Question

Determine whether the equation is exact. If it​ is, then solve it. e Superscript t Baseline (7 y minus 3 t )dt plus (2 plus 7 e Superscript t Baseline )dy equals 0et(7y−3t)dy2+7et dy=0 Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.

Answer 1

Answer:

$F(t,y)=(2+7e^t)y+3(1-t)e^t +C$

Step-by-step explanation:

You have the following differential equation:

$e^t(7y-3t)dt+(2+7e^t)dy=0$

This equation can be written as:

$Mdt+Ndy=0$

where

$M=e^t(7y-3t)\\\\N=(2+7e^t)$

If the differential equation is exact, it is necessary the following:

$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial t}$

Then, you evaluate the partial derivatives:

$\frac{\partial M}{\partial y}=\frac{\partial}{\partial t}e^t(7y-3t)\\\\\frac{\partial M}{\partial t}=7e^t\\\\\frac{\partial N}{\partial t}=\frac{\partial}{\partial t}(2+7e^t)\\\\\frac{\partial N}{\partial t}=7e^t\\\\\frac{\partial M}{\partial t} = \frac{\partial N}{\partial t}$

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:

$F(t,y)=\int N \partial y = (2 +7e^t)y+g(t)$        (1)

Next, you derive the last equation respect to t:

$\frac{\partial F(t,y)}{\partial t}=7ye^t+g'(t)$

however, the last derivative must be equal to M. From there you can calculate g(t):

$\frac{\partial F(t,y)}{\partial t}=M=(7y-3t)e^t=7ye^t+g'(t)\\\\g'(t)=-3te^t\\\\g(t)=-3\int te^tdt=-3[te^t-\int e^tdt]=-3[te^t-e^t]$

Hence, by replacing g(t) in the expression (1) for F(t,y) you obtain:

$F(t,y)=(2+7e^t)y+3(1-t)e^t +C$

where C is the constant of integration