Answer to Question #255953 in Differential Equations for qaz

Question

Verify that the given functions form the fundamental set of solution of the differential equation on the given indicated interval.

2x^2y" +5xy' + y = x^2-x

y = C1x^-1/2 + C2x^-1 + 15x^2-1/6x, (-0, infinity)

Answer 1

$$2x^2y'' +5xy' + y = x^2-x$$

$$y = C_1x^{-1/2} + C_2x^{-1} +\dfrac{1}{15}x^2-\dfrac{1}{6}x, (0, \infin)$$

$$y'=-\dfrac{1}{2}C_1x^{-3/2} - C_2x^{-2} +\dfrac{2}{15}x-\dfrac{1}{6}$$

$$y''=\dfrac{3}{4}C_1x^{-5/2} +2 C_2x^{-3} +\dfrac{2}{15}$$

$$2x^2y'' +5xy' + y = x^2-x$$

$$2x^2(\dfrac{3}{4}C_1x^{-5/2} +2 C_2x^{-3} +\dfrac{2}{15})$$

$$+5x(-\dfrac{1}{2}C_1x^{-3/2} - C_2x^{-2} +\dfrac{2}{15}x-\dfrac{1}{6})$$

$$+C_1x^{-1/2} + C_2x^{-1} +\dfrac{1}{15}x^2-\dfrac{1}{6}x$$

$$=\dfrac{3}{2}C_1x^{-1/2} +4 C_2x^{-1} +\dfrac{4}{15}x^2$$

$$-\dfrac{5}{2}C_1x^{-1/2} -5 C_2x^{-1} +\dfrac{2}{3}x^2-\dfrac{5}{6}x$$

$$+C_1x^{-1/2} + C_2x^{-1} +\dfrac{1}{15}x^2-\dfrac{1}{6}x$$

$$=x^2-x, True\ x\in(0, \infin)$$