Question

дифф.уравнения срочно

Answer 1

$1)\ \ y'+2y-y^2=0\ \,\ \ \int \dfrac{dy}{y(y-2)}=\int dx\ \ ,\\\\\\-\dfrac{1}{2}\int \dfrac{dy}{y}+\dfrac{1}{2}\int \dfrac{dy}{y-2}=\int dx\ \ ,\ \ -\dfrac{1}{2}\, ln|y|+\dfrac{1}{2}\, ln|y-2|=x+C\\\\\\ln\Big|\dfrac{y-2}{y}\Big|=2(x+C)$

$2)\ \ (2x-y)\, dx+(x+y)\, dy=0\ \ ,\ \ \ \dfrac{dy}{dx}=\dfrac{y-2x}{x+y}\ \ ,\\\\y=ux\ ,\ \ y'=u'x+u\ \ ,\ \ \ u'x+u=\dfrac{u-2}{1+u}\ \ ,\ \ u'x=\dfrac{u-2-u-u^2}{1+u}\\\\\\\dfrac{du}{dx}=-\dfrac{u^2+2}{x\, (u+1)}\ \ ,\ \ \ \int \dfrac{(u+1)\, du}{u^2+2}=-\int \dfrac{dx}{x}\ \ ,\\\\\\\int \dfrac{u\, du}{u^2+2}+\int \dfrac{du}{u^2+2}=-ln|x|+C_1\\\\\\\dfrac{1}{2}\, ln|u^2+2|+\dfrac{1}{\sqrt2}\, arctg\dfrac{u}{\sqrt2}=-ln|x|+C$

$\dfrac{1}{2}\, ln\Big|\dfrac{y^2}{x^2}+2\Big|+\dfrac{1}{\sqrt2}\, arctg\dfrac{y}{\sqrt2\, x}=-ln|x|+C\\\\\\ln\sqrt{\dfrac{y^2}{x^2}+2}+\dfrac{1}{\sqrt2}\, arctg\dfrac{y}{\sqrt2\, x}=-ln|x|+C$

$3)\ \ xy'+y=sinx\ \ ,\ \ \ y'+\dfrac{y}{x}=\dfrac{sinx}{x}\\\\y=uv\ \ ,\ \ \ u'v+uv'+\dfrac{uv}{x}=\dfrac{sinx}{x}\ \ ,\ \ \ u'v+u\, \Big(v'+\dfrac{v}{x}\Big)=\dfrac{sinx}{x}\ ,\\\\a)\ \ \dfrac{dv}{dx}=-\dfrac{v}{x}\ \ ,\ \ \int \dfrac{dv}{v}=-\int \dfrac{dx}{x}\ \ ,\ \ ln|v|=-ln|x|\ \ ,\ \ v=-\dfrac{1}{x}\\\\b)\ \ -\dfrac{du}{dx}\cdot \dfrac{1}{x}=\dfrac{sinx}{x}\ \ ,\ \ \int du=-\int sinx\, dx\ \ ,\ \ u=cosx+C\ \ ,\\\\c)\ \ y=-\dfrac{1}{x}\cdot (cosx+C)$

$4)\ \ y''=cos4x\\\\y'=\int cos4x\, dx=\dfrac{1}{4}\, sin4x+C_1\\\\y=\int \Big(\dfrac{1}{4}\, sin4x+C_1\Big)\, dx=-\dfrac{1}{16}\, cos4x+C_1x+C_2\\\\\\5)\ \ y''-10y'+25y=0\ \ ,\ \ \ k^2-10k+25=0\ \ ,\ \ (k-5)^2=0\ \ ,\ k=5\\\\y=e^{5x}\, (C_1+C_2x)$