Question

Дифференциальные уравнения помогите !!!

Answer 1

Ответ:

$1)\ \ 2x\sqrt{y^2+1}\, dx-x^2\, dy=4\, dy\\\\2x\sqrt{y^2+1}\, dx=(x^2+4)\, dy\ \ ,\ \ \ \int \dfrac{2x\, dx}{x^2+4}=\int \dfrac{dy}{\sqrt{y^2+1}}\ \ ,\\\\ln(x^2+4)=ln|y+\sqrt{y^2+1}|+lnC_1\\\\y+\sqrt{y^2+1}=C(x^2+4)\ \ ,\ \ \ C=\dfrac{1}{C_1}$

$2)\ \ 6y(x^2+y^2)\, dx=(5xy^2+3x^3)\, dy\\\\\dfrac{dy}{dx}=\dfrac{6yx^2+6y^3}{5xy^2+3x^3}\ \ ,\ \ \ u=\dfrac{y}{x}\ ,\ \ y=ux\ ,\ \ y'=u'x+u\\\\\\u'x+u=\dfrac{6ux^3+6u^3x^3}{5u^2x^3+3x^3}\ \ ,\ \ u'x+u=\dfrac{6u+6u^3}{5u^2+3}\ \ ,\ \ u'x=\dfrac{6u+6u^3}{5u^2+3}-u\\\\\\u'x=\dfrac{6u+6u^3-5u^3-3u}{5u^2+3}\ \ ,\ \ \ u'x=\dfrac{u^3+3u}{5u^2+3}\ ,\\\\\\\int \dfrac{(5u^2+3)\, du}{u(u^2+3)}=\int \dfrac{dx}{x}$

$\dfrac{5u^2+3}{u(u^2+3)}=\dfrac{A}{u}+\dfrac{Bu+C}{u^2+3}\\\\5u^2+3=Au^2+3A+Bu^2+Cu\\\\u^2\ |\ A+B=5\\{}\ u\ |\ C=0\\u^0\ |\ 3A=3\ \ \ \ ,\ \ A=1\ \ ,\ \ \ B=5-A=4\\\\\\\int \dfrac{(5u^2+3)\, du}{u(u^2+3)}=\int \Big(\dfrac{1}{u}+\dfrac{4u}{u^2+3}\Big)\, du=ln|u|+2\, ln|u^2+3|+C_1\\\\\int \dfrac{dx}{x}=ln|x|+C_2\\\\\\ln\Big|\, \dfrac{y}{x}\, \Big|+2\, ln\Big|\, \dfrac{y^2}{x^2}+3\, \Big|=ln|x|+ln|C|\\\\\\\dfrac{y}{x}\cdot \Big(\dfrac{y^2}{x^2}+3\Big)^2=Cx\ \ ,\ \ \ \dfrac{y\cdot (y^2+3x^2)^2}{x^5}=Cx\ \ ,$

$(y^2+3x^2)^2=Cx^6$

$3)\ \ xy'=\dfrac{y}{x+1}+x\\\\y'-\dfrac{1}{x(x+1)}\, y=1\ \ ,\ \ \ y=uv\ ,\ \ y'=u'v =uv'\\\\u'v+uv'-\dfrac{1}{x(x+1)}\, uv=1\\\\ u'v+u\Big(v'-\dfrac{1}{x(x+1)}\, v\Big)=1\\\\a)\ \ v'-\dfrac{1}{x(x+1)}\, v=0\ \ ,\ \ \dfrac{dv}{dx}=\dfrac{1}{x(x+1)}\, v\ \ ,\ \ \ \int \dfrac{dv}{v}=\int \dfrac{dx}{x(x+1)}\\\\ \int \dfrac{dv}{v}=\int \Big(\dfrac{1}{x}-\dfrac{1}{x+1}\Big)\, dx\ \ ,\ \ \ \ ln|v|=ln|x|-ln|x+1|\ \ ,\\\\v=\dfrac{x}{x+1}$

$b)\ \ u'v=1\ \ ,\ \ \ u'=\dfrac{1}{v}\ \ ,\ \ \ \dfrac{du}{dx}=\dfrac{x+1}{x}\ \ ,\ \ \ du=(1+\dfrac{1}{x})\, dx\ \ ,\\\\\int du=\int \Big(1+\dfrac{1}{x}\Big)\, dx\ \ ,\ \ \ u=x+ln|x|+C\\\\c)\ \ y=\dfrac{x}{x+1}\cdot (x+ln|x|+C)$