The most general form of the linear second order differential equation with constant coefficients is: P (dy2/dt2 )+ Q ( dy/dt) + R (y) = f(t) (1)
Where P, Q, R are the constants.
The other form of linear second order differential equation is
a(t). (dy2/dt2 ) + b(t). ( dy/dt) + c(t).y = f(t) (2)
Equations 1st and 2nd both are the form of linear second order differential equation with the only difference that in the second form coefficients are not constant, they are the time variant.
The simplest differential equation is f(t) = 0. If f(t) = 0 then it is called as homogeneous and if f(t ) ≠ 0 then the differential equation is called as non-homogeneous. The problems, which we discuss here, are homogeneous linear second order Differential Equations.
P (dy2/dt2 )+ Q ( dy/dt) + R (y) = 0
It would be better to start with the linear, constant coefficient, homogeneous second order differential equation
Py” + Qy’ + Ry = 0
This is another way to write the linear homogeneous second order differential equation.
Some important points to define the linearity or non linearity of the differential equations are:
· The function f1(t) = sin t and f2(t) = t are linearly independent for the values of k1 and k2 for which, k1sint + k2 t = 0.
For all t values k1 = 0 and k2 = 0.
k1.t + k2.3t = 0
Will have the nonzero solution for k1 = -3 and k2 = 1.
· The characteristic equation (also named as auxiliary equation) of a linear second order differential equation
y” + Py’ + Qy = 0 is
p2 + a.p + q = 0
Have a look on the examples.
Example 1: Find out the solution of the given differential equation (dy/dt) – 9y = 0?
Solution: Solution of this problem can be obtained only by inspection. Here those Functions are needed whose second derivative is 9 times of the original function.
Two following functions can be considered as:
y(t) = e3t and the other one is y(t) = e-3t
Here one thing should be noticed that these are not the only solutions for this problem. Others can also be like,
y(t) = -9e3t y(t) = 123e3t
y(t) = 7e3t – 6e-3t y(t) = -92e3t -16e-3t
So it can be said that any function in the below form would be the solution for this problem
y(t) = k1 e3t + k2 e-3t
This example proved a very important thing about linear homogeneous second order differential equation that if y1(t) and y2(t) are the two linearly independent solutions for a homogeneous second order linear differential equation
p.y” + Q.y’ + R.y = 0
y(t) = k1 y1(t) + k2 y2(t) (3)
This is known as principle of superposition.
To determine the value of the constants k1 and k2 various methods are present like:
· One way is that specify the value of the function at two distinct points i.e. at t1 and t1
y(t0) = y0 y(t1) = y1
· The other way, specify the value of the function and its derivative at a particular Point like:
y(t1) = y0 y’(t1) = y1
These are called boundary values.
If the function cannot be guess then they can find with help of characteristic equation. For the given problem the characteristic equation is:
p2 – 9 = 0
p = ±3
These are the two roots so the solution is:
y1(t) = e3t and y2(t) = e-3t
This is the same as before.
Let’s solve the above problem with the following conditions.
y’ – 9y = 0
While the given conditions are y(0) = 2 and y’(0) = -1
Let’s have the two functions
y(t) = e3t and y(t) = e-3t
The general solution of this problem as said above is
y(t) = k1 y1(t) + k2 y2(t)
y(t) = k1 e3t + k2 e -3t
Now we just need to calculate the value of k1 and k2
The first derivative of this equation is
y’(t) = 3 k1 e3t +(-3) k2 e -3t
At t = 0
2 = k1 + k2
-1 = 3k1 + (-3)k2
Thus k1 = 5 / 6 and k2 = 7 / 6
So the solution of the given problem is
y(t) = 5/6 e3t +7/6 e -3t
Now if the coefficients are not constant then the differential equation is
a(t)y” + b(t)y’ + c(t)y = g(t)
y(t0) = y0 and y’(t0) = y’0
Here one theorem should be noticed for the question to find out a set of fundamental solutions will exist for a given problem
Let’s have a differential equation
y” + a(t)y’ + b(t) y = 0
Where a(t) and b(t) are continuous functions on some interval. In that interval select a point let it be t0.
y1(t) would be a solution which satisfies the initial condition,
y(t0) = 1 and y’(t0) = 0,
y2(t) would be a solution of a differential equation that satisfies the initial condition,
y(t0) = 0 and y’(t0) = 1,
Then y1(t) and y2(t) would form a fundamental set of solutions for the differential equation.
One other important thing about the linear differential equation is that the general solution os a linear Combination of two linearly independent solutionsy1 and y2 that implies that neither y1 nor y2 is a constant multiple of the other like if f(t) = t2 and g(t) = 9.t2 are linearly dependent but f(t) = et and
g(t) = t. et are linearly independent.
This can be understood better with the help of an example.
Example: With the help of the theorem defined above find a set of solutions for
(dy2/dx2) + 4 (dy/dx) + 3y = 0 (at t0 = 0)
Solution: The two solutions for this problem are
y(t) = e-3t and y(t) = e-t
The solution of this problem is same as discussed before
y(t) = k1 y1(t) + k2 y2(t),
y(t) = k1 e-3t + k2 e-t
On applying the initial condition as discussed in the theorem
y(0) = 1 and y’(0) = 0
After solving it the value of k1 and k2 can be obtained i.e. -1/2 and 3/2.
So the equation become
y(t) = (-1/2) e-3t + (3/2) e-t
With the same on applying the second set of the initial condition we get
y(0) = 0 and y’(0) = 1
y(t) = (-1/2) e-3t + (1/2) e-t
This can be further solved with the help of Wroskian theorem and the solution can be obtained which will not be equal to zero.
One another example is here to understand differential equation better
Example: Solve the differential equation given below dy2/dt2 + 6.(dy/dt) + 5y = 0 while the given conditions are y(0) = 4 and y’(0) = -1?
Solution: The characteristic equation for the given problem, p2 + 6p + 5 = 0 and roots are p = -1 and p = -5. So the solutions are e-t and e-5t.
y(t) = k1 e-t + k2 e-5t
y’(t) = -k1 e-t -5 k2 e-5t
On solving these equations k1 = 19/4 and k2 = -3/4.
y(t) = 19/4 e-t + -3/4 e-5t
There is an another theorem to solve second order differential equation which states
The solution of the differential equation y” + Py’ + Qy = 0 can be determined into one of the following three cases which will definitely depend on the solution of the characteristic equation
p2 + a.p + q = 0.
· Distinct real roots (if p1 ≠ p2) are of the characteristic equation then the general solution is
Y = k1ep1t + k2 ep2t
· If roots are equal i.e. if p1 = p2 of the auxiliary equation then the general solution is
Y = k1ep1t + k2 .t ep1t = (k1 + k2.t) ep1t
· If p1 = α + iβ and p2 = α – iβ are complex roots then the solution is
Y = k1eat . cos βt+ k2 eat sinβt
Various application of second order differential equations are present in daily life and in science. One of the many application of linear differential equation is describing the motion of an oscillating spring. Suppose that a mass of m is placed on the end of a vertically hanging spring and then set it to the motion then how its motion can be described. This can be done with the help of the use of differential equation. According to the second law of motion of Newton force f = ma where a is the acceleration and m is the mass of the object. Let r is the downward displacement from equilibrium then a = r”
mr” = -kx
r”+ (k/m)x = 0
Here Ω = √(k/m) where k is the spring constant.
Differential equations are useful in describing the force on the spring either it is in vertical or horizontal Position on a level surfaces. According to Hooke’s law which states that if a spring is stretched x units then it exerts a force on spring that is proportional to x, i.e.
F = -kx
This type of motion is called simple harmonic motion. If the spring is in the position of damping vibration like explained above then the force works opposite to the motion of the spring. Thus
Damping force = -C(dx/dt )
Here k is the damping constant.
md2x/dt2 = restoring force + damping force of the spring = (-kx) – C.(dx/dt)
This can be better understood with the help of below theory:
On considering a spring of length l in its natural position, a weight of mass m is hanging up with this spring and its length varies from l to L. let’s assume all forces, velocities and displacements in the downward direction positive and in the upward direction let’s assume negative. All measurements are considered from the equilibrium position of the spring. According the second law of Newton force
F = ma
Now a differential equation is required so that displacement of the object at any time t can be calculated. There are four forces that are working upon the object which we are going to calculate. Two forces will always act upon the object and other two are conditionals means will depend upon the situation.
The forces working upon the object are
1. Gravity force (fg)
The gravity force will always act upon the object i.e. fg = mg
2. Spring force (fs): The second force that is working upon the object is the spring force that can be calculated with the help of Hook’s law according to which spring exerts a force on the object.
fs = -k(L+ x)
Which is equal to spring constant k>0 times the displacement of the spring from its natural length. For the setup we are working upon the spring’s natural length is L+x.
If the object is at rest in its equilibrium position means it’s in rest then the displacement is L and the force will be fs = -kL where negative sign shows that it is in upward direction. If the spring will be stretched further from the rest position then L+x will be positive.
3. Damping force (fd): The third force which is acting upon the object is the damping force fd which is conditional that is may or may not be present.
Basically dampers work to counteract any movement. It’s better to discuss about it but first mathematically it’s defined as
fd = -βx’
Where β > 0 is a damping coefficient. If the object is moving downward then the velocity will be positive and fd will work to pull the object above again and it will be negative. If the object is moving upward then the velocity will be negative and fd will be positive i.e. in the opposite direction. Means it always act to counter the current motion of the object.
4. External force (f(t): If working on the object then force will be considered here and referred as f(t).
mx’ = mg – k(L+x)- βx’+ f(t)
mx’+βx’+kx = mg-Lk+ f(t)
If the object is at rest then two forces will be working on that object, the gravity force and the force due to spring. And these two forces cancel each other i.e.
mg = kL
Using it in the above equation
mx’+βx’+kx = f(t)
The initial conditions are obtained
X(0) = x(0)
X’(0) = x’(0)
First condition is the initial displacement from the equilibrium position and second condition is the initial velocity.
Some particular cases are here to discuss about
· Free undamped vibrations
· This is the simplest case, free vibration means f(t) = 0 means
mx’ + ku = 0
The characteristics equation has the roots
R = ±√(k/m)
And Ω0 = √(k/m)
Here Ω0 is called the natural frequency.
Skydiving is another application of the differential equation. The basic quantities to describe the motion of an object are position (x), velocity (v) and acceleration (a). Velocity is first time derivative of the position and acceleration is the first time derivative of the velocity thus acceleration is the second time derivative of the position(x). So position of an object can be determined with help of Newton’s second law of motion.
f = ma
f = m.(d2x/dt2)
Differential equations are useful in electronic circuits and in mechanical vibrations which are the important parts of science and engineering.
Application of Differential equations can be better understand with help of examples. Here some examples are presented regarding second order differential equation.
Example: Determine the general solution of the differential equation
(dy2/dt2 )+ 6 ( dy/dt) + 12y = 0 ?
Solution: This is an example of homogeneous differential equation. The characteristic equation is
p2 + 6p + 12 = 0
Two complex roots are there of this differential equation as follows
P = -6 ± √(36 - 48)/2
P = -6 ± √(-12)/2
P = -3 ± √(-3)
P = -3 ± √ (3i)
Thus α = -3 and β = √ (3i).
This provides the general solution i.e.
Y = k1e3x . cos √(3x)+ k2 e-3x sin√(3x)
Here in this example should be noticed that roots are complex but the solution of this differential equation is real.
Example: Determine the solution of the differentials equation given below with the given conditions y(0) = 2 and y’(0) = 1.
(dy2/dx2 )+ 4 ( dy/dtx) + 4y = 0?
Solution: First find out the auxiliary equation of the given differential equation which is:
p2 + 4p + 4 = (p+2)2 = 0
This equation has two equal roots at p= 2
Thus the general solution is given as
Y = k1e-2x + k2 xe-2x
Now because of the values y = 2 and x = 0, we got
2 = k1(1) + k2(0)(1) = k1
On solving the values y’ = 1 and x = 0, we got
k2 = 5
y = 2e-2x + 5 xe-2x
Let a1, a2, a3…….an and f be the function of x with a common Domain. An equation of the form
yn +a1(x)yn-1+ a2 (x) y n-2+…..+an-1(x)y’+ gn(x)y = g(x)
This is linear differential equation of order n. if g(x) = 0 the equation is homogeneous otherwise non homogeneous.