Answer:
![F(t,y)=(2+7e^t)y+3(1-t)e^t +C](https://tex.z-dn.net/?f=F%28t%2Cy%29%3D%282%2B7e%5Et%29y%2B3%281-t%29e%5Et%20%2BC)
Step-by-step explanation:
You have the following differential equation:
![e^t(7y-3t)dt+(2+7e^t)dy=0](https://tex.z-dn.net/?f=e%5Et%287y-3t%29dt%2B%282%2B7e%5Et%29dy%3D0)
This equation can be written as:
![Mdt+Ndy=0](https://tex.z-dn.net/?f=Mdt%2BNdy%3D0)
where
![M=e^t(7y-3t)\\\\N=(2+7e^t)](https://tex.z-dn.net/?f=M%3De%5Et%287y-3t%29%5C%5C%5C%5CN%3D%282%2B7e%5Et%29)
If the differential equation is exact, it is necessary the following:
![\frac{\partial M}{\partial y}=\frac{\partial N}{\partial t}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpartial%20M%7D%7B%5Cpartial%20y%7D%3D%5Cfrac%7B%5Cpartial%20N%7D%7B%5Cpartial%20t%7D)
Then, you evaluate the partial derivatives:
![\frac{\partial M}{\partial y}=\frac{\partial}{\partial t}e^t(7y-3t)\\\\\frac{\partial M}{\partial t}=7e^t\\\\\frac{\partial N}{\partial t}=\frac{\partial}{\partial t}(2+7e^t)\\\\\frac{\partial N}{\partial t}=7e^t\\\\\frac{\partial M}{\partial t} = \frac{\partial N}{\partial t}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpartial%20M%7D%7B%5Cpartial%20y%7D%3D%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial%20t%7De%5Et%287y-3t%29%5C%5C%5C%5C%5Cfrac%7B%5Cpartial%20M%7D%7B%5Cpartial%20t%7D%3D7e%5Et%5C%5C%5C%5C%5Cfrac%7B%5Cpartial%20N%7D%7B%5Cpartial%20t%7D%3D%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial%20t%7D%282%2B7e%5Et%29%5C%5C%5C%5C%5Cfrac%7B%5Cpartial%20N%7D%7B%5Cpartial%20t%7D%3D7e%5Et%5C%5C%5C%5C%5Cfrac%7B%5Cpartial%20M%7D%7B%5Cpartial%20t%7D%20%3D%20%5Cfrac%7B%5Cpartial%20N%7D%7B%5Cpartial%20t%7D)
The partial derivatives are equal, then, the differential equation is exact.
In order to obtain the solution of the equation you first integrate M or N:
(1)
Next, you derive the last equation respect to t:
![\frac{\partial F(t,y)}{\partial t}=7ye^t+g'(t)](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpartial%20F%28t%2Cy%29%7D%7B%5Cpartial%20t%7D%3D7ye%5Et%2Bg%27%28t%29)
however, the last derivative must be equal to M. From there you can calculate g(t):
![\frac{\partial F(t,y)}{\partial t}=M=(7y-3t)e^t=7ye^t+g'(t)\\\\g'(t)=-3te^t\\\\g(t)=-3\int te^tdt=-3[te^t-\int e^tdt]=-3[te^t-e^t]](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpartial%20F%28t%2Cy%29%7D%7B%5Cpartial%20t%7D%3DM%3D%287y-3t%29e%5Et%3D7ye%5Et%2Bg%27%28t%29%5C%5C%5C%5Cg%27%28t%29%3D-3te%5Et%5C%5C%5C%5Cg%28t%29%3D-3%5Cint%20te%5Etdt%3D-3%5Bte%5Et-%5Cint%20e%5Etdt%5D%3D-3%5Bte%5Et-e%5Et%5D)
Hence, by replacing g(t) in the expression (1) for F(t,y) you obtain:
![F(t,y)=(2+7e^t)y+3(1-t)e^t +C](https://tex.z-dn.net/?f=F%28t%2Cy%29%3D%282%2B7e%5Et%29y%2B3%281-t%29e%5Et%20%2BC)
where C is the constant of integration