Answer:
I believe the lines are the 2nd, 4th, and 5th
Step-by-step explanation:
I don't have a clear explani=ation and you might get confused when I try to explain it :/
Answer: fgm
Step-by-step explanation:
Answer: The required solution is ![y=50e^{0.1386t}.](https://tex.z-dn.net/?f=y%3D50e%5E%7B0.1386t%7D.)
Step-by-step explanation:
We are given to solve the following differential equation :
![\dfrac{dy}{dt}=ky~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)](https://tex.z-dn.net/?f=%5Cdfrac%7Bdy%7D%7Bdt%7D%3Dky~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~%28i%29)
where k is a constant and the equation satisfies the conditions y(0) = 50, y(5) = 100.
From equation (i), we have
![\dfrac{dy}{y}=kdt.](https://tex.z-dn.net/?f=%5Cdfrac%7Bdy%7D%7By%7D%3Dkdt.)
Integrating both sides, we get
![\int\dfrac{dy}{y}=\int kdt\\\\\Rightarrow \log y=kt+c~~~~~~[\textup{c is a constant of integration}]\\\\\Rightarrow y=e^{kt+c}\\\\\Rightarrow y=ae^{kt}~~~~[\textup{where }a=e^c\textup{ is another constant}]](https://tex.z-dn.net/?f=%5Cint%5Cdfrac%7Bdy%7D%7By%7D%3D%5Cint%20kdt%5C%5C%5C%5C%5CRightarrow%20%5Clog%20y%3Dkt%2Bc~~~~~~%5B%5Ctextup%7Bc%20is%20a%20constant%20of%20integration%7D%5D%5C%5C%5C%5C%5CRightarrow%20y%3De%5E%7Bkt%2Bc%7D%5C%5C%5C%5C%5CRightarrow%20y%3Dae%5E%7Bkt%7D~~~~%5B%5Ctextup%7Bwhere%20%7Da%3De%5Ec%5Ctextup%7B%20is%20another%20constant%7D%5D)
Also, the conditions are
![y(0)=50\\\\\Rightarrow ae^0=50\\\\\Rightarrow a=50](https://tex.z-dn.net/?f=y%280%29%3D50%5C%5C%5C%5C%5CRightarrow%20ae%5E0%3D50%5C%5C%5C%5C%5CRightarrow%20a%3D50)
and
![y(5)=100\\\\\Rightarrow 50e^{5k}=100\\\\\Rightarrow e^{5k}=2\\\\\Rightarrow 5k=\log_e2\\\\\Rightarrow 5k=0.6931\\\\\Rightarrow k=0.1386.](https://tex.z-dn.net/?f=y%285%29%3D100%5C%5C%5C%5C%5CRightarrow%2050e%5E%7B5k%7D%3D100%5C%5C%5C%5C%5CRightarrow%20e%5E%7B5k%7D%3D2%5C%5C%5C%5C%5CRightarrow%205k%3D%5Clog_e2%5C%5C%5C%5C%5CRightarrow%205k%3D0.6931%5C%5C%5C%5C%5CRightarrow%20k%3D0.1386.)
Thus, the required solution is ![y=50e^{0.1386t}.](https://tex.z-dn.net/?f=y%3D50e%5E%7B0.1386t%7D.)
C
All u have to do is add 23.5 to 23.5