Question

Помогите пожалуйста. Срочно нужно!

Answer 1

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

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

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

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

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

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