Question

Solve the following initial-value problem:

(ex+y)dx+(3+x+yey)dy=0, y(0)=1

0=

Answer 1

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

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

`(\partial M)/(\partial y)=(\partial N)/(\partial x)`

So

`(\partial M)/(\partial y)=1`

`(\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`