Question

помогите решить дифференциальные уравнения первого порядка
(y-1)dx +(1 – x)dy = 0
xdx – y(4 + x2) dy = 0
cos x cos y dx – sin x sin y dy = 0
x sin y dx + x cos y dy = 0

Answer 1

Ответ:

$\displaystyle 1)\ \ (y-1)\, dx+(1-x)\, dy=0\\\\\int \frac{dx}{1-x}=-\int \frac{dy}{y-1}\ \ \Rightarrow \ \ \ \ -ln|1-x|=-ln|y-1|+lnC\ \ ,\\\\\\C(1-x)=y-1\ \ ,\ \ \ \boxed{\ y=C(1-x)+1\ }\\\\\\2)\ \ x\, dx-y(4+x^2)\, dy=0\\\\\\\int \frac{x\, dx}{4+x^2}=\int y\, dy\ \ \Rightarrow \ \ \ \dfrac{1}{2}\cdot ln|4+x^2|=\dfrac{y^2}{2}-\frac{C}{2}\ \ ,\\\\\\ln(4+x^2)=y^2-C\ \ ,\ \ \ \boxed{\ y^2=ln(4+x^2)+C\ }$

$\displaystyle 3)\ \ cosx\cdot cosy\, dx-sinx\cdot siny\, dy=0\\\\\int \frac{cosx\, dx}{sinx}=\int \frac{siny\, dy}{cosy}\ \ \Rightarrow \ \ \ \int ctgx\, dx=\int tgy\, dy\ \ ,\\\\\\ln|sinx|=-ln|cosy|-lnC\ \ ,\ \ ln|sinx|=ln\, \frac{1}{|C\, cosy|}\ \ ,\ \ sinx=\frac{1}{C\, cosy}\\\\\boxed{\ cosy=\frac{1}{C\, sinx}\ }\\\\\\4)\ \ x\, siny\, dx+x\, cosy\, dy=0\\\\\int \frac{x\, dx}{x}=-\int \frac{cosy\, dy}{siny}\ \ \Rightarrow \ \ \ \ \int dx=-\int ctgy\, dy\ \ ,$

$\displaystyle x=-ln|siny|+lnC\ \ ,\ \ x=ln\frac{C}{siny}\ \ ,\ \ \frac{C}{siny}=e^{x}\ \ .\ \ \ siny=\frac{C}{e^{x}}\\\\\\\boxed{\ y=arcsin\frac{C}{e^{x}}\ }$