Question

Решить дифференциальные уравнения,

Answer 1

$6)\; \; dy=\sqrt{y^2+4}\, dx-x\, dy\\\\(1+x)\, dy=\sqrt{y^2+4}\, dx\\\\\int \frac{dy}{\sqrt{y^2+4}}=\int \frac{dx}{x+1}\\\\ln|y+\sqrt{y^2+4}|=ln|x+1|+lnC\\\\y+\sqrt{y^2+4}=C(x+1)\\\\\\7)\; \; \sqrt{x^2+4}\, dy-dx=y\, dx\\\\\sqrt{x^2+4}\, dy=(y+1)\, dx\\\\\int \frac{dy}{y+1}=\int \frac{dx}{\sqrt{x^2+4}}\\\\ln|y+1|=ln|x+\sqrt{x^2+4}|+lnC\\\\y+1=C\, (x+\sqrt{x^2+4})$

$8)\; \; y^2\, dx-2xy\, dy=4y\, dy-dx\\\\(y^2+1)\, dx=(2xy+4y)\, dy\\\\(y^2+1)\, dx=2y(x+2)\, dy\\\\\int \frac{dx}{x+2}=\int \frac{2y\, dy}{y^2+1}\\\\ln|x+2|=ln|y^2+1|+lnC\\\\x+2=C(y^2+1)\\\\\\9)\; \; 2x\sqrt{4-y^2\, }dx-dy=x^2\, dy\\\\2x\sqrt{4-y^2}\, dx=(x^2+1)\, dy\\\\\int \frac{2x\, dx}{x^2+1}=\int \frac{dy}{\sqrt{4-y^2}}\\\\ln|x^2+1|=arcsin\frac{y}{2}+C$

$10)\; \; x^2\, dy=\sqrt{y^2+1}\, dx-4\, dy\\\\(x^2+4)\, dy=\sqrt{y^2+1}\, dx\\\\\int \frac{dy}{\sqrt{y^2+1}}=\int \frac{dx}{x^2+4}\\\\ln|y+\sqrt{y^2+1}|=\frac{1}{2}\cdot arctg\frac{x}{2}+C$