Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point.

Explanation

Let the moving point be `(x,y)`
Given that, `(dy)/(dx)=x+yimplies (dy)/(dx)-y=x`
Here, `P=-1`, `Q=x`
`:. I.F.=e^(int-1dx)=e^(-x)`
and general solution : `y(e^(-x))=intxe^(-x)dx+c`
`implies ye^(-x)=-xe^(-x)-int1(-e^(-x))dx+c`
`=-xe^(-x)-e^(-x)+c`
`implies y=-x-1+ce^(x)`
This curve passes through `(0,0)`
`:. 0=0-1+cimpliesc=1`
`:.`Curve is `y=-x-1+e^(x)`
`implies x+y+1=e^(x)`