Question

Помогите, пожалуйста, решить дифференциальное уравнение

(x/(sqrt(x^2-y^2)-1))dx-(y/(sqrt(x^2-y^2)))dy=0

Answer 1

Смотри полученное решение

Answer 2

$Big (frac{x}{sqrt{x^2-y^2}}-1Big ), dx-frac{y}{sqrt{x^2-y^2}}, dy=0\\frac{x-sqrt{x^2-y^2}}{sqrt{x^2-y^2}}, dx=frac{y}{sqrt{x^2-y^2}}, dy\\frac{dy}{dx}=frac{(x-sqrt{x^2-y^2})cdot sqrt{x^2-y^2}}{ycdot sqrt{x^2-y^2}}, Big |frac{:x}{:x}\\y'=frac{(1-sqrt{1-(frac{y}{x})^2}, )cdot sqrt{1-(frac{y}{x})^2}}{frac{y}{x}cdot sqrt{1-(frac{y}{x})^2}}\\t=frac{y}{x}; ,; ; y'=t'x+t\\t'x+t=frac{(1-sqrt{1-t^2})cdot sqrt{1-t^2}}{tcdot sqrt{1-t^2}}=frac{sqrt{1-t^2}-(1-t^2)}{tcdot sqrt{1-t^2}}$

$t'x+t=frac{sqrt{1-t^2}-1+t^2}{tcdot sqrt{1-t^2}}\\t'x=frac{sqrt{1-t^2}-1+t^2}{t, sqrt{1-t^2}}-t=frac{sqrt{1-t^2}}{t, sqrt{1-t^2}}-frac{1-t^2}{t, sqrt{1-t^2}}-t\\t'x=frac{1}{t}-frac{sqrt{1-t^2}}{t}-t; ; to ; ; t'=frac{1}{x}cdot Big (frac{1}{t}-frac{sqrt{1-t^2}}{t}-tBig )\\frac{dt}{dx}=frac{1}{x}cdot (frac{1}{t}-frac{sqrt{1-t^2}}{t}-t)\\ frac{dt}{frac{1}{t}-frac{sqrt{1-t^2}}{t}-t}=frac{dx}{x}\\frac{t, dt}{1-sqrt{1-t^2}-t^2}=frac{dx}{x}$

$a); int frac{t, dt}{(1-t^2)-sqrt{1-t^2}}=[, u=1-t^2; ,; du=-2t, dt, ]=-frac{1}{2}int frac{du}{u-sqrt{u}}=\\=[, u=z^2; ,; du=2z, dz; ,; z=sqrt{u}, ]=-frac{1}{2}int frac{2z, dz}{z^2-z}=\\=-int frac{z, dz}{z(z-1)}=-int frac{dz}{z-1}=-ln|z-1|+lnC_1=lnfrac{C_1}{|z-1|}=\\=lnfrac{C_1}{|sqrt{u}-1|}=lnfrac{C_1}{|sqrt{1-t^2}-1|}=lnfrac{C_1}{|sqrt{1-frac{y^2}{x^2}}-1|}=lnfrac{C_1, |x|}{|sqrt{x^2-y^2}-x|}; ;\\b); ; int frac{dx}{x}=ln|x|+ln C_2=ln(C_2|x|); ;$

$lnfrac{C_1|x|}{|sqrt{x^2-y^2}, -x|}=ln(C_2|x|)\\frac{C_1|x|}{|sqrt{x^2-y^2}, -x|}=C_2|x|\\sqrt{x^2-y^2}, -x=C; ; ; ; (, C= frac{C_1}{C_2}, )$