Question

Решить дифференциальное уравнение (1+x^2)dy+ydx=0 y(1)=1

Answer 1

Ответ:

$(1+x^2)\, dy+y\, dx=0\ \ ,\ \ \ y(1)=1\ \ ,\\\\\\\dfrac{dy}{dx}=-\dfrac{y}{1+x^2} \\\\\\\displaystyle \int \frac{dy}{y}=-\int \frac{dx}{1+x^2}\\\\\\ln|y|=-arctgx+C\\\\\\y(1)=1:\ \ \ ln1=-arctg1+C\ \ ,\ \ 0=-\frac{\pi}{4}+C\ \ ,\ \ \ C=\frac{\pi}{4}\\\\\\\boxed{\ ln|y|=-arctgx+\frac{\pi }{4} \ }$