Question

Решить задачу Коши по математике

Answer 1

$(x - y)dx + (x + y)dy = 0 \\ (x + y)dy = - (x - y)dx \\ y'= - \frac{x - y}{x + y}$

Однородное ДУ

$y = ux \\ y'= u'x + u$

$u'x + u = - \frac{x - ux}{x + ux} \\ u'x + u = - \frac{1 - u}{1 + u} \\ \frac{du}{dx} x = \frac{u - 1 - u(1 + u)}{u + 1} \\ \frac{du}{dx} x = \frac{u - 1 - u - u {}^{2} }{u + 1} \\ \frac{du}{dx} x = \frac{ - {u}^{2} - 1}{u + 1} \\ \frac{du}{dx} x = - \frac{ {u}^{2} + 1}{u + 1} \\ \int\limits \frac{u + 1}{u {}^{2} + 1 } du = - \int\limits \frac{dx}{x} \\ \int\limits \frac{udu}{u { }^{2} + 1} + \int\limits \frac{du}{u {}^{2} + 1} = - ln(x) + C \\ \frac{1}{2} \int\limits \frac{2udu}{u {}^{2} + 1} + arctgu = - ln(x) + C\\ \frac{1}{2} \int\limits \frac{d(u {}^{2} + 1) }{u {}^{2} + 1} + arctgu = - ln(x) + C \\ \frac{1}{2} ln( {u}^{2} + 1) + arctgu = - ln(x) + C \\ \frac{1}{2} ln( \frac{ {y}^{2} }{ {x}^{2} } + 1 ) + arctg \frac{y}{x} = - ln(x) + C$

общее решение

$y(1) = 0$

$\frac{1}{2} ln(1) + arctg0 = - ln(1) + C \\ C = 0$

$\frac{1}{2} ln( \frac{ {y}^{2} }{ {x}^{2} } + 1) + arctg \frac{y}{x} = - ln(x) \\ ln( \frac{ {y}^{2} }{ {x}^{2} } + 1) + 2arctg \frac{y}{x} = - 2 ln(x) \\ ln( \frac{ {y}^{2} }{ {x}^{2} } + 1) + ln( {x}^{2} ) = - 2arctg \frac{y}{x} \\ ln( {y}^{2} + {x}^{2} ) = - 2arctg \frac{y}{x}$

частное решение

Answer 2

$(x-y) dx+(x+y)dy =0; \ \ \ \ y(1)=0 \\ \\ (x-y)+(x+y)\frac{dy}{dx} =0 \\ \\ y=vx; \ \ \ y'=v'x+v \\ \\ (x-vx)+(x+vx)\cdot (v'x+v)=0 \\ \\ x-vx+v'x^2+vx+v'vx^2+v^2x=0 \\ \\ v'(x^2+vx^2)+x+v^2x=0 \\ \\ v'x^2\cdot (1+v)+x\cdot (1+v^2)=0 \\ \\ v'x^2\cdot (1+v)=-x\cdot(1+v^2) \\ \\ v'\cdot \frac{1+v}{1+v^2}=\frac{-x}{x^2} \\ \\ v'\cdot \frac{1+v}{1+v^2}=-\frac{1}{x} \\ \\ \int {(\frac{1}{1+v^2}+\frac{v}{1+v^2})} \, dv =- \int {\frac{1}{x}} \, dx$

$arctg \, (v)+\frac{1}{2}\int {\frac{d(1+v^2)}{1+v^2}} =-\ln{|x|}+C \\ \\ arctg \, (v)+\frac{1}{2} \ln{|1+v^2|} =-\ln{|x|}+C \\ \\ arctg \, (\frac{y}{x})+\frac{1}{2} \ln{|1+(\frac{y}{x})^2|} =-\ln{|x|}+C \\ \\$

$y(1)=0 \\\\ arctg \, (\frac{0}{1})+\frac{1}{2} \ln{(1+(\frac{0}{1})^2)} =-\ln{|1|}+C \\ \\ 0+\frac{1}{2} \ln{1} =-0+C\\ \\ C=0 \\ \\$

$arctg \, (\frac{y}{x})+\frac{1}{2} \ln{|1+(\frac{y}{x})^2|} =-\ln{|x|}+0 \\\\ arctg \, (\frac{y}{x})+\frac{1}{2} \ln{|1+(\frac{y}{x})^2|} +\ln{|x|}=0$