Question

y’’ = y’/x + x
2(xy’+y) = xy^2

Answer 1

$2(xy'+y)=xy^2;\ 2(xy)'=\frac{(xy)^2}{x};\ xy=u;\ \int \frac{d u}{u^2}=\int\frac{d x}{2x};\ -\frac{1}{u}=\frac{1}{2}\ln|x|+C;$

найдем С, используя начальные условия: x=1; y=2;

$-\frac{1}{1\cdot 2}=\frac{1}{2}\ln 1+C;\ C=-\frac{1}{2};\ -\frac{1}{u}=\frac{1}{2}\ln |x|-\frac{1}{2};\ u=\frac{2}{1-\ln |x|};\ y=\frac{2}{x(1-\ln |x|)}$

$y''=\frac{y'}{x}+x;\ xy''-y'=x^2; \frac{(y')'x-y'x'}{x^2}=1;\ \left(\frac{y'}{x}\right)'=1; \frac{y'}{x}=x+C_1; \ y'=x^2+C_1x;$

$y=\frac{x^3}{3}+\frac{C_1x^2}{2}+C_2$