Question

Помогите пожалуйста. Срочно нужно!

Answer 1

$1); ; y'=dfrac{y}{x}+dfrac{x^2}{y^2}\\u=frac{y}{x}; ; ,; ; y=ux; ,; ; y'=u'x+u\\u'x+u=u+dfrac{1}{u^2}; ; ,; ; ; u'x=dfrac{1}{u^2}; ; ,; ; dfrac{du}{dx}cdot x=dfrac{1}{u^2}; ,\\int u^2, du=int dfrac{dx}{x}\\dfrac{u^3}{3}=ln|x|+C; ; ,; ; dfrac{y^3}{3x^3}=ln|x|+C; ; ,; ; underline {; y^3=3x^3cdot (ln|x|+C); }$

$2); ; xy'=y+dfrac{x}{cosfrac{y}{x}}\\u=frac{y}{x}; ; ,; ; y=ux; ,; ; y'=u'x+u\\u'x^2+ux=ux+dfrac{x}{cosu}; ; ,; ; u'x^2=dfrac{x}{cosu}; ; ,; ; u'=dfrac{1}{xcdot cosu}; ,\\int cosu, du=int dfrac{dx}{x}; ; ,; ; sinu=ln|x|+C; ,\\underline {; sinfrac{y}{x}=ln|x|+C; }; ; to ; ; underline {; y=xcdot arcsin(ln|x|+C); }$

$3); ; xy'=sqrt{y^2-9x^2}+y\\u=frac{y}{x}; ; ,; ; y=ux; ,; ; y'=u'x+u\\x, (u'x+u)=sqrt{u^2x^2-9x^2}+ux; ; ,; ; u'x+u=dfrac{sqrt{x^2(u^2-9)}}{x}+u\\u'x=dfrac{x, sqrt{u^2-9}}{x}; ; ,; ; dfrac{du}{dx}=dfrac{sqrt{u^2-9}}{x}; ; ,; ; int dfrac{du}{sqrt{u^2-9}}=int dfrac{dx}{x}; ,\\ln|u+sqrt{u^2-9}|=ln|x|+lnC\\underline {; frac{y}{x}+sqrt{frac{y^2}{x^2}-9}=Cx; }$

$4); ; y'=e^{-frac{y}{x}}+dfrac{y}{x}\\t=frac{y}{x}; ; ,; ; y=tx; ,; ; y'=t'x+t\\t'x+t=e^{-t}+t; ; ,; ; t'=dfrac{e^{-t}}{x}; ; ,; ; int e^{t}, dt=int dfrac{dx}{x}; ,\\e^{t}=ln|x|+C; ; ,; ; underline {; e^{-frac{y}{x}}=ln|x|+C; }; ; ; ili; ; ; -frac{y}{x}=ln, (ln|x|+C); ,\\underline {; y=-xcdot ln, (ln|x|+C); }$

$5); ; xy'=y+x, sin^2frac{y}{x}\\u=frac{y}{x}; ; ,; ; y=tx; ,; ; y'=t'x+t\\x, (t'x+t)=tx+x, sin^2t; ; ,; ; t'x+t=t+sin^2t; ; ,; ; t'=dfrac{sin^2t}{x}; ,\\int dfrac{dt}{sin^2t}=int dfrac{dx}{x}; ; ,; ; -ctgt=ln|x|+C; ; ,; ; underline {; ctgfrac{y}{x}=-ln|x|-C; }$

$6); ; xy'=y+sqrt{2x^2-y^2}\\y'=frac{y}{x}+sqrt{2-(frac{y}{x})^2}\\u=frac{y}{x}; ; ,; ; y=ux; ,; ; y'=u'x+u\\u'x+u=u+sqrt{2-u^2}; ; ,; ; u'=dfrac{sqrt{2-u^2}}{x}; ,; ; int dfrac{du}{sqrt{2-u^2}}=int dfrac{dx}{x}; ,\\arcsinfrac{u}{sqrt2}=ln|x|+C; ; ,; ; underline {; arcsinfrac{y}{xsqrt2}=ln|x|+C; }; ; ; ili\\underline {; y=xsqrt2cdot sin, (ln|x|+C); }$