Answer to Question #218075 in Calculus for Vic

Question

Find the general solution given y₁=x² is a particular solution of x^2y"-3xy'+4y=0.

Answer 1

Let $y_2(x)$ be  a second solution of $x^2y''-3xy'+4y=0.$

$$y''-\dfrac{3}{x}y'+\dfrac{4}{x^2}y=0.$$

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

$$y_2=x^2\int\dfrac{e^{\int {3 \over x}dx}}{x^4}dx=x^2\int\dfrac{x^3}{x^4}dx=x^2\ln x$$

The general solution on $(0, \infin)$ is given by

$$y=c_1y_1+c_2y_2$$

$$y=c_1x^2+c_2x^2\ln x$$