Differential operator takes an action of finding derivative of a function.If we have y = f(x) then the derivative of y with respect to xis represented as, y’ = f’(x)
Or d/dx y = d/dx f(x) here d/dx is a differential operator and it takes an action of getting derivative of f(x) with respect to x.
Various forms to represent a differential operator:
d/dx , D, Dx, delx all these operators are used for first order derivatives of a function with respect to ‘x’.
in case of different variable i.e. t we can represent these operators as, d/dt, D, Dt, delt.
for nth order derivatives we mathematicsrepresent the differential operator using such notations: dn/dxn, Dn, Dxn
Derivative of a function f(x) can be written as: d/dx f(x) = f ‘(x) or [f(x)]’.
Linear differential operators with constant coefficient:
Let us consider an ordinary differential equation:
yn + a1yn-1 + a2yn-2 + ………..+an = q(x) where y is a function of x.
using differential operator d/dx we can write this equation as:
dn/dxn y + a1 dn-1/dxn-1 y + ……. + an = q(x)
or usinf D we get,
(Dn + a1Dn-1 + ………….+ an) y = q(x)
Now let us assume, Dn + a1 Dn-1 + ……. + an = p(D)
Where, p(D) represents a polynomial of differential operators with constant coefficient.
Hence we get more simple form as: p(D) y = q(x)
Here p(D) acts as a operator and operates on a function y = f(x).
Sum rule: differentiation is linear
D(f + g) = D(f) + D(g)
D(f – g) = D(f) – D(g)
Constant factor rule:
D(k * f) = k* D(f)
(D1 o D2)(f) = D1(D2(f))
xD is not equal to Dx.
D(c1f + c2g) = c1D(f) + c2D(g)
Solved examples using polynomial differential operator.
Characteristic equation for the above ordinary differential equation is: r2 – 5r + 6 = 0
Factorizing characteristic function we get,
(r – 3)(r – 2) = 0
Gives, r = 3 and r = 2
Hence general solution we get as: c1 * e^3x + c2 * e^2x
(D2 – 3D + 2)y = sin 2x
Characteristic equation: r2 – 3r + 2 = 0
Factorizing this equation we get,
(r – 2)(r – 1) = 0
Gives, r = 2 and r =1
Hence general solution we get as: c1 * e^x + c2 * e^2x
Particular solution for q(t) = sin t is of the form: A sin t + B cos t
Hence particular solution: c3 sin 2x + c4 cos 2x
Solution for the above give ordinary differential equation:
c1 * e^x + c2 * e^2x + c3 * sin 2x + c4 * cos 2x
Differential operator ‘del’ is an important vector differential operator.Del (triangular) = x del/ del x + y del/ del y + z del/ delz. Operator del is used to find divergence, curl and gradient of various objects.
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