Question

Solve

x ^ 2 * (d ^ 2 * y)/(dx) + x * (dy)/(dx) - 9y = 48x ^ 5

Answer 1

$x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}-9y=48x^5$

$D(D-1)y+Dy-9y=48x^5$

$(D^2-D+D-9)y=48x^5$

$(D^2-9)y=48x^5$

A.E;

$m^2-9=0\implies m=3,-3$

Put $x=e^z\therefore z=ln\ x$

C.F=$C_1e^{-3z}+C_2e^{3z}$

$P.I=\frac{1}{(D^2-9)}\cdot48e^{5z}$

$=\frac{1*48}{((5)^2-9)}\cdot e^{5z}=\frac{48}{16}\cdot e^{5z}$

$P.I=3e^{5z}$

Complete solution;

y=C.F+P.I

$y=C_1e^{-3z}+C_2e^{3z}+3e^{5z}$

$y=C_1e^{-3ln\ x}+C_2e^{3ln\ x}+3e^{5ln\ x}$

$y=\frac{C_1}{x^3}+C_2x^3+3x^{5}$ , where C₁ and C₂ are arbitrary constants