Find the general solution of x y''-(2x+1) y'+(x+1) y=0 . An…

Find the general solution of $x\,y''-(2x+1)\,y'+(x+1)\,y=0$. An exponential particular solution exists.

Step 1: Find the exponential particular solution

Try $y_1=e^{ax}$, treating $a$ as an unknown constant: $y_1'=ae^{ax}$, $y_1''=a^2e^{ax}$. Substituting and dividing by $e^{ax}$, \[a^2x - a(2x+1) + (x+1) = x(a^2-2a+1) - (a-1) = x(a-1)^2-(a-1) = (a-1)\bigl[x(a-1)-1\bigr] = 0.\]

For this to hold for every $x$, the bracket $x(a-1)-1$ cannot vanish identically, so we need $a-1=0$, i.e. $a=1$.

Step 2: Verify $y_1=e^x$

\[ xe^x-(2x+1)e^x+(x+1)e^x = e^x[x-2x-1+x+1]=0.\ \checkmark \]

Step 3: Substitute $y=y_1v$ and reduce the order

Set $y=e^xv(x)$: $y'=e^x(v+v')$, $y''=e^x(v+2v'+v'')$. Substituting into $xy''-(2x+1)y'+(x+1)y=0$ and dividing by $e^x$, the $v$ terms cancel (Step 2's identity) and

\[ x(v+2v'+v'') - (2x+1)(v+v') + (x+1)v = 0 \implies xv'' - v' = 0. \]

Step 4: Solve the first-order equation for $w=v'$

\[ xw'-w=0 \implies \frac{dw}{w}=\frac{dx}{x} \implies \ln|w|=\ln|x|+\mathrm{const} \implies w=Ax. \]

Step 5: Integrate for $v$

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

Step 6: General solution

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

Step 7: Verify $y_2=x^2e^x$

\[ y_2'=(x^2+2x)e^x,\qquad y_2''=(x^2+4x+2)e^x. \]

\[ x(x^2+4x+2)e^x - (2x+1)(x^2+2x)e^x + (x+1)x^2e^x = e^x\bigl[(1{-}2{+}1)x^3+(4{-}5{+}1)x^2+(2{-}2)x\bigr]=0.\ \checkmark \]

Step 8: Linear independence

\[ W(y_1,y_2) = \begin{vmatrix} e^x & x^2e^x \\ e^x & (x^2+2x)e^x\end{vmatrix} = e^{2x}(x^2+2x-x^2) = 2xe^{2x} \neq 0 \quad (x\neq0), \] so $y_1=e^x$ and $y_2=x^2e^x$ form a fundamental system.