Question

Solve xd^2y/dx^2-(2x-1)dy/dx+(x-1)y=0

Answer 1

$x\frac{d^2y}{dx^2}-(2x-1)\frac{dy}{dx} +(x-1)y=0\\ \frac{d^2y}{dx^2}-(2-\frac{1}{x})\frac{dy}{dx}+(1-\frac{1}{x})y=0\\ Here, p=-2+\frac{1}{x}, q=1-\frac{1}{x}\\ Then,\\ 1+p+q=0 \implies y_1=e^x \, \text{is a solution of given differential equation.}\\ \text{ Now, put} \,y=vy_1=ve^x, \\ \text{ and corresponding }\,y' and \,y'' \text{ values in the given differential equation, we get}\\ \frac{d^2v}{dx^2}+{1}{x}\frac{dv}{dx}=0\\ Put\\ u=\frac{dv}{dx}.\\\text{ Therefore, the above differential equation is} \, \frac{du}{dx}+\frac{u}{x}=0\\\text{ Using method of separation of variables},\\ \frac{du}{u}=-\frac{dx}{x}\\ logu=-logx+logc_1\\ u=\frac{c_1}{x}\\ \frac{dv}{dx}=\frac{c_1}{x}\\ v=c_1logx+c_2\\ ye^{-x}=c_1logx+c_2\\ y(x)=e^{x}(c_1logx+c_2)\\ \text{This is the required solution.}$