Use the substitution $y = ux$, where $u$ is a function of $x$, to show that the solution of the differential equation $$ \frac{\mathrm{d} y}{\mathrm{d} x} = \frac x y + \frac y x \qquad (x> 0, \ y> 0) $$ that satisfies $y = 2$ when $x = 1$ is $$ y = x \sqrt{ 4+2\ln x } \qquad \left( x > \mathrm{e}^{-2} \right). $$
Use a substitution to find the solution of the differential equation $$ \frac{\mathrm{d} y}{\mathrm{d} x} = \frac x y + \frac {2y} x \qquad (x> 0, \ y> 0) $$ that satisfies $y = 2$ when $x = 1$.
Find the solution of the differential equation $$ \frac{\mathrm{d} y}{\mathrm{d} x} = \frac {x^2} y + \frac {2y} x \qquad (x> 0, \ y> 0) $$ that satisfies $y = 2$ when $x = 1$.
