Question

Мат анализ 2 срочно!
100 баллов

Answer 1

Ответ:

$( {x}^{2} - 3 {y}^{2} )dx + 2xydy = 0 \\ 2xydy = - ( {x}^{2} - 3 {y}^{2} )dx \\ 2xyy' = 3 {y}^{2} - {x}^{2}$

разделим на х^2

$2 \frac{y}{x} y = 3 \frac{ {y}^{2} }{ {x}^{2} } - 1$

Это однородное ДУ.

Замена:

$\frac{y}{x} = U \\ y = U'x + U$

$2u(U'x + U) = 3{U}^{2} - 1 \\ U'x + U = \frac{3 {U}^{2} - 1}{2U} \\ U'x = \frac{3 {U}^{2} - 1 - 2 {U}^{2} }{2U} \\ \frac{dU}{dx} x = \frac{ {U}^{2} - 1 }{2U} \\ \int\limits \frac{2U}{ {U}^{2} - 1} dU = \int\limits \frac{dx}{x} \\ \int\limits \frac{d( {U}^{2} - 1)}{ {U}^{2} - 1 } = ln(x) + ln(C) \\ ln( {U}^{2} - 1 ) = ln(Cx) \\ { U}^{2} - 1 = Cx \\ \frac{ {y}^{2} }{ {x}^{2} } = Cx + 1 \\ {y}^{2} = C {x}^{3} + {x}^{2}$

общее решение

Answer 2

$(x^2-3y^2)\, dx+2xy\, dy=0\\\\\dfrac{dy}{dx}=-\dfrac{x^2-3y^2}{2xy}\ \ ,\ \ y'=-\dfrac{x}{2y}+\dfrac{3y}{2x}\ \ ,\\\\t=\dfrac{y}{x}\ \ ,\ \ y=tx\ \ ,\ \ y'=t'x+t\\\\t'x+t=-\dfrac{1}{2t}+\dfrac{3t}{2}\ \,\ \ t'x+t=\dfrac{3t^2-1}{2t}\ \ ,\ \ t'x=\dfrac{3t^2-1-2t^2}{2t}\ \ ,\\\\t'x=\dfrac{t^2-1}{2t}\ \ ,\ \ \dfrac{dt}{dx}=\dfrac{t^2-1}{2tx}\ \ ,\ \ \int \dfrac{2t\, dt}{t^2-1}=\int \dfrac{dx}{x}\ \ ,\\\\ln|t^2-1|=ln|x|+lnC\ \ ,\ \ t^2-1=Cx\ \ ,\ \ \dfrac{y^2}{x^2}=Cx+1\ \ ,$

$y^2=x^2\, (Cx+1)\ \ ,\ \ \ y=\pm \sqrt{x^2(Cx+1)}\ \ ,\ \ y=\pm |x|\, \sqrt{Cx+1}$