Ordinary Differential Equations

This document provides solutions for ordinary differential equations and explains how to solve them step by step. It also offers access to solved assignments, essays, and dissertations on ordinary differential equations at Desklib.

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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