Question

помогите с дифференциалами
a) y'=(2y+1)×tg x
b) xy'-y=x×tg(y/x)
c) y'+y=x/y^2

Answer 1

$displaystyle frac{dy}{dx}=(2y+1)tgx|*frac{dx}{2y+1}\frac{dy}{2y+1}=tgxdx\frac{1}{2}intfrac{d(2y+1)}{2y+1}=-intfrac{d(cosx)}{cosx}\frac{1}{2}ln|2y+1|=-ln|cosx|+C|*2\ln|2y+1|=-2ln|cosx|+ln|C^*|\2y+1=frac{C^*}{cos^2x}\y=frac{C^*-cos^2x}{2cos^2x}$

$xy'-y=x*tgfrac{y}{x}\y=tx;y'=t'x+t\x(t'x+t)-tx=xtgt|:x\t'x+t-t=tgt\frac{dt}{dx}x=tgt|*frac{ctgtdx}{x}\ctgtdt=frac{dx}{x}\intfrac{d(sint)}{sint}=intfrac{dx}{x}\ln|sint|=ln|x|+ln|C|\sint=Cx\frac{1}{x}sinfrac{y}{x}=C$

$y'+y=frac{x}{y^2}|*y^2\y^2y'+y^3=x\z=y^3;z'=3y^2y';y^2y'=frac{z'}{3}\frac{z'}{3}+z=x\z'+3z=3x\z=uv;z'=u'v+v'u\u'v+v'u+3uv=3x\u'v+u(v'+3v)=3x\begin{cases}v'+3v=0\u'v=3xend{cases}\frac{dv}{dx}+3v=0|*frac{dx}{v}\frac{dv}{v}=-3dx\intfrac{dv}{v}=-3int dx\ln|v|=-3x\v=e^{-3x}\frac{du}{dx}e^{-3x}=3x\du=3xe^{3x}\int du=int 3xe^{3x}\u=xe^{3x}-frac{e^{3x}}{3}+C\z=y^3=x-frac{1}{3}+Ce^{-3x}\y=sqrt[3]{x-frac{1}{3}+Ce^{-3x}}$