Question

Решить дифференциальные уравнения, найти общее решение

Answer 1

1)Однородное дифференциальное уравнение.
$\displaystyle \frac{dy}{dx}= \frac{y^{3} + x^{2} y}{ x^{3} }\\y=tx=\ \textgreater \ t=\frac{y}{x};y'=t'x+t\\t'x+t=\frac{t^3x^3+x^3t}{x^3}\\t'x+t=t^3+t\\t'x=t^3\\\frac{dt}{dx}x=t^3|*\frac{x}{dx*t^3}\\\frac{dt}{t^3}=\frac{dx}{x}\\\int\frac{dt}{t^3}=\int\frac{dx}{x}\\-\frac{1}{2t^2}=ln|x|+C|*-2\\\frac{1}{t^2}=-2ln|x|+C\\\frac{x^2}{y^2}=-2ln|x|+C\\\frac{y^2}{x^2}=\frac{1}{-2ln|x|+C}\\y^2=\frac{x^2}{-2ln|x|+C}\\y=^+_-\frac{x}{\sqrt{C-2ln|x|}}$

$y'+y*tgx=14x^6cosx\\y=uv=\ \textgreater \ y'=u'v+v'u\\u'v+v'u+uv*tgx=14x^6cosx\\\begin{cases}v'+vtgx=0\\u'v=14x^6cosx\end{cases}\\\frac{dv}{dx}+vtgx=0\\\frac{dv}{dx}=-vtgx|*\frac{dx}{v}\\\frac{dv}{v}=-tgxdx\\\int\frac{dv}{v}=-\int tgxdx\\ln|v|=ln|cosx|\\v=cosx\\u'cosx=14x^6cosx\\\frac{du}{dx}=14x^6\\du=14x^6dx\\\int du=14\int x^6dx\\u=2x^7+C\\y=cosx(2x^7+C)$

$y'-y*tgx=3\\y=uv=\ \textgreater \ y'=u'v+v'u\\u'v+v'u-uvtgx=3\\\begin{cases}v'-vtgx=0\\u'v=3\end{cases}\\\frac{dv}{dx}-vtgx=0|*\frac{dx}{v}\\\frac{dv}{v}=tgxdx\\\int\frac{dv}{v}=\int tgxdx\\ln|v|=-ln|cosx|\\v=\frac{1}{cosx}\\\frac{u'}{cosx}=3\\\frac{du}{dx}=3cosx\\du=3cosxdx\\\int du=3\int cosxdx\\u=3sinx+C\\y=\frac{3sinx+C}{cosx}$