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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 $$w$$

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

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

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

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

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

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

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

$$y=v e^{x}$$

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

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