Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.

Let F(x,y) be the curve 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 - 5)

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 = 4 – x + Ce^x

⇒ x + y – 4 = Ce^x

Now, it is given that curve passes through (0,2).

Thus, equation (2) becomes:

0 + 2 – 4 = C e⁰

⇒ - 2 = C

⇒ C = -2

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

x + y – 4 =-2e^x

⇒ y = 4 – x – 2e^x

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

y = 4 – x – 2e^x