Since, the slope of tangent to a curve at any point (x,y) is x+y. ∴
dy
dx
=x+y⇒
dy
dx
−y=x This is a linear differentiation equation So, P=−1 and Q=x Now, IF=e∫Pdx=e∫−dx=e−x ⇒e−x⋅
dy
dx
−e−xy=xe−x ⇒
dy
dx
(y⋅e−x)=xe−x Integrating both sides, we get y⋅e−x=∫xe−xdx =−xe−x−∫−e−xdx =−xe−x+(−e−x)+c where C= constant of integration ⇒ye−x=−xe−x−e−x+c ⇒y=−x−1+cex, where c is arbitrary constant.
