2. 0,32.
3. 0,25
4. 0,06
5. 0,85
1. -
Может так
1)6*3=18 (д)- июль
2)18+6=24 (д)- июль и июнь вместе
3)24-9=15(д)-август
4)24+15=39 (д)
Ответ:39 дней было в течение лета
1) y''-2y'-3y=0 ⇒ β²-2β-3=0 ⇒β=-1,β=3
ФСР:
![\Phi .C.P.:\ e^{-x}; e^{3x}](https://tex.z-dn.net/?f=%5CPhi+.C.P.%3A%5C+e%5E%7B-x%7D%3B+e%5E%7B3x%7D)
![y(x)=c_1e^{-x}+c_2e^{3x}](https://tex.z-dn.net/?f=y%28x%29%3Dc_1e%5E%7B-x%7D%2Bc_2e%5E%7B3x%7D)
![y'(x)=c'_1e^{-x}-c_1e^{-x}+c'_2e^{3x}+3c_2e^{3x}.](https://tex.z-dn.net/?f=y%27%28x%29%3Dc%27_1e%5E%7B-x%7D-c_1e%5E%7B-x%7D%2Bc%27_2e%5E%7B3x%7D%2B3c_2e%5E%7B3x%7D.)
Полагаем, что
![c'_1e^{-x}+c'_2e^{3x}=0](https://tex.z-dn.net/?f=c%27_1e%5E%7B-x%7D%2Bc%27_2e%5E%7B3x%7D%3D0)
, тогда
![y'(x)=-c_1e^{-x}+3c_2e^{3x}\\ y''(x)=-c'_1e^{-x}+c_1e^{-x}+3c'_2e^{3x}+9c_2e^{3x}.](https://tex.z-dn.net/?f=y%27%28x%29%3D-c_1e%5E%7B-x%7D%2B3c_2e%5E%7B3x%7D%5C%5C+%0Ay%27%27%28x%29%3D-c%27_1e%5E%7B-x%7D%2Bc_1e%5E%7B-x%7D%2B3c%27_2e%5E%7B3x%7D%2B9c_2e%5E%7B3x%7D.)
Подставим выражения для y, y' и y'' в исходное уравнение:
![(-c'_1e^{-x}+c_1e^{-x}+3c'_2e^{3x}+9c_2e^{3x})-2(-c_1e^{-x}+3c_2e^{3x})-\\ -3(c_1e^{-x}+c_2e^{3x})=e^{4x}](https://tex.z-dn.net/?f=%28-c%27_1e%5E%7B-x%7D%2Bc_1e%5E%7B-x%7D%2B3c%27_2e%5E%7B3x%7D%2B9c_2e%5E%7B3x%7D%29-2%28-c_1e%5E%7B-x%7D%2B3c_2e%5E%7B3x%7D%29-%5C%5C+-3%28c_1e%5E%7B-x%7D%2Bc_2e%5E%7B3x%7D%29%3De%5E%7B4x%7D)
Раскрываем скобки, приводим подобные слагаемые, получим:
![-c'_1e^{-x}+3c'_2e^{3x}=e^{4x}](https://tex.z-dn.net/?f=-c%27_1e%5E%7B-x%7D%2B3c%27_2e%5E%7B3x%7D%3De%5E%7B4x%7D)
Решаем систему уравнений:
![\begin{cases} c'_1e^{-x}+c'_2e^{3x}=0 \\ -c'_1e^{-x}+3c'_2e^{3x}=e^{4x} \end{cases} \ \textless \ =\ \textgreater \ \begin{cases}4c'_2e^{3x}=e^{4x} \\ c'_1=-c'_2e^{4x} \end{cases} \ \textless \ =\ \textgreater \ \\ \begin{cases}c'_2= \frac{1}{4} e^x \\ c'_1=-\frac{1}{4}e^{5x} \end{cases} =\ \textgreater \ \begin{cases}c_2= \frac{1}{4} e^x + \acute {C_2} \\ c_1=-\frac{1}{20}e^{5x}+ \acute {C_1} \end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D+c%27_1e%5E%7B-x%7D%2Bc%27_2e%5E%7B3x%7D%3D0+%5C%5C+-c%27_1e%5E%7B-x%7D%2B3c%27_2e%5E%7B3x%7D%3De%5E%7B4x%7D+%5Cend%7Bcases%7D+%5C+%5Ctextless+%5C+%3D%5C+%5Ctextgreater+%5C++%5Cbegin%7Bcases%7D4c%27_2e%5E%7B3x%7D%3De%5E%7B4x%7D+%5C%5C+c%27_1%3D-c%27_2e%5E%7B4x%7D+%5Cend%7Bcases%7D+%5C+%5Ctextless+%5C+%3D%5C+%5Ctextgreater+%5C++%5C%5C+%5Cbegin%7Bcases%7Dc%27_2%3D+%5Cfrac%7B1%7D%7B4%7D+e%5Ex+%5C%5C+c%27_1%3D-%5Cfrac%7B1%7D%7B4%7De%5E%7B5x%7D+%5Cend%7Bcases%7D+%3D%5C+%5Ctextgreater+%5C++%5Cbegin%7Bcases%7Dc_2%3D+%5Cfrac%7B1%7D%7B4%7D+e%5Ex+%2B+%5Cacute+%7BC_2%7D+%5C%5C+c_1%3D-%5Cfrac%7B1%7D%7B20%7De%5E%7B5x%7D%2B+%5Cacute+%7BC_1%7D++%5Cend%7Bcases%7D)
Полученные для с1 и с2 выражения подставляем в формулу решения:
![y(x)=c_1e^{-x}+c_2e^{3x}=(-\frac{1}{20}e^{5x}+ \acute {C_1})e^{-x}+(\frac{1}{4} e^x + \acute {C_2})e^{3x}=\\ = \acute {C_1}e^{-x}+ \acute {C_2}e^{3x}+\frac{1}{5}e^{4x}.](https://tex.z-dn.net/?f=y%28x%29%3Dc_1e%5E%7B-x%7D%2Bc_2e%5E%7B3x%7D%3D%28-%5Cfrac%7B1%7D%7B20%7De%5E%7B5x%7D%2B+%5Cacute+%7BC_1%7D%29e%5E%7B-x%7D%2B%28%5Cfrac%7B1%7D%7B4%7D+e%5Ex+%2B+%5Cacute+%7BC_2%7D%29e%5E%7B3x%7D%3D%5C%5C+%3D+%5Cacute+%7BC_1%7De%5E%7B-x%7D%2B+%5Cacute+%7BC_2%7De%5E%7B3x%7D%2B%5Cfrac%7B1%7D%7B5%7De%5E%7B4x%7D.)
Итак, найдено общее решение исходного уравнения:
![y=\acute {C_1}e^{-x}+ \acute {C_2}e^{3x}+\frac{1}{5}e^{4x}.](https://tex.z-dn.net/?f=y%3D%5Cacute+%7BC_1%7De%5E%7B-x%7D%2B+%5Cacute+%7BC_2%7De%5E%7B3x%7D%2B%5Cfrac%7B1%7D%7B5%7De%5E%7B4x%7D.)
Из условий у(0)=0 и у'(0)=0 найдем С1 и С2:
![y(0)=0\ =\ \textgreater \ \acute {C_1} + \acute {C_2}+\frac{1}{5}=0.\\ y'=(\acute {C_1}e^{-x}+ \acute {C_2}e^{3x}+\frac{1}{5}e^{4x})'=-\acute {C_1}e^{-x}+ 3\acute {C_2}e^{3x}+\frac{4}{5}e^{4x}.\\ y'(0)=0\ =\ \textgreater \ -\acute {C_1}}+ 3\acute {C_2}+\frac{4}{5}=0.](https://tex.z-dn.net/?f=y%280%29%3D0%5C+%3D%5C+%5Ctextgreater+%5C++%5Cacute+%7BC_1%7D+%2B+%5Cacute+%7BC_2%7D%2B%5Cfrac%7B1%7D%7B5%7D%3D0.%5C%5C+%0Ay%27%3D%28%5Cacute+%7BC_1%7De%5E%7B-x%7D%2B+%5Cacute+%7BC_2%7De%5E%7B3x%7D%2B%5Cfrac%7B1%7D%7B5%7De%5E%7B4x%7D%29%27%3D-%5Cacute+%7BC_1%7De%5E%7B-x%7D%2B+3%5Cacute+%7BC_2%7De%5E%7B3x%7D%2B%5Cfrac%7B4%7D%7B5%7De%5E%7B4x%7D.%5C%5C+%0Ay%27%280%29%3D0%5C+%3D%5C+%5Ctextgreater+%5C++-%5Cacute+%7BC_1%7D%7D%2B+3%5Cacute+%7BC_2%7D%2B%5Cfrac%7B4%7D%7B5%7D%3D0.)
Решаем последнюю систему:
![\begin{cases} \acute {C_1} + \acute {C_2}=-\frac{1}{5} \\ \acute {C_1}}- 3\acute {C_2}=\frac{4}{5} \end{cases} \ \textless \ =\ \textgreater \ \begin{cases} 4 \acute {C_2}=-1 \\ \acute {C_1}}=-\acute {C_2}-\frac{1}{5} \end{cases} \ \textless \ =\ \textgreater \ \begin{cases} \acute {C_2}=-\frac{1}{4} \\ \acute {C_1}}=\frac{1}{20} \end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D+%5Cacute+%7BC_1%7D+%2B+%5Cacute+%7BC_2%7D%3D-%5Cfrac%7B1%7D%7B5%7D+%5C%5C+%5Cacute+%7BC_1%7D%7D-+3%5Cacute+%7BC_2%7D%3D%5Cfrac%7B4%7D%7B5%7D+%5Cend%7Bcases%7D+%5C+%5Ctextless+%5C+%3D%5C+%5Ctextgreater+%5C++%5Cbegin%7Bcases%7D+4+%5Cacute+%7BC_2%7D%3D-1+%5C%5C+%5Cacute+%7BC_1%7D%7D%3D-%5Cacute+%7BC_2%7D-%5Cfrac%7B1%7D%7B5%7D+%5Cend%7Bcases%7D+%5C+%5Ctextless+%5C+%3D%5C+%5Ctextgreater+%5C++%5Cbegin%7Bcases%7D+%5Cacute+%7BC_2%7D%3D-%5Cfrac%7B1%7D%7B4%7D++%5C%5C+%5Cacute+%7BC_1%7D%7D%3D%5Cfrac%7B1%7D%7B20%7D+%5Cend%7Bcases%7D+)
Найденные числа подставим в полученное общее решение:
![y=\frac{1}{20} e^{-x}-\frac{1}{4}e^{3x}+\frac{1}{5} e^{4x}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B1%7D%7B20%7D+e%5E%7B-x%7D-%5Cfrac%7B1%7D%7B4%7De%5E%7B3x%7D%2B%5Cfrac%7B1%7D%7B5%7D+e%5E%7B4x%7D)
- это ответ!
Пусть стоимость кисточки х, тогда стоимость краски 9х. По условию за кисточку заплатили на 20 руб меньше, чем за краски. Составим уравнение
9х-х=20
8х=20
х=20:8
х=2,5 т.е. 2 руб 50 коп за кисточку
9*2,5=22,5 22 руб 50 коп за краски