General solution of the Differential Equation $y'' - (\frac{2}{x^2} + \frac{1}{x})(xy' - y) = 0$?

Question

If $y(x) = x$ is a solution of the Differential Equation $y'' - (\frac{2}{x^2} + \frac{1}{x})(xy' - y) = 0$.

Then the General Solution is ?

As its a competitive problem we have to do in short time!.

Is this a special differential equation like Legendre differential equation or do we have to use some other methods of variation of parameters?

The answer is given to be $y = (\alpha + \beta e^x) x$ for parameters $\alpha , \beta$

Answer 1

$$y''=(\frac{2}{x^2}+\frac{1}{x})(xy'-y)=(x+2)(\frac{xy'-y}{x^2})=(x+2)(\frac{y}{x})'$$Take $y=xu$ then:$$y''=xu''+2u'=(x+2)u'\to u''=u'\to u=Ae^x+B\to\\y=Axe^x+Bx$$

Answer 2

Hint: We rewrite the equation by multiplying both sides by $x^2$ to get $x^2y''-(x+2)(xy'-y)=0$. The substitution $u=xy'-y$ gives $u'=xy''$. Inserting $u$ and $u'$ in the equation we obtain a first order separable (or linear) equation in the variables $x$ and $u$:$$xu'-(x+2)u=0.$$