Question

помогите плиз с дифференциальным уравнением x^2y'+y^2=0 A(1;-1)

Answer 1

$\displaystyle x^2y'+y^2=0\\y=tx;y'=t'x+t\\x^2(t'x+t)+t^2x^2=0|:x^2\\\frac{dt}{dx}x+t+t^2=0|*\frac{dx}{x(t+t^2)}\\-\frac{dx}{x}=\frac{dt}{t^2+t}-\int\frac{dx}{x}=\int(\frac{1}{t}-\frac{1}{t+1})dt\\-ln|x|=ln|t|-ln|t+1|+C\\-ln|x|=ln|\frac{y}{x}|-ln|\frac{y+x}{x}|+C\\-ln|x|=ln|y|-ln|x|-ln|\frac{y+x}{x}|+C\\ln|\frac{x+y}{xy}|=ln|C|\\\frac{x+y}{xy}=C\\y(1)=-1;\\\frac{0}{-1}=C\\C=0\\\frac{x+y}{xy}=0;\\y=-x$