Question

Solve the following initial-value problem: (ex+y)dx+(3+x+yey)dy=0, y(0)=1 0=

Answer 1

Answer:

$e^x+xy+3y+(y-1)e^y=4$

Step-by-step explanation:

Given that

$(e^x+y)dx+(3+x+ye^y)dy=0$

Here

$M=e^x+y$

$N=3+x+ye^y$

We know that

M dx + N dy=0 will be exact if

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

So

$\frac{\partial M}{\partial y}=1$

$\frac{\partial N}{\partial x}=1$

it means that this is a exact equation.

$\int d\left(e^x+xy+3y+(y-1)e^y\right)=0$

Noe by integrating above equation

$e^x+xy+3y+(y-1)e^y=C$

Given that

x= 0 then y= 1

$e^0+0+3+(1-1)e^1=C$

C=4

So the our final equation will be

$e^x+xy+3y+(y-1)e^y=4$