Question

Решите 2 диф уравнения
Только не спамте пж!

Answer 1

Ответ:

$a) \,y = \frac{9}{112}(2x-5)^{\frac{7}{3}} + C_1x + C_2,\,\,\, C_1,\, C_2 \in R\\\\b)y = \frac{x^2}{4} + C_1ln(x) + C_2,\,\,\, C_1,\,C_2 \in R$

Пошаговое объяснение:

$a)\;y''=\sqrt[3]{2x-5}\\\\y'=\int\sqrt[3]{2x-5}\;dx = \int (2x-5)^{\frac{1}{3}}\cdot \frac{1}{2} \; d(2x - 5) = \frac{3}{4}\cdot\frac{1}{2}(2x-5)^{\frac{4}{3}} + C_1 = \\\\ = \frac{3}{8}(2x-5)^{\frac{4}{3}} + C_1\\\\y = \int (\frac{3}{8}(2x-5)^{\frac{4}{3}} + C_1) \; dx = \int \frac{3}{8}(2x-5)^{\frac{4}{3}} \frac{1}{2} \; d(2x - 5) + \int C_1 \; dx = \\\\ =\frac{3}{16}\cdot\frac{3}{7}(2x-5)^{\frac{7}{3}} + C_1x + C_2 = \frac{9}{112}(2x-5)^{\frac{7}{3}} + C_1x + C_2$

$b)\ y'' = (1 - \frac{y'}{x})\\\\y'' = \frac{x - y'}{x}\\\\xy''+y'-x=0\\\\--------------------\\(xy')' = (x)'y' + x(y')'=y'+xy''\\\\ (-\frac{x^2}{2})' = -x\\--------------------\\\\d(xy'-\frac{x^2}{2}) = 0\\\\xy'-\frac{x^2}{2} = C_1\\\\y'=\frac{C_1}{x} + \frac{x}{2}\\\\y = \int (\frac{C_1}{x} + \frac{x}{2}) \, dx = C_1ln(x) + \frac{x^2}{4} + C_2$