Question

(x+2)y"+xy'-y=0

Answer 1

Answer on Question #77342, Math / Differential Equations

$$(x + 2) y ^ {\prime \prime} + x y ^ {\prime} - y = 0$$

We can notice that $y = x$ — the solution. Then, according to Liouville's formula

$$\left| \begin{array}{l l} y & x \\ y ^ {\prime} & 1 \end{array} \right| = C e ^ {- \int \frac {x}{x + 2} d x}$$ $$y - x y ^ {\prime} = C e ^ {- x + 2 \log | x + 2 |}$$ $$\frac {y - x y ^ {\prime}}{x ^ {2}} = \frac {C (x + 2) ^ {2} e ^ {- x}}{x ^ {2}}$$ $$- \left(\frac {y}{x}\right) ^ {\prime} = \frac {C (x + 2) ^ {2} e ^ {- x}}{x ^ {2}}$$ $$- \frac {y}{x} = C \left(- \frac {e ^ {- x} (x + 4)}{x} + D\right)$$ $$y = C e ^ {- x} (x + 4) + D x$$