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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, D_{x}, del_{x} 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, D_{t}, del_{t}.
for nth order derivatives we mathematicsrepresent the differential operator using such notations: d^{n}/dx^{n}, D^{n}, D_{x}^{n}
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:
y^{n} + a_{1}y^{n-1} + a_{2}y^{n-2} + ………..+a_{n} = q(x) where y is a function of x.
using differential operator d/dx we can write this equation as:
d^{n}/dx^{n} y + a_{1} d^{n-1}/dx^{n-1} y + ……. + a_{n} = q(x)
or usinf D we get,
(D^{n} + a_{1}D^{n-1} + ………….+ a_{n}) y = q(x)
Now let us assume, D^{n} + a_{1} D^{n-1} + ……. + a_{n} = 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)
Composition rule:
(D_{1} o D_{2})(f) = D_{1}(D_{2}(f))
xD is not equal to Dx.
Linearity rule:
D(c_{1}f + c_{2}g) = c_{1}D(f) + c_{2}D(g)
Solved examples using polynomial differential operator.
Characteristic equation for the above ordinary differential equation is: r^{2} – 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: c_{1} * e^3x + c_{2} * e^2x
(D^{2} – 3D + 2)y = sin 2x
Characteristic equation: r^{2} – 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: c_{1} * e^x + c_{2} * e^2x
Particular solution for q(t) = sin t is of the form: A sin t + B cos t
Hence particular solution: c_{3} sin 2x + c_{4} cos 2x
Solution for the above give ordinary differential equation:
c_{1} * e^x + c_{2} * e^2x + c_{3} * sin 2x + c_{4} * 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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