Solve the following initial-value problem:(ex+y)dx+(3+x+yey)dy=0, y(0)=10=
Accepted Solution
A:
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 thatM 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 thatx= 0 then y= 1[tex]e^0+0+3+(1-1)e^1=C[/tex]C=4So the our final equation will be[tex]e^x+xy+3y+(y-1)e^y=4[/tex]