Analytical solutions of some mechanics problems by Elzaki transform

Objectives: To discuss the applicability of new kind of transform technique named as Elzaki transform in solving the higher order ordinary linear differential equations occurred in various categories of vibrations in the field of engineering mechanics. Methods: We have selected some problems in the field of engineering mechanics to study the solutions with the help of Elzaki transform technique. We have used Elzaki transform method and also MATLAB to draw graphs for the obtained solutions of selected problems. Findings: We have successfully applied this new transform technique in finding the analytical solutions and compared with the graphs drawn with MATLAB. Novelty/Applications: We may apply this new kind of transform technique named as Elzaki transform to get the analytical solutions of differential equations.


Introduction
The differential equations have played a central role in every aspect of applied mathematics for long time and also the linear differential equations with constant coefficients find their important applications in the study of mechanical systems (1)(2)(3) .
Elzaki transform technique is particularly useful for finding solutions of mechanical problems and this technique is very effective for the solution of the response of a linear ordinary differential equation to the given initial data. The Elzaki transform may be used to solve intricate problems in engineering, mathematics and applied science without resorting to a new frequency domain (4)(5)(6) .

Definition of Elzaki transform
The Elzaki transform is defined for function of exponential order, we consider function where the constant M must be a finite number, k 1 , k 2 may be finite or infinite. The Elzaki transform of the function f (t) is defined as E{ f (t)} = T (v) = v . (1)(2)(3)(4)(5)(6) (1,2)

Ordinary linear differential equation with constant coefficients:
The general form of n th order linear differential equation with constant coefficients is d n y dt n +C 1 d n−1 y dt n−1 +C 2 d n−2 y dt n−2 + · · · +C n−1 dy dt + C n y = Q (t) where C 1 ,C 2 . . . C n real constants are and Q (t) is a function of the independent variable t (2,3) .

Elzaki transform to an ordinary differential equation
The Elzaki transform may be used to solve ordinary linear differential equations with initial or boundary conditions. Let y (0) = a 0 , y ′ (0) = a 1 , y ′′ (0) = a 2 , . . . . . . , y (n−1) (0) = a n−1 be the given initial or boundary conditions n th ordinary differential equation, where a 0 , a 1 , a 2 , . . . . . . , a n−1 are the constants. On taking the Elzaki transform on both sides of differential equation and also using these conditions, we obtain an algebraic equation known as subsidiary equation. The solution is the obtained by applying the inverse Elzaki transform (2,3) .

Solve the differential equation y
Now substituting the values given in the above, we have ( Further it can be written as Now taking inverse Elzaki transform on both sides of (2.2.2), we get

Solve
The given equation can be written as Using the derivatives of Elzaki transform, we get Now substituting the values in conditions at t=0, then After further simplification, it can be written as Now by taking inverse Elzaki transform of both sides of (2.3.2), we get

Solve the differential equation
Using the derivatives of transform, we have Substituting the values given in conditions, we get Further it can be written as Taking inverse Elzaki transform of both sides (2.4.2), we get https://www.indjst.org/

Applications to Mechanics
Here in this section we discuss solutions of some mechanics problems by using Elzaki transform technique.
3.1. Example : A particle with mass of 2 grams, which moves on X-axis and gets attracted towards the origin with a force 8X. If Q is initially at rest at X=10 then find its position at any time with an assumptions (i) No other external force acts on it (ii) A damping force equal to 8 times of the instantaneous velocity acts.
Now applying Elzaki transform to (3.1.1), we haveE{ d 2 X dt 2 } + 4E{X} = 0. Using the derivatives of Elzaki transform, then From the values given in conditions Taking inverse transform either sides of (3.1.2), we get , Further gives the solution as X (t) = 10 cos2t. (ii) In this case the equation of motion of a particle is 2 d 2 X dt 2 = −8X − 8 dX dt . It can be written as d 2 X dt 2 + 4 dX d t + 4X = 0 (3.1.3), with the initial conditions X (0) = 10 and X ′ (0) = 0.
Now by taking the Elzaki transform of both sides of (3.1.3), we have E{ d 2 X dt 2 } + 4E{ dX dt } + 4E{X} = 0. Using the derivatives of Elzaki transform, we get Substituting the values of initial conditions, Now by taking inverse Elzaki transform on either side of (3.1.4), we get the solution as X (t) = 10 e −2t + 20 te −2t .

Example :
A body of mass m moves along the X-axis, it is under the influence of a force which is directly proportional to its instantaneous speed and in a direction opposite to the direction of motion. Assuming that at t = 0 the particle is located at x = a and moving to the right with a speedV 0 , find the position where the mass comes to rest.

Solution:Here the equation of motion is m d
Substituting the values given in conditions, we have This can also be written as Now taking inverse Elzaki transform of both sides of (3.2.2), we get After simplification, the solution of the given differential equation is Hence the mass m comes to rest at a distance mV 0 µ + a from the center O.

Example :
A spring hangs vertically. A weight of 10 lb. is attached and the spring stretches 2 inches. The weight is replaced with a 60 lb. weight and allowed to come to rest in equilibrium. It is then pulled down 6 inches and released with velocity zero. The mass of the 60 lb. and weight is = , where g = the acceleration due to gravity.
Substituting these values in (3.3.1), we get This can be written as d 2 s dt 2 + gs = 0 (3.3.2) with the initial conditions are Taking the Elzaki transform of both sides of (3.3.2), we have https://www.indjst.org/ Substituting the initial conditions, we have Now applying inverse Elzaki transform of both sides of (3.3.3), we get is the solution.

Example :
A 10-kilogram mass is attached to a spring which is thereby stretched 0.7 meters from its natural length. The mass is started in motion from the equilibrium position with an initial velocity of 1 meter/sec in the upward direction. Find the resulting motion if the force due to air resistance is −90 dx dt newtons. Solution: The motion is damped (air resistance) and free (no external force). From Newton's Second law,m d 2 x dt 2 = −90 dx dt − kx (3.4.1), where m = the mass (10 kg), and k is the spring constant.
Substitute these values in (3.4.1), https://www.indjst.org/ with the initial conditionals x (0) = 0 and x ′ (0) = −1 (the initial velocity is in the negative x-direction). Now taking the Elzaki transform to (3.4.2), we have Substituting the values of conditions, we get Further it can be expressed as Applying inverse transform to (3.4.3), we get x (t) = − 1 5 e −2t + 1 5 e −7t , is the required solution.
https://www.indjst.org/ Taking the Elzaki transform to (3.5.2), we get E{ d 2 x dt 2 } + 4E{ dx dt } + 4E{x} = 0. Using the derivative of Elzaki transform substituting the values of conditions, Further it can be rewritten as Now by taking inverse Elzaki transform to above (3.5.3), we get is the solution. where x(t) is the position of the mass at t, and v (t) = dx(t) dt is its velocity. If the vibration is free, we mean there is no external force, F (t) . If the vibration is undamped, we mean there is no air or medium resistance force. Thus, the only force acting on our system is the force of the spring which by Hooke's law is kx. By Newton's second law, our differential equation is Now taking the Elzaki transform to (3.6.2), we have E  Now applying inverse Elzaki transform rewritten to (3.6.3), we get x 0 is called the amplitude of the motion, and

Conclusion
This new kind of Elzaki Transform technique is useful in finding the analytical solutions of differential equations occurred in the discipline of Engineering Mechanics.