Уравнение Бернулли
![y'+xy=xy^2|:y^2\\\frac{y'}{y^2}+\frac{x}{y}=x\\z=\frac{1}{y};z'=-\frac{y'}{y^2};\frac{y'}{y^2}=-z'\\-z'+xz=x\\z=uv;z'=u'v+v'u\\-u'v-v'u+xuv=x\\-u'v+u(-v'+xv)=x\\\begin{cases}-v'+xv=0\\u'v=-x\end{cases}\\-\frac{dv}{dx}+xv=0|*\frac{dx}{v}\\\frac{dv}{v}=xdx\\\int\frac{dv}{v}=\int xdx\\ln|v|=\frac{x^2}{2}\\v=e^\frac{x^2}{2}\\\frac{du}{dx}e^\frac{x^2}{2}=-x\\du=-xe^{-\frac{x^2}{2}}\\\int du=\int e^{-\frac{x^2}{2}}d(-\frac{x^2}{2})\\u=e^{-\frac{x^2}{2}}+C\\z=\frac{1}{y}=1+Ce^\frac{x^2}{2}](https://tex.z-dn.net/?f=y%27%2Bxy%3Dxy%5E2%7C%3Ay%5E2%5C%5C%5Cfrac%7By%27%7D%7By%5E2%7D%2B%5Cfrac%7Bx%7D%7By%7D%3Dx%5C%5Cz%3D%5Cfrac%7B1%7D%7By%7D%3Bz%27%3D-%5Cfrac%7By%27%7D%7By%5E2%7D%3B%5Cfrac%7By%27%7D%7By%5E2%7D%3D-z%27%5C%5C-z%27%2Bxz%3Dx%5C%5Cz%3Duv%3Bz%27%3Du%27v%2Bv%27u%5C%5C-u%27v-v%27u%2Bxuv%3Dx%5C%5C-u%27v%2Bu%28-v%27%2Bxv%29%3Dx%5C%5C%5Cbegin%7Bcases%7D-v%27%2Bxv%3D0%5C%5Cu%27v%3D-x%5Cend%7Bcases%7D%5C%5C-%5Cfrac%7Bdv%7D%7Bdx%7D%2Bxv%3D0%7C%2A%5Cfrac%7Bdx%7D%7Bv%7D%5C%5C%5Cfrac%7Bdv%7D%7Bv%7D%3Dxdx%5C%5C%5Cint%5Cfrac%7Bdv%7D%7Bv%7D%3D%5Cint+xdx%5C%5Cln%7Cv%7C%3D%5Cfrac%7Bx%5E2%7D%7B2%7D%5C%5Cv%3De%5E%5Cfrac%7Bx%5E2%7D%7B2%7D%5C%5C%5Cfrac%7Bdu%7D%7Bdx%7De%5E%5Cfrac%7Bx%5E2%7D%7B2%7D%3D-x%5C%5Cdu%3D-xe%5E%7B-%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%5C%5C%5Cint+du%3D%5Cint+e%5E%7B-%5Cfrac%7Bx%5E2%7D%7B2%7D%7Dd%28-%5Cfrac%7Bx%5E2%7D%7B2%7D%29%5C%5Cu%3De%5E%7B-%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%2BC%5C%5Cz%3D%5Cfrac%7B1%7D%7By%7D%3D1%2BCe%5E%5Cfrac%7Bx%5E2%7D%7B2%7D)
![y=\frac{1}{1+Ce^\frac{x^2}{2}}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B1%7D%7B1%2BCe%5E%5Cfrac%7Bx%5E2%7D%7B2%7D%7D)
ДУ с разделяющимися переменными.
![\frac{xtgydy}{dx}=1-x^2|*\frac{dx}{x}\\tgydy=(\frac{1}{x}-x)dx\\\int tgydy=\int(\frac{1}{x}-x)dx\\-ln|cosy|=ln|x|-\frac{x^2}{2}+C\\\frac{x^2}{2}-ln|xcosy|=C](https://tex.z-dn.net/?f=%5Cfrac%7Bxtgydy%7D%7Bdx%7D%3D1-x%5E2%7C%2A%5Cfrac%7Bdx%7D%7Bx%7D%5C%5Ctgydy%3D%28%5Cfrac%7B1%7D%7Bx%7D-x%29dx%5C%5C%5Cint+tgydy%3D%5Cint%28%5Cfrac%7B1%7D%7Bx%7D-x%29dx%5C%5C-ln%7Ccosy%7C%3Dln%7Cx%7C-%5Cfrac%7Bx%5E2%7D%7B2%7D%2BC%5C%5C%5Cfrac%7Bx%5E2%7D%7B2%7D-ln%7Cxcosy%7C%3DC)
ДУ с разделяющимися переменными.
![(1+x^2)dy-2x(y+3)dx=0|*\frac{1}{(y+3)(1+x^2)}\\\frac{dy}{y+3}=\frac{2xdx}{1+x^2}\\\int\frac{d(y+3)}{y+3}=\int\frac{d(1+x^2)}{1+x^2}\\ln|y+3|=ln|1+x^2|+ln|C|\\y+3=C(1+x^2)\\y=C(1+x^2)-3\\y(0)=-1\\-1=C-3\\C=2\\y=2(1+x^2)-3](https://tex.z-dn.net/?f=%281%2Bx%5E2%29dy-2x%28y%2B3%29dx%3D0%7C%2A%5Cfrac%7B1%7D%7B%28y%2B3%29%281%2Bx%5E2%29%7D%5C%5C%5Cfrac%7Bdy%7D%7By%2B3%7D%3D%5Cfrac%7B2xdx%7D%7B1%2Bx%5E2%7D%5C%5C%5Cint%5Cfrac%7Bd%28y%2B3%29%7D%7By%2B3%7D%3D%5Cint%5Cfrac%7Bd%281%2Bx%5E2%29%7D%7B1%2Bx%5E2%7D%5C%5Cln%7Cy%2B3%7C%3Dln%7C1%2Bx%5E2%7C%2Bln%7CC%7C%5C%5Cy%2B3%3DC%281%2Bx%5E2%29%5C%5Cy%3DC%281%2Bx%5E2%29-3%5C%5Cy%280%29%3D-1%5C%5C-1%3DC-3%5C%5CC%3D2%5C%5Cy%3D2%281%2Bx%5E2%29-3)
Площ.квад=а² а=√72 -сторона квад.
диагональ квад. это диаметр круга
b²=a²+a² b=√72+72=√144=12-диаметр круга
r=b\2=6-радиус
S=πr²=6²π=36π
ответ; площадь круга 36π
3 и 8/85 сократили дробь на 3(разделили числитель и знаменатель на 3)
1)65*2=130 км проедет пассажир за 2 часа
2)45+130=175 км расстояние от города
===============================================