Question

Проинтегрировать однородное уравнение. С полным решением $(5x+y)dy-(x-5y)dx=0$
+65б

Answer 1

$(5x+y)\, dy-(x-5y)\, dx=0\\\\\\\dfrac{dy}{dx}=\dfrac{x-5y}{5x+y}\ \ \ ,\ \ \ \ \ \ \ u=\dfrac{y}{x}\ \ ,\ \ y=ux\ ,\ \ y'=u'x+u\\\\\\u'x+u=\dfrac{x-5ux}{5x+ux}\ \ ,\ \ \ u'x+u=\dfrac{1-5u}{5+u}\ \ ,\ \ u'x=\dfrac{1-5u-5u-u^2}{5+u}\ \ ,\\\\\\\dfrac{du}{dx}=\dfrac{1-10u-u^2}{x(5+u)}\ \ ,\ \ \int \dfrac{(5+u)\, du}{u^2+10u-1}=-\int \dfrac{dx}{x}$

$\int \dfrac{(5+u)\, du}{u^2+10u-1}=\int \dfrac{(5+u)\, du}{(u+5)^2-26}=\dfrac{1}{2}\int \dfrac{2(5+u)\, du}{(u+5)^2-26}=\\\\\\=\dfrac{1}{2}\cdot ln\Big|(u+5)^2-26\Big|+C_1=\dfrac{1}{2}\cdot ln\Big|\, \dfrac{y^2}{x^2}+\dfrac{10y}{x} -1\, \Big|+C_1\ ;\\\\\\\dfrac{1}{2}\cdot ln\Big|\, \dfrac{y^2}{x^2}+\dfrac{10y}{x} -1\, \Big|=ln|x|+lnC\\\\\\\sqrt{\dfrac{y^2}{x^2}+\dfrac{10y}{x} -1}=Cx$