Find the general solution of xy'' + 2y' - xy = 0 , given…

Find the general solution of $xy'' + 2y' - xy = 0$, given that $y_1 = \dfrac{e^x}{x}$ is a particular solution.

Step 1: Verify $y_1$

\[ y_1' = \frac{e^x(x-1)}{x^2}, \qquad y_1'' = \frac{e^x(x^2-2x+2)}{x^3}. \]

\[ xy_1''+2y_1'-xy_1 = e^x\!\left[\frac{x^2-2x+2}{x^2}+\frac{2x-2}{x^2}-1\right] = e^x\!\left[\frac{x^2}{x^2}-1\right] = 0.\ \checkmark \]

Step 2: Substitution $y=y_1v$

\[ y' = \frac{e^x(x-1)}{x^2}v + \frac{e^x}{x}v', \qquad y'' = \frac{e^x(x^2-2x+2)}{x^3}v + \frac{2e^x(x-1)}{x^2}v' + \frac{e^x}{x}v''. \]

Step 3: Reduce the order

Substituting into $xy''+2y'-xy=0$, the coefficient of $v$ reproduces Step 1's identity and vanishes; the coefficient of $v'$ is

\[ e^x\!\left[\frac{2(x-1)}{x}+\frac{2}{x}\right] = 2e^x, \] so the reduced equation is

\[ e^xv''+2e^xv'=0 \implies v''+2v'=0. \]

Step 4: Solve for $w=v'$

\[ w'+2w=0 \implies \frac{dw}{w}=-2\,dx \implies w=Ae^{-2x}. \]

Step 5: Integrate for $v$

\[ v' = Ae^{-2x} \implies v = -\frac{A}{2}e^{-2x}+B. \]

Step 6: General solution

Then $y=\dfrac{e^x}{x}\left(-\dfrac{A}{2}e^{-2x}+B\right)=B\dfrac{e^x}{x}-\dfrac{A}{2}\cdot\dfrac{e^{-x}}{x}$. Renaming $B=C_1$, $-A/2=C_2$:

Step 7: Verify $y_2=\dfrac{e^{-x}}{x}$

\[ y_2' = -\frac{e^{-x}(x+1)}{x^2}, \qquad y_2'' = \frac{e^{-x}(x^2+2x+2)}{x^3}. \]

\[ xy_2''+2y_2'-xy_2 = e^{-x}\!\left[\frac{x^2+2x+2}{x^2}-\frac{2x+2}{x^2}-1\right]=e^{-x}[1-1]=0.\ \checkmark \]

Step 8: Linear independence

\[ W(y_1,y_2) = \frac{e^x}{x}\!\left(-\frac{e^{-x}(x+1)}{x^2}\right) - \frac{e^{-x}}{x}\cdot\frac{e^x(x-1)}{x^2} = -\frac{x+1}{x^3}-\frac{x-1}{x^3} = -\frac{2}{x^2}\neq0, \]

confirming that $y_1,y_2$ form a fundamental system.