1. Suppose thatz1is a solution ofL(y) =d1 and thatz2is a solution ofL(y) =d2. Then show thatz1+z2 is a solution of the equation
L(y(t)) =d1(t) +d2(t).
2. If a complex valued functionz is a solution of the equationL(x) = 0 then, show that the real and imaginary parts ofz are also solutions ofL(x) = 0.
3. (Reduction of the order) Consider an equation
L(x) =a0(t)x00+a1(t)x0+a2(t)x= 0, a0(t)6= 0, t∈I.
wherea0, a1 anda2 are continuous functions defined onI. Letx16= 0 be a solution of this equation. Show thatx2 defined by
x2(t) =x1(t) Z t
− Z s
ds, t0 ∈I,
is also a solution. In addition, show thatx1 and x2 are linearly independent onI.
x0p =x01u1+x02u2+ (x1u01+x2u02).
We do not wish to end up with second order equations for u1, u2 and naturally we choose u1 and u2 to satisfy
x1(t)u01(t) +x2(t)u02(t) = 0 (2.17) Added to it, we already known how to solve first order equations. With (2.17) in hand we now have
x0p(t) =x01(t)u1(t) +x02(t)u2(t). (2.18) Differentiation of (2.18) leads to
x00p =u01x01+u1x001+u02x02+u2x002. (2.19) Now we substitute (2.16),(2.18) and (2.19) in (2.14) to get
[a0(t)x001(t) +a1(t)x01(t) +a2(t)x1(t)]u1+ [a0(t)x002(t) +a1(t)x02(t) +a2(t)x2(t)]u2+ u01a0(t)x01+u02a0(t)x02 =d(t),
and since x1 and x2 are solutions of (2.15), hence
x01u01(t) +x02u02(t) = d(t)
We solve foru01 and u02 from (2.17) and (2.20), to determinexp.It is easy to see u01(t) = a −x2(t)d(t)
u02(t) = a0(t)W[xx1(t)d(t)1(t),x2(t)]
whereW[x1(t), x2(t)] is the Wronskian of the solutionsx1andx2. Thus,u1 andu2 are given by
u1(t) =−R x
2(t)d(t) a0(t)W[x1(t),x2(t)]dt u2(t) =R x1(t)d(t)
Now substituting the values of u1 and u2 in (2.16) we get a desired particular solution of the equation (2.14). Indeed
xp(t) =u1(t)x1(t) +u2(t)x2(t), t∈I is completely known. To conclude, we have :
Theorem 2.4.1. Let the functions a0, a1, a2 and d in (2.14) be continuous functions on I.
Further assume that x1 and x2 are two linearly independent solutions of (2.15). Then, a particular solution xp of the equation (2.14)is given by (2.16).
Theorem 2.4.2. The general solution x(t) of the equation(2.14) onI is x(t) =xp(t) +xh(t),
where xp is a particular solution given by (2.16)and xh is the general solution of L(x) = 0.
Also, we note that we have an explicit expression for xp which was not so while proving Theorem 2.3.15. The following example is for illustration.
Example 2.4.3. Consider the equation
Note that x1 = t and x2 = t2 are two linearly independent solutions of the homogeneous equation on [1,∞). Now
W[x1(t), x2(t)] =t2.
Substituting the values of x1, x2, W[x1(t), x2(t)], d(t) = tsint and a0(t) ≡ 1 in (2.21), we have
u1(t) =tcost−sint u2(t) = cost
and the particular solution isxp(t) =−tsint. Thus, the general solution is x(t) =−tsint+c1t+c2t2,
wherec1 and c2 are arbitrary constants.
The method of variation of parameters has an extension to equations of ordern(n > 2) which we state in the form of a theorem, the proof of which has been omitted. Let us consider an equation of then-th order
L(x(t)) =a0(t)xn(t) +a1(t)xn−1(t) +· · ·+an(t)x(t) =d(t), t∈I. (2.22) Theorem 2.4.4. Let a0, a1,· · · , an, d:I →Rbe continuous functions. Let
c1x1+c2x2+· · ·+cnxn
be the general solution ofL(x) = 0. Then, a particular solution xp of (2.22) is given by xp(t) =u1(t)x1(t) +u2(t)x2(t) +· · ·+un(t)xn(t),
where u1, u,· · · , un satisfy the equations
u01(t)x1(t) +u02(t)x2(t) +· · ·+u0n(t)xn(t) = 0 u01(t)x01(t) +u02(t)x02(t) +· · ·+u0n(t)x0n(t) = 0
· · · · u01(t)x(n−2)1 (t) +u02(t)x(n−2)2 (t) +· · ·+u0n(t)x(n−2)n (t) = 0 a0(t)£
u01(t)x(n−1)1 (t) +u02(t)x(n−1)2 (t) +· · ·+u0n(t)x(n−1)n (t)¤
The proof of the Theorem 2.4.4 is similar to the previous one with obvious modifications.
1. Find the general solution ofx000+x00+x0+x= 1 given that cost,sintande−tare three linearly independent solutions of the corresponding homogeneous equation. Also find the solution when x(0) = 0, x0(0) = 1, x00(0) = 0.
2. Use the method of variation of parameter to find the general solution ofx000−x0 =d(t) where
(i) d(t) =t, (ii) d(t) =et, (iii) d(t) = cost, and (iv) d(t) =e−t.
In all the above four problems assume that the general solution of x000 −x0 = 0 is c1+c2e−t+c3et.
3. Assuming that cosRt and sinRtR form a linearly independent set of solutions of the homogeneous part of the differential equation x00 +R2x = f(t), R 6= 0, t ∈ [0,∞), where f(t) is continuous for 0 ≤ t < ∞ show that a solution of the equation under consideration is of the form
RsinRt+ 1 R
where A and B are some constants. Show that particular solution of (2.14) is not unique. (Hint : Ifxpis a particular solution of (2.14) andxis any solution of (2.15) then show that xp+c xis also a particular solution of (2.14) for any arbitrary constantc.) Two Useful Formulae
Two formulae proved below are interesting in themselves. They are also useful while studying boundary value problems of second order equations. Consider an equation
L(y) =a0(t)y00+a1(t)y0+a2(t)y= 0, t∈I,
wherea0, a1, a2 :I →Rare continuous functions in additiona0(t) 6= 0 fort∈I. Let u and v be any two twice differentiable functions onI. Consider
uL(v)−vL(u) =a0(uv00−vu00) +a1(uv0−vu0). (2.23) The Wronskian of uand v is given by W(u, v) =uv0−vu0 which shows that
dtW(u, v) =uv00−vu00.
Note that the coefficients of a0 and a1 in the relation (2.23) are W0(u, v) and W(u, v) respectively. Now we have
Theorem 2.4.5. If u andv are twice differential functions onI, then uL(v)−vL(u) =a0(t)d
dtW[u, v] +a1(t)W[u, v], (2.24) where L(x) is given by (2.7). In particular, if L(u) =L(v) = 0 then W satisfies
dt [u, v] +a1W[u, v] = 0. (2.25)
Theorem 2.4.6. (Able’s Formula) Ifuand vare solutions of L(x) = 0given by (2.7), then the Wronskian of u and v is given by
W[u, v] =k exp h
Z a1(t) a0(t)dt
i , where k is a constant.
Proof. Since u and v are solutions of L(y) = 0, the Wronskian satisfies the first order equation (2.25) and Solving we get
W[u, v] =kexp h
Z a1(t) a0(t)dt
(2.26) wherek is a constant.
The above two results are employed to obtain a particular solution of a non-homogeneous second order equation.
Example 2.4.7. Consider the general non-homogeneous initial value problem given by L(y(t)) =d(t), y(t0) =y0(t0) = 0, t, t0∈I, (2.27) whereL(y) is as given in (2.14). Assume thatx1andx2are two linearly independent solution ofL(y) = 0. Letx denote a solution ofL(y) =d. Replaceu and v in (2.24) byx1 and x to get
dtW[x1, x] +a1(t)
a0(t)W[x1, x] =x1 d(t)
which is a first order equation forW[x1, x]. Hence W[x1, x] = exp
− Z t
i Z t t0
a0(s) ds (2.29)
While deriving (2.29) we have used the initial conditionsx(t0) =x0(t0) = 0 in view of which W[x1(t0), x(t0)] = 0. Now using the Able’s formula, we get
x1x0−xx01 =W[x1, x2] Z t
a0(s)W[x1(s), x2(s)]ds. (2.30) The equation (2.30) as well could have been derived withx2 in place of x1 in order to get
x2x0−xx02 =W[x1, x2] Z t
a0(s)W[x1(s), x2(s)]ds. (2.31) From (2.30) and (2.31) one easily obtains
x(t) = Z t
a0(s)W[x1(s), x2(s)] ds. (2.32) It is time for us to recall that a particular solution in the form of (2.32) has already been derived while discussing the method of variation of parameters.