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Question Number 128646 by Algoritm last updated on 09/Jan/21

Answered by Olaf last updated on 09/Jan/21

$$y^{″}-\frac{1+x}{x}{y}^{\prime }+\frac{y}{x}=0\left(1\right)$$$$y^{″}-y^{\prime }-\frac{y^{\prime }-y}{x}=0\left(2\right)$$$$\mathrm{Let}u=y^{\prime }-y\left(3\right)$$$$\left(2\right):u^{\prime }-\frac{u}{x}=0$$$$\frac{u^{\prime }}{u}=\frac{1}{x}$$$$u={\mathrm{C}}_{1}x$$$$\left(3\right):y^{\prime }-y={\mathrm{C}}_{1}x$$$$y_0=-{\mathrm{C}}_{1}x-{\mathrm{C}}_1=-{\mathrm{C}}_1(x+1)$$$$\frac{y_{\mathrm{H}}^{\prime }}{y_{\mathrm{H}}}=1\Rightarrow y_{\mathrm{H}}={\mathrm{C}}_2{e}^x$$$$y=y_{\mathrm{H}}+y_0={\mathrm{C}}_2{e}^x-{\mathrm{C}}_1(x+1)$$

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