A curve passes through the origin and the slope of the tangent to the curve at any point (

Formula :

i)

ii)

iii)

iv)

v) General solution :

For the differential equation in the form of

General solution is given by,

Where, integrating factor,

Answer :

The slope of the tangent to the curve

The slope of the tangent to the curve is equal to the sum of the coordinates of the point.

Therefore differential equation is

………eq(1)

Equation (1) is of the form

Where, and

Therefore, integrating factor is

………

General solution is

………eq(2)

Let,

Let, u=x and v= e^-x

………

………

………

Substituting I in eq(2),

Dividing above equation by e^-x,

Therefore, general solution is

The curve passes through origin , therefore the above equation satisfies for x=0 and y=0,

Substituting c in general solution,

Therefore, equation of the curve is