Hallo,
\( \begin{array}{l} y^{\prime}+\frac{y}{2 x}=\frac{5}{x y} |-\frac{y}{2 x} . \\ y^{\prime}=\frac{5}{x y}-\frac{y}{2 x} \\ y^{\prime}=\frac{1}{x}\left(\frac{5}{y}-\frac{y}{2}\right) \\ \frac{d y}{d x}=\frac{1}{x}\left(\frac{5}{y}-\frac{y}{2}\right)=\frac{1}{x}\left(\frac{10-y^{2}}{2 y}\right) \\ \frac{2 y}{10-y^{2}}=\frac{d x}{x} \\ -\ln \left(10-y^{2}\right)=\ln |x|+c \quad \mid \cdot(-1) \\ \ln \left(10-y^{2}|=-\ln | x\right)-c \quad \mid e^{h o c h} \\ 10-y^{2}=\frac{1}{x}* C_{1} \cdot |-10 \\ -y^{2}=\frac{c_{1}}{x}-10 \quad \text { |. }(-1) \\ y^{2}=\frac{-c_{1}}{x}+10 \\ y= \pm \sqrt{\frac{-c_{1}}{x}+10} \\ \end{array} \)
Aus: \( y(1)=2 \sqrt{2} \Rightarrow \) neg. Lsg entfällt
\( \begin{array}{l} \Rightarrow 2\sqrt{2}=\sqrt{-c_{1}+10 } \\ 8=-C_{1}+10 \quad |-10 \\ -2=-C_{1} \quad \Rightarrow C_{1}=2 \\ \Rightarrow \quad y=\sqrt{\frac{-2}{x}+10} \end{array} \)
