Question

Диференциальные уравнения, высшая математика
Помогите пожалуйста ​

Answer 1

Ответ:

$1)\ \ \displaystyle (xy^2+x)\, dx-(y-x^2y)\, dy=0\\\\x(y^2+1)\, dx=y(1-x^2)\, dy\ \ ,\ \ \int \frac{x\, dx}{1-x^2}=\int \frac{y\, dy}{y^2+1}\ \ ,\\\\-\frac{1}{2}\, ln|1-x^2|=\frac{1}{2}\, ln|y^2+1|-\frac{1}{2}\, lnC \ \ ,\\\\ln|y^2+1|=lnC-ln|1-x^2|\ \ ,\ \ \ y^2+1=\frac{C_1}{1-x^2}\ \ ,\\\\y=\pm \sqrt{\frac{C_1}{1-x^2}-1}\ \ ,\ \ y=\sqrt{\frac{C_1-1+x^2}{1-x^2}}\ \ ,\\\\y=\sqrt{\frac{C+x^2}{1-x^2}}\ \ ,\ \ C=C_1-1\ .$

$2)\ \ \displaystyle \frac{dy}{dx}=\frac{y}{x}+tg\frac{y}{x}\\\\u=\frac{y}{x}\ \ ,\ \ y=ux\ \ ,\ \ y'=u'x+u\\\\u'x+u=u+tgu\ \ ,\ \ \ u'x=tgu\ \ ,\ \ \frac{du}{tgu}=\frac{dx}{x}\ \ ,\ \ \int ctgu\, du=\int \frac{dx}{x}\ \ ,\\\\ln|sinu|=ln|x|+lnC\ \ ,\ \ sin\frac{y}{x} =Cx\ \ ,\ \ \frac{y}{x}=arcsin(Cx)\ \ ,\\\\y=x\cdot arcsin(Cx)$

$3)\ \ \displaystyle y'-\frac{y}{x}+y^2=0\ \ ,\ \ \ y=uv\ ,\ y'=u'v+uv'\ \ ,\\\\u'v+uv'-\frac{uv}{x}=(uv)^2\ \ ,\ \ \ u'v+u\cdot \Big(v'-\frac{v}{x}\Big)=u^2v^2\ \ ,\\\\a)\ \ v'-\frac{v}{x}=0\ \ ,\ \ \int \frac{dv}{v}=\int \frac{dx}{x}\ \ ,\ \ \ ln|v|=ln|x|\ \ ,\ \ \ v=x\\\\b)\ \ u'\cdot x=u^2x^2\ \ ,\ \ \ \frac{du}{dx}=u^2x\ \ ,\ \ \int \frac{du}{u^2}=\int x\, dx\ \ ,\\\\-\frac{1}{u}=\frac{x^2}{2}+\frac{C}{2}\ \ ,\ \ u=-\frac{2}{x^2+C}\ \ ,\\\\c)\ \ y=-\frac{2x}{x^2+C}$