MA5512 Ordinary Differential Equations: Assignment on ODE Solutions

This document presents a comprehensive solution to an Ordinary Differential Equations (ODE) assignment, addressing key concepts and problem-solving techniques. The solution begins by tackling a Riccati equation, transforming it into a second-order ODE and deriving its general non-constant and two constant solutions. It then proceeds to solve Cauchy-Euler equations using both standard methods and the method of undetermined coefficients, demonstrating the application of these techniques to find general solutions. The assignment also explores modeling oscillations of a spring, deriving the general solution for the ODE and finding the particular integral using variation of constants. Furthermore, the solution utilizes the Frobenius method to solve a differential equation, providing a detailed step-by-step approach. This resource is valuable for students studying ODEs, offering clear explanations and practical examples to enhance understanding and improve problem-solving skills.

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Ordinary Differential Equations
Ordinary Differential Equations

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Ordinary Differential Equations
1.(a). Given Differential equation : dy
dx = y2 + 3y – 4
Substituting y(x) = −v' ( x)
v (x) ,
And also, differentiating y(x) with respect to x-
dy
dx = y’(x) = −v v'' +¿ ¿,
Put it in differential equation : −v v'' +¿ ¿ = (- v'
v )2 + 3(- v'
v ) – 4
-vv’’ + (v’)2 = (v’)2 - 3v’v -4v2
-vv’’ = -3v’v – 4v2
v’’ – 3v’ – 4v = 0 ...(1)
equation (1) is the desired Second Order ODE.
Now, from the differential equation, we have, dy
y2+3 y −4 = dx
dy
( y+4 ) ( y−1) = dx
dy
5 ( y + 4 ) − dy
5( y −1) = dx
Integrating both sides-
ln(5(y + 4))1/5 - ln(5(y – 1))1/5 = x + C (Using the property ln(xy) = yln(x) )
Here, C is the constant of integration. So,
y + 4
y−1 = e5(x + C) (Using the property ln(x) – ln(y) = ln
(x/y) )
Applying Componendo and Dividendo Rule-
2 y +3
5 = e5(x +C)+1
e5 (x+C )−1
Taking e5/ 2( x+C) common from Numerator and denominator-

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Ordinary Differential Equations
2 y +3
5 = e5/ 2(x +C)+e−5 /2( x+C)
e5 / 2(x+C )−e−5 /2 (x+C)
2 y +3
5 = coth(5/2(x + C))
y = ½( 5(coth(5/2(x + C))) – 3) is the required non constant solution.
Also,
Solving the above equation (1) by putting v’’ = D2 operator and v’= D, then, the
auxiliary equation will be-
(D2 – 3D – 4)v = 0
Solving this quadratic equation,
D = 3± √9+16
2 = 3± 5
2
D = 4, -1
Therefore, v(x) of the above second order ODE (eq.1) is ,
v(x) = Ae4x + Be-x
v’(x) = 4Ae4x –Be-x
Therefore,
y(x) = - 4 A e4 x−Be− x
Ae4 x + Be− x is the required two constant solution.
1.(b). Given Cauchy-Euler Equation : 2x2 d2 y
d x2 + 7x dy
dx - 18y = 0
Assuming the solution of this equation to be y(x) = xr only for x > 0, then,
We will get the equation in terms of r as-
2r(r – 1) + 7r – 18 = 0
Solving it-
2r2 + 5r – 18 = 0

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Ordinary Differential Equations
r = −5 ± √52 +4∗2∗18
2∗2 = 2, 9/2
so, solution of the given Cauchy Euler’s Equation is the linear combination of
individual solutions-
y(x) = Ax2 + Bx9/2
1.(c). Given Cauchy-Euler Equation : 2x2 d2 y
d x2 + 7x dy
dx - 18y = 11x + 9
Solving for the Non Homogeneous Part-
Substitute x = et, so, we have RHS as 11et + 9
From this form, we can guess the form of the solution as Y = Cet + F (C and F are
constants)
Therefore, we have,
2 d2 y
d t2 + 7 dy
dt – 18y = 11et + 9
Making the substitution d2 Y
d t2 = Cet and dY
dt = Cet for calculation of C and F,
2Cet + 7Cet – 18Cet – 18F = 11et + 9
Comparing the equivalent coefficients,
F = -0.5 and C = -11/9 So,
Total Solution of the whole equation is ytotal = y + Y
ytotal = Ae2t + Be9t/2 - 11
9 et – 1
2
2.(a). For f(t) = 0 in the given equation d2 x
d t2 + x = f(t) , according to Arfken, Weber &
Harris (2012), the general solution of the oscillatory differential equation takes form-
x = Asin(t) + Bcos(t)

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Ordinary Differential Equations
2.(b). For f(t) = sin(t), According to Sharma (2018), we can find the particular
integral(PI) as-
xp(t) = -x1(t)∫ x2 ( t ) f (t )
W ( x1 , x2 ) dt+ x2 (t )∫ x1 ( t ) f (t)
W ( x1 x2) dt
Here, W is the Wronskian : W(x1,x2) = x1(t)x2’(t) – x1’(t)x2(t)
From the Complementary function, x1 = Asint, x2 = Bcost
And W = -AB[(sint)2 + (cost)2] = -AB
So, xp(t) = sint∫ cost sint dt−cost∫¿ ¿
= ( si n3 t)
2 - tcost
2 + cost sin 2 t
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Complete solution is : x(t) = Asin(t) + Bcos(t) + ( si n3 t)
2 - tcost
2 + cost sin 2 t
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2.(c). The given Equation is : d2 x
d t2 + x = 0
According to Frobenious Method-
Let us assume the solution as x(t) = ∑
i=0
∞
ai ti+ s
Substituting into the equation will yield-
∑
i=0
∞
ai[(i + s)(i + s – 1) ti+s −2 + ti+s] = 0
Considering lowest degree coefficients-
aos(s – 1)ts – 2 + a1s(s + 1)ts – 1 + ∑
i=2
∞
ai(i + s)(i + s – 1) ti+s −2 + ∑
i=0
∞
ai ti+ s = 0
Shifting the indices in the first summation-

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Ordinary Differential Equations
aos(s – 1)ts – 2 + a1s(s + 1)ts – 1 + ∑
i=0
∞
¿¿ ¿(i + s + 2)(i + s + 1) + a i ]ti+s = 0
Here, all the coefficients in the power series must equal to zero. So,
s(s-1) = 0 i.e., s = 0 or s = 1
Considering 2nd term, if s = 0, regardless of a1, the coefficient of term a0s(s + 1)
vanishes. But if s = 1, we must put a1 = 0. Hence we set a1 = 0 regardless of the value
of s. Remaining coefficients are calculated using Recurrence Relation from the third
term-
a i + 2 = −ai
( i+s+ 2 ) (i+s+ 1) , i ≥ 0
So, we only have to take even numbered coefficients in the series
For s = 0, we can take i = 1, 2, 3,..
a2 = - a0
2
a4 = - a2
12 = a0
24 and so on,
Putting all these coefficients in ∑
i=0
∞
ai ti+ s will give the cosine series i.e.,
x(t) = a0cost
For s = 1, we can have i = 1, 2, 3,..
a2 = - a0
6
a4 = - a2
20 = a0
120 and so on,
Putting all these coefficients in ∑
i=0
∞
ai ti+ s will give the sine series i.e.,
x(t) = a0sint

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Ordinary Differential Equations
References-
Arfken G., Weber H. & Harris F.E.,(2012), Mathematical Methods for Physicists, 7th
edition. Cambridge: Academic Press
Sharma R.D. (2018). Mathematics (2019 edition). Dhanpat Rai Publication, New Delhi