Question

помогите решить задачу Коши

Answer 1

Ответ.

$y'=\dfrac{y^2-3xy}{y^2-x^2}\ \ ,\ \ \ y(1)=2\\\\\\y'=\dfrac{\dfrac{y^2}{x^2}-3\cdot \dfrac{y}{x}}{\dfrac{y^2}{x^2}-1}\ \ ,\ \ \ t=\dfrac{y}{x}\ \ ,\ \ y=tx\ \ ,\ \ y'=t'x+t\ \ ,\\\\\\t'x+t=\dfrac{t^2-3t}{t^2-1}\ \ ,\ \ \ t'x=\dfrac{t^2-3t}{t^2-1}-t\ \ ,\ \ t'x=\dfrac{t^2-3t-t^3+t}{t^2-1}\ \ ,\\\\\\\displaystyle t'x=\dfrac{-t^3+t^2-2t}{t^2-1}\ \ ,\ \ \ \int \dfrac{(t^2-1)dt}{-t\, (t^2-t+2)}=\int \dfrac{dx}{x}\ \ ,$

$\displaystyle \star \ \int \dfrac{(t^2-1)\, dt}{t\cdot (t^2-t+2)}=-\frac{1}{2}\int \frac{dt}{t}+\int \frac{\dfrac{3}{2}\, t-\dfrac{1}{2}}{t^2-t+2}\, dt=\\\\\\=-\frac{1}{2}\, ln|t|+\frac{1}{2}\int \frac{(3t-1)\, dt}{(t-\frac{1}{2})^2+\frac{7}{4}}=-\frac{1}{2}\, ln|t|+\frac{1}{2}\int \frac{3(t-\frac{1}{2})+\frac{1}{2}}{(t-\frac{1}{2})^2+\frac{7}{4}}\, dt=$

$\displaystyle =-\frac{1}{2}\, ln|t|+\frac{3}{4}\int \frac{2(t-\frac{1}{2})\, dt}{(t-\frac{1}{2})^2+\frac{7}{4}}+\frac{1}{2}\int \frac{dt}{(t-\frac{1}{2})^2+\frac{7}{4}}=$

$\displaystyle =-\frac{1}{2}\, ln|t|+\frac{3}{4}\, ln\Big|(t-\frac{1}{2})^2+\frac{7}{4}\Big|+\frac{1}{2}\cdot \frac{2}{\sqrt7}\, arctg\frac{2\, (t-\frac{1}{2})}{\sqrt7}-C\ ;$

$\displaystyle -\frac{1}{2}\ ln\Big|\, \frac{y}{x}\, \Big|+\frac{3}{4}\, ln\Big|\frac{y^2}{x^2}-\frac{y}{x}+2\Big|+\frac{1}{\sqrt7}\, arctg\frac{2\, (\frac{y}{x}-\frac{1}{2})}{\sqrt7}-C=-ln|x|\\\\\\-\frac{1}{2}\ ln\Big|\, \frac{y}{x}\, \Big|+\frac{3}{4}\, ln\Big|\frac{y^2}{x^2}-\frac{y}{x}+2\Big|+\frac{1}{\sqrt7}\, arctg\frac{2y-x}{\sqrt7\, x}-C=-ln|x|\ \ -\ obshee\ reshenie$

$\displaystyle y(1)=2:\ \ -\frac{1}{2}\ ln2+\frac{3}{4}\, ln4+\frac{1}{\sqrt7}\, arctg\frac{3}{\sqrt7}-C=-ln1\ \ \Rightarrow \\\\\\C=-\frac{1}{2}\ ln2+\frac{3}{4}\, ln4+\frac{1}{\sqrt7}\, arctg\frac{3}{\sqrt7}\\\\\\-\frac{1}{2}\ ln\Big|\, \frac{y}{x}\, \Big|+\frac{3}{4}\, ln\Big|\frac{y^2}{x^2}-\frac{y}{x}+2\Big|+\frac{1}{\sqrt7}\, arctg\frac{2y-x}{\sqrt7\, x}+\frac{1}{2}\ ln2+\frac{3}{4}\, ln4+\frac{1}{\sqrt7}\, arctg\frac{3}{\sqrt7}=\\\\=-ln|x|\ \ \ \ \ -\ chastnoe\ reshenie$