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Given that $y_{1}=x$ and $y_{2}=x \ln x$ are solutions of the differential equation
$x^{2} y^{\prime \prime}-x y^{\prime}+y=0 ; x>0$
Show that $y 1$ and $y 2$ are linearly independent on the interval $(0, \infty),$ then find
the particular solution that satisfies the initial conditions $y(1)=7$ and $y^{\prime}(1)=2$

$w\left(y_{1}, y_{2}\right)=| \begin{array}{l}{x} \\ {1}\end{array} \text{ } \begin{array}{l}{x\ln (x)} \\ {\ln (x+1)}\end{array} |$

$=x \ln (x)+x-x \ln (x)=x \neq 0 \quad [x>0]$

$y_{1}, y_{2} \text { are } L\cdot {I} \text { on }(0, \infty)$

$y=c_{1} y_{1}+c_{2} y_{2}=c_{1} x+c_{2} x \ln (x)$

$y(1)=7 \Rightarrow c_{1}+0=7 \rightarrow c_{1}=7$

$y^{\prime}=c_{1}+c_{2}(\ln (x)+1) \rightarrow y{\prime}(1)=2 \rightarrow 7+c_{2}(0+1)=2$

$c_{2}=-5$

$y=7 x+-5 x \ln (x)$

Find all differentiable functions $f(x)$ on some interval $|$ such that $w(x, x f(x))=1+2 f(x)$

$W=\left|\begin{array}{ll}{x} & {x f(x)} \\ {1} & {f+f^{\prime} x}\end{array}\right|$

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

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

$x^{2} f^{\prime}-2 f=1 \quad \div x^{2}$

$\frac{d f}{d x}-\frac{2}{x^{2}} f=\frac{1}{x^{2}} \Rightarrow$ Linear in $f$

$\mu=e^{\int-\frac{2}{x^{2}}d x}=e^{-2\left(\frac{-1}{x}\right)}=e^{2 / x}$

$\mu {f}=\int \mu {Q}(x) d x+c$

$\rightarrow e^{2 / x} f=\int e^{2 / x} \cdot \frac{1}{x^{2}}+c$

$e^{2 / x} f=-\frac{1}{2} e^{\frac{2}{x}}+c$

$\rightarrow f(x)=\frac{-1}{2}+c \: e^{-2 / x}$