Question

Решить дифференциальные уравнения

Answer 1

$1)\; \; 6x^2=\frac{dy}{dx}\; ,\; \; y(3)=5\; ,\\\\\int 6x^2\, dx=\int dy\\\\2x^3+C=y\\\\y(3)=5:\; 5=2\cdot 3^3+C\; \; \to \; \; C=-49\\\\y=2x^3-49\\\\\\2)\; \; (3x^2+1)=6\cdot \frac{dy}{dx}\\\\\int (3x^2+1)\, dx=6\int dy\\\\x^3+x+C=6y\\\\y=\frac{x^3+x+C}{6}\\\\y())=0:\; \; 0=\frac{C}{6}\; \; \to \; \; C=0\\\\y=\frac{x^3+x}{6}$