Question

дифференциальные уравнения второго порядка, поподробнее будьте добры

Answer 1

$xy''-y'+\frac{1}{x}=0 \\ y'=z; y''=z' \\ z'-\frac{z}{x}=-\frac{1}{x^2} \\ z=uv \\ u'v+uv'-\frac{uv}{x}=-\frac{1}{x^2} \\ \left \{ {{v'-\frac{v}{x}=0} \atop {u'v=-\frac{1}{x^2}}} \right. \\ \frac{dv}{dx}=\frac{v}{x} \\ \frac{dv}{v}=\frac{dx}{x} \\ lnv=lnx \\ v=x \\\\ u'x=-\frac{1}{x^2} \\ \frac{du}{dx}=-\frac{1}{x^3} \\ du=-\frac{dx}{x^3} \\ u=\frac{1}{2x^2}+C_1 \\ z=uv=\frac{1}{2x}+C_1x \\ y'=\frac{1}{2x}+C_1x \\ y=\int \frac{1}{2x}+C_1x=\frac{1}{2}lnx+\frac{C_1x^2}{2}+C_2=\frac{1}{2}lnx+C_1x^2+C_2$

$\left \{ {{y=\frac{1}{2}ln1+C_1*1^2+C_2=2} \atop {y'=\frac{1}{2*1}+C_1*1=\frac{3}{2}}} \right. \\\\ \left \{ {{C_1+C_2=2} \atop {\frac{1}{2}+C_1=\frac{3}{2}}} \right. \\\\ \left \{ {{C_1+C_2=2} \atop {C_1=1}} \right. \\\\ \left \{ {{1+C_2=2} \atop {C_1=1}} \right. \\\\ \left \{ {{C_2=1} \atop {C_1=1}} \right. \\\\ y^*=\frac{1}{2}lnx+x^2+1$