The solution of the differential equation, `e^x(x +1)dx + (ye^y -xe^x)dy = 0` with initial condition f(0) = 0, is

Options

A. `2xe^(y)+y(2)=C`
B. `2xe^(x)e^(-y)+y^(2)=C`
C. `ye^(x)+x^(2)=C`
D. `ye^(y)+x^(2)=C`

Answer

B

Explanation

Given differential equation is
`e^(x)dy+(ye^(x)+2x)dx=0`
`implies e^(x)(dy)/(dx)+ye^(x)+2x=0`
`implies (dy)/(dx)+y=-2xe^(-x)`
which is a linear differential equation
`(dy)/(dx)+Py=Q`
On comparing, `P=1` and `Q=-2xe^(-x)`
`:. I.F.=e^(int1dx)=e^(x)`
Therefore, the general solution of the given differential equation is
`y*I.E=int(-2xe^(-x)xxe^(x))dx+C`
`implies ye^(x)=-int2xdx+C`
`implies ye^(x)=-x^(2)+Cimpliesye^(x)+x^(2)=C`