A second order differential equation is called linear if it can be written as
and nonlinear if it cannot be written in this form.
The above equation is linear in the unknown function y and its derivatives, whereas p and q as well as r may be given function of x .
If r (x) = 0 , then
y" + p (x) y' q (x) y = 0 is called homogeneous.
If r (x) ≠ 0 , then
y" + p (x) y' + q (x) y = r(x) is called nonhomogeneous.
Fundamental Theorem for homogeneous Equation
For a homogeneous linear differential equation any linear combination of two solutions is a again a solution.
If y_{1} and y_{2} are solutions of homogeneous linear differential equation, then
y = c_{1}y_{1} + c_{2}y_{2} (Linear combination)
is also a solution of the homogeneous linear differential equation.
Note: It does not hold for nonhomogeneous linear equations or nonlinear equations.
General Solution and Initial Value Problem
For second order homogeneous linear differential equations a general solutions is of the form
y = c_{1}y_{1} + c2y2 (c_{1} ,c_{2} are arbitrary constants)
An initial value problem now consists of
y" + p (x) y'+ q(x) y = 0 and two initial conditions y (x_{0}) = K_{0} and y'(x_{0}) = K_{1} .
Using these conditions we can find constants c_{1} and c_{2}.
A basis of solutions is a pair y_{1} and y_{2} of linearly independent solutions.
Two functions y_{1 }(x) and y_{2} (x) are linearly independent if
k_{1}y_{1 }(x) + k_{2}y_{2} (x) = 0 ⇒ k_{1 }= 0,k_{2} = 0 .
and we call them linearly dependent if k1y1 (x) + k_{2} y_{2} (x) = 0 for some constant k_{1} ,k_{2} not both zero.
How to Obtain Basis if One Solution Is Known (Reduction of Order)
To get y_{2} , set y_{2} = uy_{1} . Then y^{'}_{2 }= u'y_{1} + uy'_{1} and y'2 = u''y'_{1} + 2u'y'_{1 }+ uy''1.
Substituting y_{2 }, y'_{2} and y''_{2 }in y'' + p(x) y' + q (x) y = 0
Collecting terms in u'',u' and u, we have
Since y_{1} is a solution then
Put u' = U and u'' = U' then
this is the desired firstorder equation.
The quotient Can not be constant (because U ≠ 0 ), so that y_{1} and y_{2} form a basis.
Example 18: If one solution of the secondorder homogeneous linear equation x^{2}y''  xy + y = 0 is y1 = x , then find a basis solution.
Example 19: If one solution of the secondorder homogeneous linear equation
y''  2xy' + 2αy = 0 is y_{1} (x) = 1 for α = 0
Then find a basis solution.
∵ y''  2xy' + 2αy = 0
Such an integral is easy to find when limits are specified.
Consider
y'' + ay' + by = 0 …….(1)
where coefficient a and b are constant.
Let us assume its solution is y = e^{λx}
Substitute y = e^{λx} , y' = λe^{λx} and y'' = λ^{2}e^{λx} in equation (1), we will get
(λ^{2} + aλ + b) e^{λx} = 0
Hence y = e^{λx} is a solution of (1) if λ is a solution of quadratic equation
This equation is called the characteristic equation (or auxiliary equation) of (1). Its roots are
This shows that y1 = e^{λ1x} and y2 = e^{λ2x} are solution of (1).
Let us consider three different cases
Case I: Two Distinct Real Roots λ^{1} and λ^{2}
In this case,
y_{1} = e^{λ1x} and y_{2} = e^{λ2x}
^{Constitute a basis of solutions of (1). The corresponding general solution is }
^{}
Example 20: Solve y'' + y'  2y = 0 with y (0) = 4 , y' (0) = 5 .
The characteristic equation is
λ^{2} + λ  2 = 0 ⇒λ = 1, 2.
Thus general solution is
y = c_{1}e^{x} + c_{2}e^{2x} .
∵ y = c_{1}e^{x} + c_{2}e 2x ⇒ y'= c_{1}e^{x } 2c_{2} e^{2x}
∵ y(0) = 4 ⇒ c_{1}+ c_{2} = 4 and _{∵} y'(0) = 5 ⇒ c_{1 } 2c_{2} = 5
Thus c_{1} = 1 and c_{2} = 3 .
Thus, general solution is
y = e^{x} + 3e ^{2x} .
Case II: Real Double Root
If the discriminant a^{2}  4b = 0, then λ = λ_{1} = λ_{2} = Hence only one solution
y_{1} = e ^{(a/2)x }.
Let us obtain a second solution y2 of y'' + ay' + by = 0,
Thus a basis of solutions of y'' + ay + by = 0 is
e ^{(a /2)x,} xe ^{(a/2)x} .
The corresponding general solution is
Example 21: Solve y"  4y' + 4y = 0 with y (0) = 3 , y' (0) = 1 .
The characteristic equation is
λ^{2 } 4λ + 4 = 0 ⇒ λ = 2, 2 .Thus general solution is
y = (c_{1} + c_{2}x) e^{2x} .
∵ y = c_{1} + c_{2}x) e^{2x} ⇒ y' = c_{2}e^{2x} + 2(c_{1} + c_{2}x) e^{2x}
∵ y (0) = 3 ⇒ c_{1} = 3 and ∵ y' (0) = 1 ⇒ c_{2 }+ 2c_{1} = 1
Thus c_{1} = 3 and c_{2} = 5 .
Thus general solution is y = (3  5x) e^{2x }.
Case III: Complex Roots
If the discriminant a^{2}  4b < 0 , then
Thus
Hence
The corresponding general solution is
Example 22: Solve y" + 0.2y' + 4.01y = 0 with y (0) = 0 , y' (0) = 2 .
The characteristic equation is
λ^{2} + 0.2λ + 4.01 = 0 ⇒ λ = 0.1 + 2i
Thus general solution is y = e ^{0.1x} (A cos 2 x + B sin 2 x) .
∵ y = e^{0.1x} (A cos 2 x + B sin 2x) ⇒ y' = B (0.1e ^{0.1x} sin 2x + 2e ^{0.1x} cos 2 x)
∵ y (0) = 0 ⇒ A = 0 ⇒ y = Be ^{0.1x} sin 2 x
and ∵y'(0) = 2 ⇒ 2B = 2 ⇒ B = 1
Thus general solution is y = e^{0.1x} sin 2x .
Example 23: Solve y'' + ω^{2}y = 0 .
The characteristic equation is λ^{2} + ω^{2} = 0 ⇒ λ = + iω
Thus general solution is y = (A cos ωx + B sin ωx) ∵ a = 0 .
Summary of Cases IIII
y'' + ay' + by = 0
Substitute x = e^{z} ,
Similarly
Thus x^{2}y"+ axy' + by = 0 ⇒ [D(D  1) + aD + b] y = 0 ⇒ [D^{2 }+ (a  1) D + b] y = 0
This is second order homogeneous equation with constant coefficients.
Example 24: Solve x^{2}y'' y"  2.5 xy'  2.0 y = 0 .
compare with x^{2}y'' + axy'+ by = 0 , then a = 2.5 and b = 2.0
Substitute x = e^{z} ⇒ y'' + (a  1)y' + by = 0 where
⇒ y''  3.5y'  2.0 y = 0
The characteristic equation is
λ^{2}  3.5λ  2.0 = 0 ⇒ λ = 0.5, 4
Thus general solution is
Example 25: Solve x^{2} y"  3xy' + 4y = 0 .
compare with x^{2}y" + axy' + by = 0 , then a = 3 and b = 4
Substitute x = e^{z} ⇒ y'' + (a  1)y' + by = 0 where
⇒ y"  4y' + 4y = 0
The characteristic equation is
λ^{2}  4λ + 4 = 0 λ = 2, 2
Thus general solution is
y = (c_{1} + c_{2}z)e^{2z} = (c_{1 }+ c_{2} In x) x^{2}.
Example 26: Solve x^{2}y" + 7 xy' + 13y = 0 .
compare with x^{2}y" + axy' + by = 0 , then a = 7 and b = 13
Substitute x = e^{z} ⇒ y'' + (a  1) y' + by = 0 where
⇒ y + 6y' +13y = 0
The characteristic equation is
λ^{2} + 6λ + 13 = 0 ⇒ λ = 3 + 2i
Thus general solution is
y = e^{ 3z} (A cos 2z + B sin 2z) = x^{3} [A cos 2 (2 ln x) + B sin (2 ln x)] .
Consider homogeneous equation
y'' + p (x) y' + q(x) y = 0
and two initial conditions y(x_{0}) = K_{0} and y' (x0) = K_{1} .
Its general solution is
y = c_{1}y_{1} + c_{2}y_{2}
Using these conditions we can find constants c_{1} and c_{2} .
If y_{1} and y_{2} are two solutions of homogeneous equation, then Wornskian is defined by
Note:
The two solutions y_{1} and y_{2} are linearly dependent if and only if their Wornskian is zero. If Wornskian is not zero then, then y_{1} and y_{2 }are linearly independent.
Example: Solution of y'' + ω^{2}y = 0 is y_{1} = cos ωx and y2 = sin ωx . Check whether they are linearly dependent or independent.
Solution: Wornskian is defined by
If ω ≠ 0 , they are linearly independent.
Example 27: Solution of y''  2y' + y = 0 is y = (c1 + c_{2}x) e^{x} . Check whether they are linearly dependent or independent.
Wornskian is defined by
They are linearly independent.
Finding a Second Solution
Consider homogeneous equation
y" + p(x) y' + q (x) y = 0
If y_{1} and y_{2} are two independent solutions, then Wronskian is defined by
By differentiating Wronskian, we will get
(y1 and y2 are two independent solutions)
Consider nonhomogeneous linear differential equation
y" + p (x) y' + q(x) y = r (x) …….(1)
General Solution
A general solution of the nonhomogeneous equation (1) is of the form
y(x) = y_{h} (x) + y_{p} (x)
where y_{h} (x) = c_{1}y_{1}(x) + c_{2}y_{2}(x) is a general solution of the homogeneous equation
y'' + p(x) y' + q(x) y = 0 and y_{p} (x) is any solution containing no arbitrary constants.
Methods of Finding y_{p} (x) :
(a)
Example 28: Find particular integral for
Example 29: Find particular integral for
Example 30: Find particular integral for
∵ f (D) = D^{2}  6D + 9 ⇒ f (3) = 9  18 + 9 = 0, f' (3) = 6  6 = 0
⇒ f'' (D) = 2
(b) and then operate.
Example 31: Find particular integral for (D^{2} + 5D + 4)y = 3  2 x .
(c)
Example 32: Find particular integral for (D^{2} + 4)y = sin 3x .
Example 33: Find particular integral for (D^{2 }+ 4)y = cos 2x .
Example 34: Find particular integral for (D^{2} + D + 1)y = cos 2 x .
(d)
Example 35: Find particular integral for (D^{2}  4D + 4)y = x^{3}e^{2x}.
Example 36: Find particular integral for (D^{2}  5D + 6)y = e^{x} cos_{ }^{2x}.
Example 37: Find particular integral for (D^{2}  6D + 13)y = 2^{x} .
Example 38: Find particular integral for
Example 39: Solve y" + 2y' + 101y = 10.4e^{x} with y(0) = 1.1 , y'(0) = 0.9 .
The characteristic equation is
λ^{2 }+ 2λ + 101 = 0 ⇒ λ = 1 + 10i .
Hence a/2 = 1 and ω = 10
Thus solution is
Thus general solution is
y = y_{h} (x) + y_{p} (x) = e^{x} (A cos10 x + B sin 10 x) + 0.1e^{x}∵ y (0) = 1.1 ⇒ A + 0.1 = 1.1 ⇒ A = 1
and ∵ y' (0) = 1 + 10B + 0.1 = 0.9 ⇒ B = 0
Thus Particular solution is
y = e^{x} cos10x + 0.1e^{x} .
Example 40: Solve differential equation y" + 4 y = 8x^{2} .
The characteristic equation is
λ^{2} + 4 = 0 ⇒ + 2i .
Hence a/2 = 0 and ω = 2
Thus solution is
⇒ y_{p} (x) = 2x^{2}  1
Thus general solution is
y = y_{h }(x) + y_{p} (x) = Acos2x + Bsin 2x + 2x^{2} + 1
Example 41: Solve differential equation y"  3y' + 2y ⇒ e^{x} .
The characteristic equation is
λ^{2}  3λ + 2 = 0 ⇒ λ = 1,2.
Thus solution is
Thus general solution is
Example 42: Solve y" + 2y' + y = e ^{x} with y(0) = 1 , y' (0) = 1 .
The characteristic equation is
λ^{2} + 2λ +1 = 0 ⇒ λ = 1, 1.
Thus solution is
Thus general solution is
and
Thus particular solution is
Example 43: Solve
y" + 2 y' + 5y = 1.25e^{0.5x} + 40 cos 4x  55 sin 4 x with y (0) = 0.2 , y' (0) = 60.1 .
The characteristic equation is
λ^{2} + 2λ + 5 = 0 ⇒ λ = 1 + 2i .
Hence,
Thus, the solution is
Hence,
Thus, general solution is
y = yh (x) + y_{p} (x) = e^{x} (Acos 2x + B sin 2x) + 0.2e^{0.5x} + 5sin 4x
∵ y(0) = A + 0.2 = 0.2 A = 0
⇒ y = Be ^{x} sin 2 x + 0.2e^{0.5x} + 5 sin 4 x
and y' = e ^{x} (B sin 2x + 2 B cos 2 x) + 0.1e^{0.5x} + 20 cos 4 x
∵ y' (0) = 2B + 0.1 + 20 = 60.1 ⇒ B = 20
Thus particular solution is y = 20e ^{x} sin 2x + 0.2e^{0.5x} + 5 sin 4 x
When a linear homogeneous secondorder ODE is written in the form
y'' + p (x) y' + q (x) y = 0 ……..(1)
points x_{0} for which p (x) and q (x) are finite are termed ordinary points of the ODE.
However, if either p (x) or q (x) diverges as x → x_{0} , the point x0 is called a singular point. Singular points are further classified as regular or irregular (or essential singularities):
These definitions hold for all finite values of x_{0} .
To analyze the behavior at x → ∞ we set x = 1/z, substitute into the differential equation, and examine the behavior in the limit z → 0.
The equation (1) is originally in the dependent variable y (x) , will now be written in terms of w (z) where w (z) = y (z^{1}) . Thus
Substituting y' and y" in equation (1), we get
From equation (4), we see that the possibility of a singularity at z = 0 depends on the behavior of
If these two expressions remain finite at z = 0 , the point x = ∞ is an ordinary point. If they diverge no more rapidly than respectively, x = ∞ is a regular singular point; otherwise it is an irregular singular point (or essential singularity).
Example 44: Consider Bessel’s equation x ^{2}y'' + xy' + (x^{2}  n^{2}) y = 0 .
which shows that the point x = 0 is a regular singularity.
By inspection we see that there are no other singularities in the finite range. Let us check at x → ∞ (z → 0) for coefficient
Since the latter expression diverges as the point x = ∞ is an irregular or essential singularity.
Singularities of Some Important ODEs
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1. What are singular points in linear differential equations of second order? 
2. How can singular points affect the solutions of linear differential equations of second order? 
3. What are the types of singular points in linear differential equations of second order? 
4. How can we determine the nature of a singular point in linear differential equations of second order? 
5. Can singular points affect the existence and uniqueness of solutions in linear differential equations of second order? 

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