Question

Решить однородное дифференциальное уравнение

Answer 1

Дифференциальное уравнение 1-го порядка, сводящееся к однородному:
$y' = \frac{x-2y-1}{x+y+1} \\\begin{cases}\alpha-2\beta-1=0\\\alpha+\beta+1=0\end{cases}\\-3\beta-2=0\\\beta=-\frac{2}{3}\\\alpha=-\frac{1}{3}\\x=\hat{x}-\frac{1}{3}\\y=\hat{y}-\frac{2}{3}\\d(\hat{x}-\frac{1}{3})=d(\hat{x})\\d(\hat{y}-\frac{2}{3})=d(\hat{y})$
$\hat{y'} = \frac{\hat{x}-2\hat{y}}{\hat{x}+\hat{y}}\\\hat{y}=t\hat{x};\hat{y'}=t'\hat{x}+t\\\frac{dt}{d\hat{x}}\hat{x}+t=\frac{\hat{x}-2t\hat{x}}{\hat{x}+t\hat{x}}\\\frac{dt}{d\hat{x}}\hat{x}=\frac{1-2t}{1+t}-t\\\frac{dt}{d\hat{x}}\hat{x}=\frac{1-3t-t^2}{1+t}\\\frac{d\hat{x}}{\hat{x}}=\frac{1+t}{-t^2-3t+1}dt\\\int\frac{d\hat{x}}{\hat{x}}=\int\frac{1+t}{-t^2-3t+1}dt\\\int\frac{d\hat{x}}{\hat{x}}=-\frac{1}{2}\int\frac{2t+2+1-1}{t^2+3t-1}dt\\$

$\int\frac{d\hat{x}}{\hat{x}}=-\frac{1}{2}\int(\frac{2t+3}{t^2+3t-1}-\frac{1}{t^2+3t-1})dt\\\int\frac{d\hat{x}}{\hat{x}}=-\frac{1}{2}\int\frac{d(t^2+3t-1)}{t^2+3t-1}+\frac{1}{2}\int\frac{d(t+\frac{3}{2})}{(t+\frac{3}{2})^2-\frac{13}{4}}\\ln|\hat{x}|=-\frac{1}{2}ln|t^2+3t-1|+\frac{1}{2\sqrt{13}}ln|\frac{2t+3-\sqrt{13}}{2t+3+\sqrt{13}}|+C\\ln|\hat{x}|=-\frac{1}{2}ln|(2t+3-\sqrt{13})(2t+3+\sqrt{13})|+\frac{1}{2\sqrt{13}}ln|\frac{2t+3-\sqrt{13}}{2t+3+\sqrt{13}}|+C$

$ln|\hat{x}|=-\frac{1}{2}ln|2t+3-\sqrt{13}|-\frac{1}{2}ln|2t+3+\sqrt{13}|+\\+\frac{1}{2\sqrt{13}}ln|2t+3-\sqrt{13}|-\frac{1}{2\sqrt{13}}ln|2t+3+\sqrt{13}|+C\\\\ln|\hat{x}|=-(\frac{1}{2}+\frac{1}{2\sqrt{13}})ln|2t+3+\sqrt{13}|+(\frac{1}{2\sqrt{13}}-\frac{1}{2})ln|2t+3+\sqrt{13}|+C\\$

$ln|\hat{x}|=(\frac{\sqrt{13}-13}{26})ln|2\frac{\hat{y}}{\hat{x}}+3+\sqrt{13}|-(\frac{\sqrt{13}+13}{26})ln|2\frac{\hat{y}}{\hat{x}}+3+\sqrt{13}|+C$
И окончательный ответ:
$ln|x+\frac{1}{3}|-(\frac{\sqrt{13}-13}{26})ln|\frac{2(3y+2)}{3x+1}+3+\sqrt{13}|+\\+(\frac{\sqrt{13}+13}{26})ln|\frac{2(3y+2)}{3x+1}+3+\sqrt{13}|=C$