⁰⁰−3y⁰ −2y= 0, y(0) =y⁰⁰(0) = 0, y⁰(0) = 9. (c) y⁽⁴⁾−y= 0, y(0) = 1, y⁰(0) =y⁰⁰(0) =y⁰⁰⁰(0) = 0. 2. Solve the following second-order linear ODEs.

(a) y⁰⁰+y⁰−2y= 2x, y(0) = 0, y⁰(0) = 1.

(b) y⁰⁰−a²y=e^bx, wherea and b are arbitrary constants.

(c) 2y⁰⁰−4y⁰−6y= 3e^2x.

(d) y⁰⁰+ 4y = 4x²+e^x, y(0) = 1, y⁰(0) = 1. (e) y⁰⁰+ 2y⁰+y =e^xcos(x).

(f) y⁰⁰−y⁰−2y= cosh(2x).

3. (a) For the second-order linear dierential equation

a₀x²y⁰⁰+a₁xy⁰+a₂y= 0, (3.65) where a₀, a₁, and a₂ are constants, show that the substitution x=e^w transforms the equation into

a₀d²y

dw² + (a₁−a₀)dy

dw +a₂y= 0, (3.66) an equation which has constant coecients.

(b) Use the result of part (a) to solve the dierential equation

2x²y⁰⁰+ 3xy⁰−15y=x⁻³. (3.67) 4. Use the method of variation of parameters to nd the solution of the

following ODEs.

(a) y⁰⁰+y= sec(x) tan(x). (b) y⁰⁰−3y⁰+ 2y= 1/(1 +e^−x).

(c) y⁰⁰−3y⁰+ 2y= cos(e^−x).

(d) (1−2x)y⁰⁰+ 4xy⁰−4y= (x−1)²e^−x. (Hint: verify that y₁(x) =x and y₂(x) = e^2x are solutions of the corresponding homogeneous equation.) (1−kx)y⁰⁰+k²xy⁰−k²y= (x−1)²e^−x. (y₁(x) =xand y₂(x) =e^kx)

5. (a) The equation

x²y⁰⁰+xy⁰ + (x²−ν²)y= 0, ν ≥0, (3.68) is an important equation in applied mathematics known as Bessel's equation. By making the substitution y(x) = x^cu(x), show that the equation reduces to

x²u⁰⁰+ (1 + 2c)xu⁰+ (c²+x²−ν²)u= 0. (3.69) (b) Ifνis not an integer, then the general solution of Bessel's equation

may be written as

y(x) =AJ_ν(x) +BJ_−ν(x), (3.70) whereA andB are arbitrary constants, andJ_ν is the Bessel func- tion of orderν. Use the result of part (a) to show that there exist constants C and D such that

A√

xJ_1/2(x) +B√

xJ−1/2(x) =Csin(x) +Dcos(x). (3.71)