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Verify that $y=e^{x}$ is a solution of the equation $x y^{\prime \prime}-(2 x+1) y^{\prime}+(x+1) y=0, x>0$ .
Use this fact and the method of reduction of order to find the general solution of
$x y^{\prime \prime}-(2 x+1) y^{\prime}+(x+1) y=e^{x}$ .

$y=e^{x} \rightarrow y^\prime=e^{x} \rightarrow y^{\prime \prime}=e^{x}$ subs it the Homo, linear, eq...

$x e^{x}-(2 x+1) e^{x}+(x+1) e^{x}=0 \longrightarrow$

$e^{x}(x-(2 x+1)+x+1]=e^{x}(0)=0$

$y=e^{x} \rightarrow$ is asolution for the eq.

put $\quad y=v e^{x}$

$\rightarrow y^{\prime}=v^\prime e^{x}+e^{x} v=e^{x}\left(v^{\prime}+v\right)$

$y^{\prime \prime}=v^{\prime \prime} e^{x}+e^{x} v^\prime+e^{x} v+v^\prime e^{x}$

$=e^{x}\left(v^{\prime \prime}+v^\prime+v^\prime+v\right)$

$y^{\prime \prime}=e^{x}\left(v^{\prime \prime}+2 v^{\prime}+v\right)$ subs in the D.E

$x\left(v^{\prime \prime}+2 v^{\prime}+v\right) e^{x}-(2 x+1)\left(v^{\prime}+v\right) e^{x}+(x+1) v e^{x}=e^{x} \quad \div e^{x}$

$x\left(v^{\prime \prime}+2 v^{\prime}\right)-(2 x+1) v^{\prime}=1$

$v^{\prime \prime}(x)+v^{\prime}(2 x-2 x+1)=1$

$\longrightarrow v^{\prime \prime} x-v^{\prime}=1 \quad \div x$

$v^{\prime \prime}-\frac{1}{x} v^{\prime}=\frac{1}{x}$

$w=v^{\prime} \quad w^{\prime}=v^{\prime \prime}$

$w-\frac{1}{x} w=\frac{1}{x}$ Linear in $w$

$w^{\prime}-\frac{1}{x} w=\frac{1}{x}$

$\mu=e^{\int {-1 / x} d x}=e^{-\ln |x|}=\frac{1}{x}$

$y w=\int Q(x) d x \rightarrow \frac{1}{x} w=\int \frac{1}{x} \frac{1}{x} d x$

$\rightarrow \frac{w}{x}=\frac{1}{-x}+c_{1}$

$\frac{w}{x}=\frac{1}{-x}+c_{1} \rightarrow w=-1+c_{1} x$

$\longrightarrow v^{\prime}=-1+c_{1} x$

$v=\int-1+c_{1} x d x+c_{2} \rightarrow v=-x+\frac{c_{1} x^{2}}{2}+c_{2}$

$y=v e^{x}$

$y=\left(-x+\frac{c_{1}}{2} x^{2}+c_{2}\right) e^{x}$