Answer to Question #105503 in Differential Equations for khushi

Question

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

Answer 1

$2x^2y'' + 3xy'-y=0, \quad x>0$

$y_1(x) = x^{-1}, \quad y_2(x)=?$

First, divide the equation by $2x^2.$ Obtain

$y'' + \frac{3}{2x}y' - \frac{y}{2x^2}=0$

Use Liouville's formula to determine a second linearly independent solution of the equation

$$y_2 = y_1\int{\frac{e^{-\int{P_1(x)}dx}}{y_1^2}dx}$$

where $P_1(x)=\frac{3}{2x}.$ Substitute $y_1 \text{ and } P_1$ into the formula.

$$y_2 = \frac{1}{x}\int{\frac{e^{-\int{3/(2x)dx}}}{\frac{1}{x^2}}dx} = \frac{1}{x}\int{x^2e^{-3/2\ln(x)}dx} = \frac{1}{x}\int{x^2\cdot \frac{1}{x^{3/2}}dx} \\ = \frac{1}{x}\int{x^{1/2}dx} = \frac{1}{x}\cdot \frac{2}{3}x^{3/2} = \frac{2}{3}\sqrt{x}$$

Therefore, the second linearly independent solution is $y_2(x) = \frac{2}{3}\sqrt{x}.$