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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 $$f$$

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

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

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

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

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

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