Solving Differential Equations
Differentialequations are constructed by observing the nature of the derivativeof the function at any point. For example, the velocity vof a moving object at any point is the derivative of the displacementfunction swith respect to timet.So, ds/dt= v is a simple case of differential equation.
Solvingdifferential equation means to find the overall function s.In this simple case, the solution is s = vt + C, where C is aconstant, referring the initial position of the object in this case.
Thereare many websites claiming solvedifferential equation onlineto give the solution instantly. Such a help is alright for someonewho needs the solution as one time requirement for some application.But in case of students of this subject, the techniques of solvingare important. let us take a closer look.
Inall cases the type of differential equations may not be as simple aswe described earlier and consequently their solutions too. Suppose adifferential equation dy/dx = f(x), then the approach to the solutionwould be different and dependent on the nature of the function f(x).
Henceto solve the differential equation, we have to study the type offunction on the right side. There are certain guidelines to followand let us discuss a few of them.
Thesolutions to first order differential equations are mostly simple andalmost intuitive especially for someone having knowledge in calculus.The only difficult case may be linear differential equations.
Thegeneral form of that type is, (dy/dx) + Py = Q, where P and Q may beconstants or functions of xonly.
Thesolution to this type is based on the concept of product rule ofdifferentiation and identifying a function g(x) in such a way, thesolution is y*g(x) = ∫Q*g(x)+ C. The function g(x) is called as the integrating factor and foundby the relation g(x) = e∫Pdx.
Nowlet us see how we could be a secondorder differential equation solver. The typical form of a secondorder differential equation is (d2y/dx2)+ P(dy/dx) + Qy = 0,where P and Q may be constants (including 0) oronly the functions of x.Nowlet us replace the derivatives by an operator ‘m’ the exponentsof which indicate the derivative position. That is, (d2y/dx2)= m2,(dy/dx) = m1= m and y = m0= 1.
Sowe form a characterized equation as m2+ Pm + Q = 0. Let m1and m2are the roots of this equation, then the solutions are,
y= C1em1x+ C2em2x,for distinct real roots.
y = (C1+ C2x)emx,for real identical roots
y= C1eaxcos(bx)+ C2embxeaxsin(bx),for imaginary roots.