Answer to Question #105666 in Differential Equations for ZEESHAN FAZAL

Question

given that y(x) =x^-1is one solution of the differential equation 2 x^2 y ′′ +3x y' − y = 0 x>0 find a second linearly independent solution of the equation

Answer 1

Let's assume that $y_2=x^{-1}v(x)$ is the second linearlyindependent solution of the equation.

Substituting it to the equation leads us to

$2x^2(v''x^{-1}-2v'x^{-2}+2vx^{-3})+3x(v'x^{-1}-vx^{-2})-vx^{-1}=0$

Since $x^{-1}$ is the solution of the equation, we get

$2x^2(v''x^{-1}-2v'x^{-2})+3xv'x^{-1}=0$

$2xv''-4v'+3v'=0$

$v''-\frac{v'}{2x}=0$

Assume

$v'=w$

We get

$w'-\frac{w}{2x}=0$

$\frac{dw}{w}=\frac{dx}{2x}$

$w=\sqrt{x}+c_1$

Therefore

$v=\int(\sqrt{x}+c_1)dx=\frac{2}{3}x^{3/2}+c_1x+c_2$

Assuming $c_1=c_2=0$ we get the second linearly independent solution in form $y_2=\frac{2}{3}x^{2/3}x^{-1}=\frac{2}{3}x^{-1/3}$