Question

Найти частное решение дифференциального уравнения.

Answer 1

$y'\cdot ctgx+y=1\\\\y'+\frac{1}{ctgx}\cdot y=\frac{1}{ctgx}\\\\y=uv,\; y'=u'v+uv'\\\\u'v+uv'+\frac{1}{ctgx}uv=\frac{1}{ctgx}\\\\u'v+u(v'+\frac{v}{ctgx})=\frac{1}{ctgx}\\\\1)v'+\frac{v}{ctgx}=0\\\\\frac{dv}{dx}=-\frac{v}{ctgx}\\\\\frac{dv}{v}=-\frac{dx}{ctgx}$

$\int \frac{dv}{v}=-\int tgx\cdot dx\\\\ln|v|=ln|cosx|\\\\v=cosx\\\\2)\; u'v=\frac{1}{ctgx}\\\\\frac{du}{dx}\cdot cosx=tgx\\\\\int du=\int \frac{sinx}{cos^2x}dx\\\\u=-\int (cosx)^{-2}\cdot d(cosx)=-\frac{(cosx)^{-1}}{-1}+C=\frac{1}{cosx}+C$

$3)\; y=uv=cosx(\frac{1}{cosx}+C)\\\\y=1+\frac{C}{cosx}\\\\4)\; y(\frac{\pi}{4})=\sqrt2\\\\\sqrt2=1+\frac{C}{cos\frac{\pi}{4}}\\\\\sqrt2-1=\frac{C}{\frac{\sqrt2}{2}}\\\\C=\frac{\sqrt2}{2}(\sqrt2-1)\\\\y_{chastn.reshenie}=1+\frac{\sqrt2(\sqrt2-1)}{2cosx}$