Question

Решите дифференциальное уравнение

Answer 1

$x^2y' + 2yx = \sin(x)$

левую часть уравнения можно свернуть

$(x^2y)' = \sin(x)$

тогда

$x^2y = \int \sin(x)\,\mathrm{d}x + C = -\cos(x) + C$

$y = \frac{C - \cos(x)}{x^2}$

$y(\pi) = \frac{C - \cos(\pi)}{\pi^2} = \frac{C-(-1)}{\pi^2} = \frac{C+1}{\pi^2} = 0$

$C+1 = 0$

$C = -1$

$y = \frac{-1-\cos(x)}{x^2} = -\frac{1+\cos(x)}{x^2}$