Answer:
[tex]F(t,y)=(2+7e^t)y+3(1-t)e^t +C[/tex]
Step-by-step explanation:
You have the following differential equation:
[tex]e^t(7y-3t)dt+(2+7e^t)dy=0[/tex]
This equation can be written as:
[tex]Mdt+Ndy=0[/tex]
where
[tex]M=e^t(7y-3t)\\\\N=(2+7e^t)[/tex]
If the differential equation is exact, it is necessary the following:
[tex]\frac{\partial M}{\partial y}=\frac{\partial N}{\partial t}[/tex]
Then, you evaluate the partial derivatives:
[tex]\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}[/tex]
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:
[tex]F(t,y)=\int N \partial y = (2 +7e^t)y+g(t)[/tex] (1)
Next, you derive the last equation respect to t:
[tex]\frac{\partial F(t,y)}{\partial t}=7ye^t+g'(t)[/tex]
however, the last derivative must be equal to M. From there you can calculate g(t):
[tex]\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][/tex]
Hence, by replacing g(t) in the expression (1) for F(t,y) you obtain:
[tex]F(t,y)=(2+7e^t)y+3(1-t)e^t +C[/tex]
where C is the constant of integration
