Question

Найти частное решение дифференциального уравнения
$\sqrt{y} dy+x^{3} dx=0 y(1)=4$

Answer 1

$\sqrt{y}dy+ x^3dx=0\\\\\int \sqrt{y}dy=-\int x^3dx\\\\\frac{y^{3/2}}{3/2}=-\frac{x^4}{4}+C\\\\\frac{2\sqrt{y^3}}{3}=-\frac{x^4}{4}+C\\\\y(1)=4\; \; \to \; \; \; \frac{2\sqrt{4^3}}{3}=-\frac{1}{4}+C\; ,\\\\\frac{2\cdot 2^3}{3}=-\frac{1}{4}+C\; \; \to \; \; C=\frac{16}{3}+\frac{1}{4}=\frac{64+3}{12}=\frac{67}{12}\\\\chastnoe\; reshenie:\; \; \frac{2\sqrt{y^3}}{3}=-\frac{x^4}{4}+\frac{67}{12} \; ,\\\\\sqrt{y^3}=-\frac{3x^4}{8}+\frac{67}{8}\\\\y^3=(\frac{67}{8}-\frac{3x^4}{8})^2\\\\y=\sqrt[3]{\frac{(67-3x^4)^2}{64}}$