Question

ТРУБЫ ГОРЯТ РЕШИТЕ ПОЖАЛУЙСТА!!!

Answer 1

a)

$xy'' = y'$

$z(y) = y'$

$x(z(y))' = z(y)\\\frac{(z(y))'}{z(y)} = \frac{1}{x}\\\int {\frac{\frac{dz(y)}{dx}}{z(y)}} \, dx = \int {\frac{1}{x}} \, dx\\\int {\frac{1}{z(y)}} \, dz(y) = \int {\frac{1}{x}} \, dx\\\ln z(y) = \ln x + c_0 \\\\z(y) = e^{c_0} x = 2c_1 x, \qquad (e^{c_0} = 2c_1)\\y' = 2c_1 x\\\int dy = \int 2c_1x \, dx\\y = c_1 x^2 + c_2$

b)

$y'' = \frac{1}{4\sqrt{y}}$

$z(y) = y', \quad y'' = (z(y))' = z'(y) y' = z' z$

$\frac{dz}{dy} z = \frac{1}{4\sqrt{y}}\\\int {z} \, dz = \int {\frac{1}{4\sqrt{y}}} \, dy \\\frac{z^2}{2} = \frac14 2\sqrt{y} + c_0\\z = \pm \sqrt{\sqrt{y} + c_0}\\y' = \pm \sqrt{\sqrt{y} + c_0}\\\int \frac{1}{\sqrt{\sqrt{y} + c_0}} \, dy = \int dx\\x = \frac43 (\sqrt{y} - 2c_0) \sqrt{c_0 + \sqrt{y}}$

Тут явно $y$ не выразить.