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.

Let F(x,y) be the curve passing through origin and let (x,y) be a point on the curve.

We know the slope of the tangent to the curve at (x,y) is .

According to the given conditions, we get,

This is equation in the form of (where, p = -1 and Q = x )

Now, I.F. =

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) =

--------(1)

Now,

Thus, from equation (1), we get,

⇒ y = -(x+1) + Ce^{x}

⇒ x + y + 1 = Ce^{x} -----(2)

Now, it is given that curve passes through origin.

Thus, equation (2) becomes:

1 = C

⇒ C = 1

Substituting C = 1 in equation (2), we get,

x + y – 1 = e^{x}

Therefore, the required general solution of the given differential equation is

x + y -1 = e^{x}

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