Answer:
[tex]e^x+xy+3y+(y-1)e^y=4[/tex]
Step-by-step explanation:
Given that
[tex](e^x+y)dx+(3+x+ye^y)dy=0[/tex]
Here
[tex]M=e^x+y[/tex]
[tex]N=3+x+ye^y[/tex]
We know that
M dx + N dy=0 will be exact if
[tex]\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}[/tex]
So
[tex]\frac{\partial M}{\partial y}=1[/tex]
[tex]\frac{\partial N}{\partial x}=1[/tex]
it means that this is a exact equation.
[tex]\int d\left(e^x+xy+3y+(y-1)e^y\right)=0[/tex]
Noe by integrating above equation
[tex]e^x+xy+3y+(y-1)e^y=C[/tex]
Given that
x= 0 then y= 1
[tex]e^0+0+3+(1-1)e^1=C[/tex]
C=4
So the our final equation will be
[tex]e^x+xy+3y+(y-1)e^y=4[/tex]
