Question

Y'=y²/x²+6y/x+6 найти общий интеграл дифференциального уравнения

Answer 1

$y '= \frac{ {y}^{2} }{ {x}^{2} } + \frac{6y}{x} + 6 \\$

Это однородное ДУ

Замена:

$\frac{y}{x} = u \\ y' = u'x + u \\ \\ ux + u = u {}^{2} + 6 u + 6 \\ \frac{du}{dx} x = {u}^{2} + 5 u + 6 \\ \int\limits \frac{du}{u {}^{2} + 5 u + 6} = \int\limits \frac{dx}{x} \\ \\ u {}^{2} + 5u + 6 = {u}^{2} + 2 \times u \times \frac{5}{2} + \frac{25}{4} - \frac{1}{4} = \\ = (u + \frac{5}{2} ) {}^{2} - ( \frac{1}{2} ) {}^{2} \\ \\ \int\limits \frac{d(u + \frac{5}{2}) }{(u + \frac{5}{2}) {}^{2} - ( \frac{1}{2}) {}^{2} } = ln( |x| ) + ln(C) \\ \frac{1}{2 \times \frac{1}{2} } ln( | \frac{u + \frac{5}{2} - \frac{1}{2} }{u + \frac{5}{2} + \frac{1}{2} } | ) = ln( |Cx| ) \\ ln( | \frac{u + 2}{u + 3} | ) = ln( |Cx| ) \\ \frac{ \frac{y}{x} + 2}{ \frac{y}{x} + 3} = Cx \\ \frac{y + 2x}{x} \times \frac{x}{y + 3x} = Cx \\ \frac{y + 2x}{y + 3y} = Cx$

общее решение

Answer 2

$y'=\dfrac{y^2}{x^2}+\dfrac{6y}{x}+6\\\\t=\dfrac{y}{x}\ ,\ \ y=tx\ \ ,\ \ y'=t'x+t\\\\t'x+t=t^2+6t+6\ \ ,\ \ t'x=t^2+5t+6\ \ ,\ \ \dfrac{dt}{dx}\cdot x=t^2+5t+6\ \ ,\\\\\int \dfrac{dt}{t^2+5t+6}=\int \dfrac{dx}{x}\\\\\\\star \ \int \dfrac{dt}{t^2+5t+6}=\int \dfrac{dt}{(t+\frac{5}{2})^2-\frac{25}{4}+6}=\int \dfrac{dt}{(t+\frac{5}{2})^2-\frac{1}{4}}=\int \dfrac{d(t+\frac{5}{2})}{(t+\frac{5}{2})^2-\frac{1}{4}}=$

$=\dfrac{1}{2\cdot \frac{1}{2}}\cdot ln\left|\, \dfrac{(t+\frac{5}{2})-\frac{1}{2}}{(t+\frac{5}{2})+\frac{1}{2}}\, \right|+C_1=ln\left|\dfrac{\frac{y}{x}+2}{\frac{y}{x}+3}\, \right|+C_1=ln\left|\dfrac{y+2x}{y+3x}\, \right|+C_1\ \ \star \\\\\\\\ln\left|\dfrac{y+2x}{y+3x}\, \right|=ln|x|+lnC\ \ ,\ \ \ \ ln\left|\dfrac{y+2x}{y+3x}\, \right|=ln|Cx|\ \ ,\\\\\\\boxed{\ \dfrac{y+2x}{y+3x}=Cx\ }$