Find the general solution of x^2(x+1) y'' - 2y = 0 , given…

Find the general solution of $x^2(x+1)\,y'' - 2y = 0$, given that $y_1 = 1+\dfrac1x = \dfrac{x+1}{x}$ is a particular solution.

Step 1: Verify $y_1$

\[ y_1' = -\frac1{x^2}, \qquad y_1'' = \frac{2}{x^3}. \]

\[ x^2(x+1)\cdot\frac{2}{x^3} - 2\cdot\frac{x+1}{x} = \frac{2(x+1)}{x} - \frac{2(x+1)}{x} = 0.\ \checkmark \]

Step 2: Substitution $y=y_1v$

With $y_1''=2/x^3$, $y_1'=-1/x^2$, $y_1=(x+1)/x$: $y' = y_1'v + y_1v' = -\frac{v}{x^2} + \frac{x+1}{x}v'$, $y'' = y_1''v + 2y_1'v' + y_1v'' = \frac{2v}{x^3} - \frac{2v'}{x^2} + \frac{x+1}{x}v''$.

Step 3: Reduce the order

Substituting into $x^2(x+1)y''-2y=0$: \[x^2(x+1)\!\left[\frac{2v}{x^3}-\frac{2v'}{x^2}+\frac{x+1}{x}v''\right] - \frac{2(x+1)}{x}v = 0,\]

\[ \frac{2(x+1)}{x}v - 2(x+1)v' + x(x+1)^2v'' - \frac{2(x+1)}{x}v = 0. \]

The $v$ terms cancel; dividing the rest by $(x+1)$ (assuming $x\neq-1$): $x(x+1)\,v'' - 2v' = 0.$

Step 4: Solve for $w=v'$

\[ x(x+1)w' - 2w = 0 \implies \frac{dw}{w} = \frac{2\,dx}{x(x+1)} = \left(\frac2x-\frac2{x+1}\right)dx. \]

\[ \ln|w| = 2\ln|x| - 2\ln|x+1| + \mathrm{const} \implies w = A\,\frac{x^2}{(x+1)^2}. \]

Step 5: Integrate for $v$

Write $\dfrac{x^2}{(x+1)^2} = 1 - \dfrac{2}{x+1} + \dfrac{1}{(x+1)^2}$ and integrate term by term: $v = A\left[x - 2\ln|x+1| - \frac{1}{x+1}\right] + B.$

Step 6: General solution

Multiply by $y_1=(x+1)/x$. The $B$ part gives $B\,y_1$. For the $A$ part, since $\dfrac{x+1}{x}\!\left(x-\dfrac1{x+1}\right) = (x+1)-\dfrac1x$: \[y = A\left[\left(x+1-\frac1x\right) - \frac{2(x+1)}{x}\ln|x+1|\right] + B\,\frac{x+1}{x}.\]

Renaming $B=C_1$, $A=C_2$:

Step 7: Verify $y_2$

Let $y_2 = x+1-\dfrac1x - \dfrac{2(x+1)}{x}\ln|x+1|$, and write $L=\ln|x+1|$. Differentiating, \[y_2' = 1 - \frac2x + \frac{1+2L}{x^2}, \qquad y_2'' = \frac{2}{x^2} - \frac{2}{x^3} + \frac{2}{x^2(x+1)} - \frac{4L}{x^3}.\]

Then $x^2(x+1)y_2''$ evaluates to $2(x+1)-\dfrac{2(x+1)}{x}+2-\dfrac{4(x+1)L}{x}$, and $2y_2 = 2(x+1)-\dfrac2x-\dfrac{4(x+1)L}{x}$. Subtracting: \[x^2(x+1)y_2'' - 2y_2 = -\frac{2(x+1)}{x}+\frac2x+2 = \frac{-2(x+1)+2}{x}+2 = -2+2 = 0.\ \checkmark\]

Step 8: Linear independence

Since the equation has no $y'$ term, $P=0$ in standard form, so by Abel's formula the Wronskian is constant: \[W(y_1,y_2) = y_1^2\,w = \frac{(x+1)^2}{x^2}\cdot\frac{x^2}{(x+1)^2} = 1 \neq 0,\]

confirming $y_1,y_2$ form a fundamental system.