General solution – linear differential equation

Question

I was wondering if anybody would be able to explain the process involved regarding the solution to the following linear differential equation:

$$ x^2 \frac{\mathrm{d}y}{\mathrm{d}x}+xy+1=0 $$

Thank You.

Answer 1

make the ansatz $y(x)=x^r$ and a special solution is given by $$y_p(x)=-\frac{\log(x)}{x}$$ or write $$y'(x)+\frac{y(x)}{x}=-\frac{1}{x^2}$$ $$\mu(x)=e^{\int \frac{1}{x}dx}=x$$ $$\frac{d}{dx}(x(y(x))=-\frac{1}{x}$$ $$\int \frac{d}{dx}(xy(x))dx=\int-\frac{1}{x}dx$$ $$xy(x)=-\log(x)+C_1$$

Answer 2

The quickest way is to rewrite the equation as

$$ \begin{align} xy' + y &= -\frac{1}{x} \\ (xy)' &= -\frac{1}{x} \\ xy &= -\ln|x| + C \\ y &= \frac{C-\ln|x|}{x} \end{align} $$

You can also follow the general method for a first-order linear ODE $$ y' + \frac{1}{x}y = -\frac{1}{x^2} $$ and compute the integrating factor $$ \mu(x) = \exp{\int \frac{1}{x}\ dx} = x $$ which leads us back to the beginning $$ xy' + y = -\frac{1}{x} $$