The elementary Bessel functions from the Bessel equation x 2 u " + [ x 2 + 1 4 − v 2 ] u = 0 in example 3. Formula used: The Bessel’s equation is x 2 y " + x y ' + ( x 2 − v 2 ) y = 0 and its solution is y " + p ( x ) y ' + q ( x ) y = 0 . Calculation: The Bessel’s equation is given as x 2 u " + u ( x 2 − v 2 + 1 4 ) = 0 . From the example 3 the Bessel function is given as J 1 / 2 ( x ) = 2 π x sin x and J − 1 / 2 ( x ) = 2 π x cos x , here the value of v = ± 1 2 . Squaring the equation v = ± 1 2 on both sides. v 2 = ( 1 2 ) 2 = 1 4 Substitute the value of v 2 = 1 4 in x 2 u " + u ( x 2 − v 2 + 1 4 ) = 0 . x 2 u " + u ( x 2 − 1 4 + 1 4 ) = 0 x 2 u " + x 2 u = 0 x 2 ( u " + u ) = 0 u " + u = 0 Setting u 1 = cos x and u 2 = sin x are independent solutions. From problem 17 the equation is y = u x − 1 2 . Therefore, the value of independent solution is y 1 = x − 1 2 u 1 and y 2 = x − 1 2 u 2 . Substitute the value of u 1 = cos x and u 2 = sin x in y 1 = x − 1 2 u 1 and y 2 = x − 1 2 u 2 respectively. Hence, the value of y 1 = x − 1 2 cos x and y 2 = x − 1 2 sin x . Substitute the value of y 1 = x − 1 2 cos x and y 2 = x − 1 2 sin x in J 1 / 2 ( x ) = 2 π x sin x and J − 1 / 2 ( x ) = 2 π x cos x respectively. Thus, the value of J 1 / 2 ( x ) = 2 π x x − 1 2 sin x and J − 1 / 2 ( x ) = 2 π x x − 1 2 cos x .

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4. Now we'll look at a nonhomogeneous example. The general form for these is y' + p(x)y = f(x). For this problem, we will find solutions of the equation +2xy= xe (a) Identify p(x) and f(x) in the equation above. p(x) = f(x) = (b) The complementary equation is y' + p(x)y = 0. Write the complementary equation. (c) Find a solution for the complementary equation. We'll call this solution y₁. (You only need one particular solution, so you can let k = 0 here.) Y1 = (d) Check that y₁ satisfies the complementary equation, in other words, that y₁+ p(x)y₁ = 0.

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