ORDINARY DIFFERENTIAL EQUATIONS based on the course-contents of MMB-352 (A course of Mathematics (Main/Subsidiary)
prescribed for B. Sc. (Hons.) III Semester) Unit-III: Differential Equations in
Three Variables
December 3, 2020
Lecture No. 01
Total Differential Equations (A topic of Unit III)
Total Differential Equations:- An equation of the form P dx+Q dy+R dz= 0
whereP, Q, Rare functions ofx, y andz,is called a total differential equation or Pfaffian differential equation. Its integral (solution) if exists, is of the form φ(x, y, z) = c.
Condition of Integrability: A necessary and sufficient condition for inte- grability of above equation is
P ∂Q
∂z − ∂R
∂y
+Q ∂R
∂x −∂P
∂z
+R ∂P
∂y −∂Q
∂x
= 0.
In this lecture, we shall discuss the solutions of following types Total Differ- ential equations (TDE):
1. Exact total differential equations
2. TDE reducible to exact forms which are solvable by inspections 3. Homogeneous TDE
4. TDE in which one variable is taken as constant 1
[RB] Example/ p.164Verify the condition of integrability for the following differential equations:
(1) (HA) (3x2y2−exz)dx+ (2x3+ sinz)dy+ (ycosz−ex)dz = 0 (2) (3xz+ 2y)dx+xdy+x2dz = 0
(3) (HA) ydx+dy+dz = 0
Solution 1. Do Yourself. [Ans: Integrable]
Solution 2. The given Differential equation is
(3xz+ 2y)dx+xdy+x2dz = 0 On comparing with
P dx+Q dy+R dz = 0, we have, P = 3xz+ 2y, Q=x and R =x2, then
∂P
∂y = 2, ∂P
∂z = 3x
∂Q
∂x = 1, ∂Q
∂z = 0
∂R
∂x = 2x, ∂R
∂y = 0 Then, from the condition of integrability,
P ∂Q
∂z − ∂R
∂y
+Q ∂R
∂x − ∂P
∂z
+R ∂P
∂y − ∂Q
∂x
= (3xz+ 2y)[0−0] +x[2x−3x] +x2[2−1]
= 0−x2+x2
⇒0
Hence, the equation is integrable.
Solution 3. Do Yourself. [Ans: Not Integrable]
3
Exact Total Differential Equations: A total differential equation P dx+Q dy+R dz= 0
is exact if the following conditions are satisfied:
∂P
∂y = ∂Q
∂x, ∂Q
∂z = ∂R
∂y, ∂R
∂x = ∂P
∂z
If a total differential equation is exact, then its solution is obtained by the regrouping of terms.
[RB] Problem 3/ p.167 Solve (x−y)dx−xdy+zdz = 0.
Solution. Here, P =x−y, Q=−x, R=z.
Therefore, ∂P
∂y =−1, ∂P
∂z = 0, ∂Q
∂x =−1, ∂Q
∂z = 0, ∂R
∂x = 0, ∂R
∂y = 0.
Now,
∂P
∂y = ∂Q
∂x = −1
∂Q
∂z = ∂R
∂y = 0
∂R
∂x = ∂P
∂z = 0 Thus, the given total differential equation is exact.
On regrouping the terms, we get
(x−y)dx−xdy+zdz = 0 x dx−(xdy+ydx) +zdz = 0 On integrating, we get
x2
2 −xy+ z2 2 =c Hence, the required solution
x2−2xy+z2 =a, where a= 2c.
[TB] Example 2.48/ p.57 Solve xdx+zdy+ (y+ 2z)dz = 0.
Solution. Here, P =x, Q =z, R=y+ 2z.
Therefore, ∂P
∂y = 0, ∂P
∂z = 0, ∂Q
∂x = 0, ∂Q
∂z = 1, ∂R
∂x = 0, ∂R
∂y = 1.
Now,
∂P
∂y = ∂Q
∂x = 0
∂Q
∂z = ∂R
∂y = 1
∂R
∂x = ∂P
∂z = 0 Thus, the given total differential equation is exact.
On regrouping the terms, we get
xdx+zdy+ (y+ 2z)dz = 0 x dx+ (ydz+zdy) + 2zdz = 0 On integrating,
x2
2 +yz+z2 =c which is the required solution.
Home Assignment
1. [RB](Q.35/ p-177) Test for integrability and solve the differential equation
(y+ 3z)dx+ (x+ 2z)dy+ (3x+ 2y)dz = 0.
[Ans : xy+2yz+3xz=c]
2. [TB](Q.62/ p-62) Solve
(y+z)dx+ (z+x)dy+ (x+y)dz = 0.
[Ans : xy+yz+zx=c]
5
Solution by Inspection: If a total differential equation is not exact, it may be possible to find an integrating factor which makes equation as exact.
[RB] Problem 4/ p.167 Solve y2dx−zdy+ydz = 0.
Solution. Here, P =y2, Q =−z, R=y, then P
∂Q
∂z −∂R
∂y
+Q ∂R
∂x −∂P
∂z
+R ∂P
∂y −∂Q
∂x
=y2[−1−1]−z[0−0] +y[2y−0]
= 0
Thus, the equation is integrable.
But it is not exact. (verify yourself )
Now, dividing the given equation by y2, we get y2dx−zdy+ydz
y2 = 0
dx+ ydz−zdy
y2 = 0 On integrating it, we get
x+ z y =c which is the required solution.
Home Assignment
1. [RB](Problem 5/ p-168) Solve
(2x3y+ 1)dx+x4dy+x2tanzdz = 0.
[Ans : x2y−x1 + log secz=c]
[Hint: Divide the equation by x2 to make it exact.]
Homogeneous Total Differential Equations (A topic of Unit III)
Homogeneous Total Differential Equations:A total differential equation P dx+Q dy+R dz= 0
is called homogeneous if P, Q, R are homogeneous functions of x, y, z of same degree.
In such equations, one of the variables can be seperated from the other two by using the transformation:
x=uz and y=vz so that
dx =z du+u dz and dy =z dv+v dz
Putting these values in given total differential equation, we find that the TDE is reduced to the form in which either the coefficient of dz is zero or not zero.
In either case, the new equation may easily be integrated.
[RB] Problem 9/ p.169 Solve the homogeneous equation yzdx−z2dy− xydz = 0.
Solution. Here, P =yz, Q =−z2, R=−xy.
Each of P, Q, R is homogeneous function of degree 2.
Also, given equation is integrable. (check yourself ) 1
2
Consider the transformation
x=uz and y=vz so that
dx=z du+u dz and dy=z dv+v dz Using above, the given equation reduces to
yzdx−z2dy−xydz = 0 (vz)z(zdu+udz)−z2(zdv+vdz)−(uz)(vz)dz = 0
⇒ z2[v(zdu+udz)−(zdv+vdz)−uvdz] = 0
⇒ z2[vzdu−zdv−vdz] = 0 Divide the equation by z2, we get
vzdu−zdv−vdz = 0 Again, divide by vz, we have
du− 1
vdv−1
zdz = 0 On integrating, we get
u−logv −logz = logc
⇒ u = log(cvz)
⇒ vz = aeu, where a= 1 c Putting the values of u and v in above equation, we get
y=aexz which is the required solution.
Home Assignment
1. [RB] Problem 8/ p.168 Solve the homogeneous equation 2(y+z)dx−(x+z)dy+ (2y−x+z)dz = 0.
[Ans : y+z=a(x+z)2] 2. [RB] Problem 10/ p.169 Solve
(2y−z)dx+ 2(x−z)dy−(x+ 2y)dz = 0.
[Ans : 2xy−xz−2yz=c]
3. [RB] Q. 40/ p.177 Solve
(x+z)2dy+y2(dx+dz) = 0.
[Ans : y(x+z) = a(x+y+z)]
One variable regarded as constant:If two terms ofP dx+Q dy+R dz = 0, say, P dx+Q dy can be easily integrable, then the third variablez is taken as constant so that dz = 0. So integrate P dx+Q dy = 0 and use φ(z) instead of constant of integration. Now take total differential of obtained integral and compare its coefficients with given TDE, to determine φ(z).
[RB] Problem 16/ p.171 Solve yzdx+ (xz−yz3)dy−2xydz = 0.
Solution. Given equation is integrable? (check yourself )
Consider y as a constant so that dy = 0. Hence, given equation reduces to yzdx+ (xz−yz3)dy−2xydz = 0
yzdx−2xydz = 0
⇒ zdx−2xdz = 0, (dividing by y)
⇒ dx
x = 2dz z On integrating, we get
logx = 2 logz+ logφ(y)
⇒ x = z2φ(y) (0.1)
4
Taking total derivative of (0.1), we get
dx = d[z2φ(y)]
dx = z2dφ+φ(y)·2zdz
⇒ dx−2φ(y)zdz−z2dφ = 0 From (0.1), φ(y) = x
z2. Hence, above equation becomes dx−2h x
z2 i
zdz−z2dφ = 0 dx−2x
zdz−z2dφ = 0 zdx−2xdz−z3dφ = 0 Multiplying the above equation byy, we get
yzdx−yz3dφ−2xydz = 0 (0.2) Comparing (0.2) with the given TDE, we get
−yz3dφ = (xz−yz3)dy
−yz3dφ = [{z2φ(y)}z−yz3]dy, (using (0.1)⇒x=z2φ(y))
−yz3dφ = [z3φ(y)−yz3]dy
−ydφ = φ(y)dy−ydy, (dividing by z3) φ(y)dy+ydφ−ydy = 0
d[φ·y]−d 1
2y2
= 0 On integrating, we get
φ·y− 1
2y2 = c y
φ− 1
2y
= c
⇒ φ = 1 2y+ c
y
Putting this value in (0.1), we get x=z2
1 2y+ c
y
i.e.,
zxy=y2z2+cz2 which is the required solution.
Home Assignment
1. [RB] Problem 15/ p.171 Solve
(exy+ez)dx+ (eyz+ex)dy+ (ey−exy−eyz)dz = 0.
[Ans : exy+eyz+ezx=cez] 2. [TB] Exercise 2.47/ p.57 Solve
(y2+yz)dx+ (z2 +zx)dy+ (y2−xy)dz = 0.
[Ans : y(z+x) = c(y+z)]
Lecture No. 03
Simultaneous Total Differential Equations (A topic of Unit III)
Simultaneous Total Differential Equations: The equations P1dx+Q1dy+R1dz = 0
P2dx+Q2dy+R2dz = 0 )
(0.1) whereP1, Q1, R1, P2, Q2, R2are functions ofx,yandz, are called simultaneous total differential equations. The solution of the system, if exists, is of the form:
φ(x, y, z) =c1, ψ(x, y, z) = c2. Using Cramer rule, the system (0.1) can also be written as:
dx
Q1R2−Q2R1 = dy
R1P2−R2P1 = dz P1Q2−P2Q1
or dx
P = dy Q = dz
R (0.2)
where
P =Q1R2−Q2R1, Q=R1P2−R2P1, R=P1Q2−P2Q1.
In this lecture, we shall discuss two methods for solving simultaneous total differential equations:
1
1. Grouping Method: If it is possible to take two fractions dxP = dyQ of Equation (0.2), wherein z can be cancelled or is absent, then the equation
dx
P = dyQ involves x and y only and hence by integrating gives rise
φ(x, y) = c1. (0.3)
In the similar manner, if the equation dxP = dzR contains x and z only (not y), then by integrating it, we get
ψ(x, z) =c2. (0.4)
Equations (0.3) and (0.4) taken together constitute the solution of Equation (0.2).
2. Multipliers Method: Let l, m, n be functions of x, y, z then by well- known principle of algebra, each fraction in (0.2) is equal to
ldx+mdy+ndz
lP +mQ+nR . (0.5)
Herel, m, n are called multipliers. In this rule, if possible, we choose multipli- ers such that the numerator of (0.5) is exact differential of its denominator.
Therefore (0.5) can be combined with a suitable fraction in (0.2) to give an integral. However, in some problems, another set of multipliersl0, m0, n0 are so chosen that the fraction
l0dx+m0dy+n0dz
l0P +m0Q+n0R (0.6)
is integrable and hence fractions (0.5) and (0.6) are combined to get an integral. To get another independent solution of (0.2), either the above method may be repeated or sometimes grouping method be applied.
If we are able to choose the multipliersl, m, nin (0.5) such thatldx+mdy+ndz is integrable and lP +mQ+nR= 0 then we have
each fraction = ldx+mdy+ndz
0 ,
which implies that
ldx+mdy+ndz = 0.
3 This gives us one independent solution of (0.2). As a special case, these multipliers can be considered as the constants. To get another independent solution of (0.2), either the above method may be repeated or sometimes foregoing rules be applied.
Example 2.49/p.58 Solve dx
z2y = dy
z2x = dz y2x Solution. Taking first two fractions, we get
dx
z2y = dy z2x
⇒ xdx−ydy = 0 On integrating, we get
x2 2 − y2
2 = c
⇒ x2−y2 = a, where a=2c (0.7)
Now taking last two fractions, we get dy
z2x = dz y2x
⇒ y2dy−z2dx = 0 On integrating it, we get
y3 3 − z3
3 = c0
y3−z3 = b, where b=3c’ (0.8)
Equation (0.7) and (0.8) taken together form the required solution.
Q.69/p.62 Solve dx
x2−y2−z2 = dy
2xy = dz 2xz Solution. Taking last two fractions, we get
dy
2xy = dz
2xz
dy
y = dz z
Integrating it, we get
logy= logz+ loga i.e.,
⇒ y=az (0.9)
Using multiplierx, y, z in given equation, we get each ratio = xdx+ydy+zdz
x(x2−y2−z2) + 2xy2+ 2xz2
= xdx+ydy+zdz x(x2−y2−z2+ 2y2+ 2z2)
= xdx+ydy+zdz x(x2 +y2+z2) Taking it with second fraction, we get
xdx+ydy+zdz
x(x2+y2+z2) = dy 2x y
⇒ 2xdx+ 2ydy+ 2zdz
x2+y2+z2 = dy y
⇒ d(x2+y2+z2)
x2+y2+z2 = dy y On integrating it, we get
log(x2+y2+z2) = logy+ logb i.e.,
x2+y2 +z2 =by (0.10)
Equations (0.9) and (0.10) together form the required solution.
Example 2.50/p.59 Solve dx
x(y2−z2) = dy
−y(z2+x2) = dz z(x2+y2)
5 Solution. Using multipliers x, y, z, we get
each ratio = xdx+ydy+zdz
x2(y2−z2)−y2(z2+x2) +z2(x2+y2)
= xdx+ydy+zdz
x2y2−x2z2−y2z2−x2y2+x2z2+y2z2
= xdx+ydy+zdz 0
⇒ xdx+ydy+zdz = 0 Integrating it, we get
x2+y2+z2 =a (0.11)
Now, using the multipliers 1x,−1y,−1z, we get each ratio =
1
xdx− 1ydy−1zdz
(y2−z2) + (z2+x2)−(x2+y2)
=
1
xdx− 1ydy−1zdz 0
⇒ 1
xdx− 1
ydy− 1
zdz = 0 Integrating it, we get
logx−logy−logz = logc x
yz = c yz = 1
cx let b= 1c, then
yz =bx, (0.12)
Equations (0.11) and (0.12) together form the required solution.
Home Assignment Solve the following differential equations:
1. Q.65/ p.62 dx x2 = dy
y2 = dz nxy
[Ans : 1x = 1y +c1, z=c2−y−xnxy logxy] 2. Q.66/ p.62 dx
mz−ny = dy
nx−lz = dz ly−mx [Ans : lx+my+nz=c1, x2+y2+z2 =c2] 3. Q.67/ p.62 dx
y−zx = dy
yz+x = dz x2+y2
[Ans : x2−y2−2xy =c1, x2−y2−z2 =c2]
Lecture No. 04
System of Linear Differential Equations (A topic of Unit III)
The system of equations
P11(D)y1+P12(D)y2+· · ·+P1n(D)yn=Q1(t) P21(D)y1+P22(D)y2+· · ·+P2n(D)yn=Q2(t)
· · · ·
Pn1(D)y1+Pn2(D)y2+· · ·+Pnn(D)yn =Qn(t)
where D= dtd and Pij(D) are polynomial operator with constant coefficients, is called system of n linear differential equations.
The solution of above system is a set of functions y1(t), y2(t),· · · , yn(t) satisfy all the equations of the system identically.
Remark. The solution is general if solution contains correct number of arbitrary constants.
Consider the pair of equations
p1(D)x(t) +p2(D)y(t) = q1(t) p3(D)x(t) +p4(D)y(t) = q2(t)
(0.1) The solution of system (0.1) is a set of functions x=x(t), y =y(t).
Define a determinant 4(D) :=
p1(D) p2(D) p3(D) p4(D)
=p1(D)p4(D)−p2(D)p3(D).
1
Then the system (0.1) is called (a) Degenerateif 4(D) = 0 (b) Non-degenrateif 4(D)6= 0.
A non-degenerate system has a unique solution, while a degenerate system has either no solution or infinite many solutions.
An important observation: If the system (0.1) is non-degenerate, then number of arbitrary constant in general solutionx=x(t),y =y(t) of (0.1) equals to the order of4(D).
Firstly, we discuss three methods of solving a non-degenerate system.
(A)Easily Solvable System:
Example 7.1/ p. 312Solve the system of first order equations (for x6= 0, t6= 0)
dx dt = t
x2 (0.1)
dy dt = y
t2 (0.2)
Solution. On using variable separable, we get from Equation (0.1) x2dx=tdt
On integration, we obtain
x3 = 3
2t2+c1 (0.3)
From Equation (0.2), we have dy
y = dt t2 and hence
logy−logc2 =−1 t i.e.,
y =c2e−1t (0.4)
3 Therefore the pair of functions, that is, x(t), y(t) given by Equations (0.3) and (0.4), is a solution of given system.
Example 7.2/P.312 Solve
dx
dt = 2e2t (0.1)
and dy
dt = x−y
t (0.2)
Solution. On using variable seperable, dx = 2e2tdt so, the solution of the Equation (0.1) is
x=e2t+c1 (0.3)
Now, substitute the value of x(t) from (0.3) to (0.2), we obtain dy
dt +y
t = e2t+c1 t
which is a linear differential equation dydt +P y =Q
and hence the solution of this equation is
y = 1
2e2t+c1t+c2
t−1 (0.4)
Thus the pair of functions defined by Equations (0.3) and (0.4) is a solution of given system.
Home Assignment
Solve the following system of equations:
1. (Que.2/P.365) dx
dt = 3e−t, dy
dt =x+y
[Ans : x=−3e−t+c1, y= 32e−t+c1tet+c2et]
2. (Que.3/P.365) dx
dt = 2t, dy
dt = 3x+ 2t, dz
dt =x+ 4y+t [Ans : x=t2 +c1, y=t3+t2+3c1t+c2,
z=t4+53t3+ 6c1+ 12
t2+ (c1+4c2)t+c3] 3. (Que.4/P.365) dx
dt =x+ sint, dy
dt =t−y
[Ans : x=c1et− 12(sint+ cost), y=t−1+c2e−t]
Lecture No. 05
System of Linear Differential Equations (Cont.)
(A topic of Unit III)
(B) Method of Elimination: In this method we solve the system of LDE’s by eliminating one of the dependent variables x(t) or y(t). Thus we get a single differential equation in which x(t) or y(t) is dependent variable and t is independent variable. Solving it we get the value of one dependent variable and hence another dependent variable is obtain by substitution. But in order to balance the number of arbitrary constants in the general solution, we use the above mentioned observation.
Example 7.3/ P. 313 Solve 2dx
dt −x+ dy
dt + 4y = 1, dx dt − dy
dt =t−1 (0.1)
Solution. Set D= dtd, we have from Eqs. (0.1), (2D−1)x+ (D+ 4)y= 1
Dx−Dy=t−1
(0.2)
To eliminate one of the dependent variable from (0.2), we multiply first equation by D and second equation by (D+ 4) of (0.2). After summing up
1
them, i.e.,
D(2D−1)x+D(D+ 4)y = D(1) = 0
D(D+ 4)x−D(D+ 4)y = (D+ 4)(t−1) = 4t−3 i.e.,
(3D2+ 3D)x= 4t−3 (0.3)
which is second order non-homogeneous LDE. Therefore the general solution of (0.3) is
x(t) =C.F +P.I =c1+c2e−t+ 2 3t2−7
3t (0.4)
To find the other dependent variable, substitute the value of x(t) to the second equation of (0.2), we get
Dx−Dy=t−1 D
c1+c2e−t+ 2 3t2− 7
3t
−Dy=t−1
0−c2e−t+4 3t− 7
3
−Dy=t−1 Dy=
−c2e−t+4 3t− 7
3
−(t−1) Dy=−c2e−t+ t
3 −4 3 y(t) = 1
D
−c2e−t+ t 3− 4
3
y(t) = c2e−t+t2 6 − 4t
3 +c3 (0.5) Since 4(D) =
(2D−1) (D+ 4)
D −D
= −3D2 −3D. Therefore the order of 4(D) is 2.
But from Eqs. (0.4) and (0.5), we have three arbitrary constants in the general solution. Therefore we need to balance the arbitrary constants in the general solution.
3 From equations (0.2), (0.4) and (0.5), we obtain
(2D−1)x+ (D+ 4)y = 1 (2D−1)
c1+c2e−t+ 2 3t2− 7
3t
+ (D+ 4)
c2e−t+t2 6 − 4
3t+c3
= 1 After simplification, we get
c3 = c1+ 7 4 Substitute the value of c3 in (0.5), we have
y(t) =c2e−t+ t2 6 − 4
3t+c1+ 7
4 (0.6)
Hence the pair of function x(t), y(t)defined by Eqs. (0.4) and (0.6), contains only two arbitrary constants and is the general solution of (0.2).
Example 7.5/P. 315 Solve
(D+ 3)x+ (D+ 1)y=et, (D+ 1)x+ (D−1)y=t (0.1) Solution. To eliminate one of the dependent variable from (0.1), we multiply first equation by −(D−1)and second equation by (D+ 1) of (0.1).
−(D−1)(D+ 3)x−(D−1)(D+ 1)y = −(D−1)(et) = 0 (D+ 1)(D+ 1)x+ (D+ 1)(D−1)y = (D+ 1)t = 1 +t After adding and performing the operator on them, we obtain
−(D−1)(D+ 3)x+ (D+ 1)(D+ 1)x = 1 +t
−(D2+ 3D−D−3)x+ (D2+ 2D+ 1)x = 1 +t
−(D2+ 2D)x+ 3x+ (D2+ 2D)x+x = 1 +t 4x = 1 +t i.e.,
x(t) = 1
4(t+ 1) (0.2)
To find the other dependent variable, substitute the value of x(t) to the first equation of (0.1), we get
(D+ 3)x+ (D+ 1)y = et (D+ 3)
1
4(t+ 1)
+ (D+ 1)y = et D
1
4(t+ 1)
+ 3 4t+3
4 + (D+ 1)y = et 1
4+ 3 4t+3
4 + (D+ 1)y = et
(D+ 1)y = et−1− 3 4t
which is first order non-homogeneous LDE with constant coefficients. There- fore the general solution is
y(t) =C.F +P.I =c1e−t+et 2 −1
4 − 3
4t (0.3)
Since 4(D) =
(D+ 3) (D+ 1) (D+ 1) (D−1)
= (D+ 3)(D−1)−(D+ 1)2 = −4.
Therefore the order of 4(D) is 0.
But from Eqs. (0.2) and (0.3), we have one arbitrary constant in the general solution. So we need to eliminate it from the general solution.
Therefore from equations (0.1), (0.2) and (0.3), we obtain (D+ 1)
t 4 +1
4
+ (D−1)
c1e−t+ et 2 − 1
4− 3 4t
=t. (0.4) On simplification, we get
t−2c1e−t =t which will be an identity in it only when c1 = 0.
Hence the pair of function x(t) = 1
4(t+ 1), y(t) = et 2 − 1
4− 3 4t
contains no arbitrary constant and is the general solution of (0.1).
5
Home Assignment
Solve the following system of equations:
1. (Example 7.4/P.314)(D−1)x+(D+1)y= 0, (2D+2)x+(2D−2)y= t
[Ans : x(t) = 16t2 + t8 +c1, y= 16t2 −8t +c1] 2. (Que.8/P.366) d2x
dt2 +x−d2y
dt2 −y =−cos 2t, 2dx dt − dy
dt −y= 0 3. (Que.10/P.366) dx
dt +dy
dt +y=t, d2x dt2 +d2y
dt2 + dy
dt +x+y=t2 [Ans : x(t) = t2+t−1, y=−t]
System of Linear Differential Equations (Cont.)
(A topic of Unit III)
(c) Triangular Method: A system either of the forms P1(D)x =Q1(t) P3(D)x+P4(D)y=Q2(t)
(0.1) or,
P2(D)y=Q1(t) P3(D)x+P4(D)y=Q2(t)
(0.2) is called a triangular system of LDE’s.
In such system, we solve first equation of (0.1) (or, (0.2)) to obtain x(t) (or, y(t)). Thereafter putting obtained value of x(t) (or,y(t)) in second equation of the system (0.1) (or, (0.2)) and then solving it, we get y(t) (or, x(t)).
Remark. In a triangular system, the number of arbitrary constants involved in the general solution x(t), y(t) are already balanced.
An Equivalent Triangular System: Consider the system p1(D)x+p2(D)y=q1(t)· · ·(a)
p3(D)x+p4(D)y=q2(t)· · ·(b)
(0.3)
1
2
Retain any one equation of system (0.3), say (a). Now multiply (a) by any arbitrary operatork(D) and add it to (b), i.e., {(a)×k(D) + (b)}. Then, the system (0.3) can be transformed into the following triangular system either
p1(D)x+p2(D)y=q1(t)
¯
p(D)x = ¯q(t)
(0.4)
or
p1(D)x+p2(D)y=q1(t)
¯
p(D)y= ¯q(t)
(0.5)
The system (0.4) or (0.5) is called an equivalent triangular system to (0.3) as both the systems has same order of determinant. We can solve it easily by using triangular method, and hence the obtained solution of (0.4) or (0.5) is also the solution of (0.3).
Example 7.6/P. 317Solve
(3D2+ 3D)x = 4t−3· · ·(1) (D−1)x−D2y=t2· · ·(2)
(0.1)
Solution. Equation (1) of the system (0.1) is second order non-homogeneous
LDE.
(3D2+ 3D)x= 4t−3 The A.E. is
3m2 + 3m= 0
⇒ m= 0,−1
∴C.F. =c1+c2e−t and
P.I.= 1
P(D)Q(t) = 1
3D2+ 3D(4t−3)
= 1 3D
1
D+ 1(4t−3)
= 1 3D
(1 +D)−1(4t−3)
= 1 3D
(1−D+D2− · · ·)(4t−3)
= 1
3D(4t−7)
= 1 3
4t2
2 −7t
= 2 3t2− 7
3t
Therefore the general solution of (1) is
x(t) =c1+c2e−t+2 3t2− 7
3t (0.2)
On substituting the value of x(t) in the equation (2) of the system (0.1), we
4 obtain
(D−1)x−D2y = t2 (D−1)
c1+c2e−t+2 3t2− 7
3t
−D2y = t2 D
c1+c2e−t+ 2 3t2− 7
3t
−
c1+c2e−t+2 3t2− 7
3t
−D2y = t2
0−c2e−t+4 3t− 7
3
−
c1+c2e−t+2 3t2− 7
3t
−D2y = t2
−c1−2c2e−t−2
3t2+ 11 3 t− 7
3
−D2y = t2 D2y=−7
3−c1−2c2e−t + 11 3 t− 5
3t2 y= 1
D2 − 7
3−c1−2c2e−t + 11 3 t− 5
3t2 and hence
y(t) = c4+c3t− 7
6+ c1 2
t2+11
18t3− 5
36t4−2c2e−t (0.3) Therefore the pair of functions x(t) and y(t) in Equations (0.2) and (0.3) is the general solution of the system (0.1).
Remark: Here 4(D) =
3D2+ 3D 0 D−1 −D2
= −3D4−3D3. Therefore the order of 4(D) is 4, which is the same as number of arbitrary constants in the general solution of the system (0.1). Therefore in triangular method, no need to balance the arbitrary constants.
Example 7.8/P. 318Solve the system
(D+ 4)x+Dy= 1· · ·(1) (D−2)x+y=t2· · ·(2)
(0.1)
Solution. We retain the second equation and change the first equation as {−D×(2)+(1)}in the system (0.1), we get the following equivalent triangular
system
−D{(D−2)x+y}+{(D+ 4)x+Dy}=−D(t2) + 1 (D−2)x+y =t2
or
(−D2+ 3D+ 4)x=−2t+ 1· · ·(10) (D−2)x+y=t2· · ·(20)
(0.2) Therefore the general solution of the equation(10) in the system (0.2) is
x(t) =c1e4t+c2e−t− t 2 +5
8 (0.3)
Substitute the value of x(t) in (20) of (0.2), we have
(D−2)x+y=t2 (D−2)
c1e4t+c2e−t− t 2+ 5
8
+y=t2 D
c1e4t+c2e−t− t 2 +5
8
−2
c1e4t+c2e−t− t 2+ 5
8
+y=t2
2c1e4t−3c2e−t+t− 7 4
+y=t2 and hence
y(t) = −2c1e4t+ 3c2e−t+t2−t+7
4 (0.4)
Therefore the pair x(t) and y(t) defined in the Equations (0.3) and (0.4) is the general solution of the system (0.1).
Home Assignment 1. (Example 7.7/P.317) Solve
(3D−1)x+ 4y=t, Dx−Dy=t−1.
[Ans : x(t) =, y=]
6
2. (Example 7.9/P.319)Solve
(D+ 1)x+ (D+ 1)y= 1 D2x−Dy=t−1.
3. (Que. 7/P.366)Solve
d2x
dt2 + 4x= 3 sint dx
dt − d2y
dt2 +y = 2 cost.
4. (Que. 9/P.366)Solve
d2x dt2 − dy
dt = 1−t dx
dt + 2dy
dt = 4et+x.
System of Linear Differential Equations (Cont.)
(A topic of Unit III)
Degenerate System of LDE’s: Consider the system p1(D)x+p2(D)y=q1(t) p3(D)x+p4(D)y=q2(t)
(0.1)
such that 4(D) =
p1(D) p2(D) p3(D) p4(D)
= 0, that is, the system is degenerate.
In such a system, we eliminate x(t) or y(t) and conclude the following ones:
(i) if R.H.S of system is nonzero, then there is no solution
(ii) if R.H.S of the system is zero, then there are infinitely many solutions.
Example 7.10/P. 319 Show that the system
Dx−Dy = t (0.1)
Dx−Dy = t2 (0.2)
is degenerate and find the number of solutions it has.
1
2
Solution. Here p1(D) = D, p2(D) = −D, p3(D) = D and p4(D) = −D.
Since
4(D) =
p1(D) p2(D) p3(D) p4(D)
=
D −D D −D
=−D2+D2
= 0 and hence the system is degenerate.
Now, we eliminate x or y using (0.1)−(0.2) =⇒ 0 =t−t2
as R.H.S of the above equation is nonzero, therefore the system has no solution.
Example 7.11/P. 320Solve the system
Dx−Dy = t (0.1)
4Dx−4Dy = 4t (0.2)
Solution. Here p1(D) =D, p2(D) =−D, p3(D) = 4D and p4(D) = −4D.
Since
4(D) =
p1(D) p2(D) p3(D) p4(D)
=
D −D
4D −4D
=−4D2+ 4D2
= 0 and hence the system is degenerate.
Now, we eliminate x or y using 4×(0.1)−(0.2) =⇒ 0 = 0.
As right hand side of the above equation is zero. Therefore the system has infinite number of solutions.
Home Assignment
Verify that each of the following systems is degenerate. Find solutions if they exist:
1. (Que. 12/P. 366) Dx+ 2Dy =et,Dx+ 2Dy=t.
[Ans : No Solutions]
2. (Que. 14/P. 366)(D2−1)x+(D2−1)y = 0, (D2+4)x+(D2+4)y= 0.
[Ans : Infinitely many Solutions]
∗ ∗ ∗ ∗ ∗End of Unit 3∗ ∗ ∗ ∗∗
