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`.

Explanation

The slope of the tangent to the curve `y=f(x)` at
`P(x,y)` is `(dy)/(dx)` ltbr According to the given information,
`(dy)/(dx)+5=x+y`
or `(dy)/(dx)-y=x-5`
This is a linear differential equation of the form
`(dy)/(dx) + Py=Q`, where `P=-1`and `Q=x-5`
Now, `I.F=e^(intPdx)= e^(int(-1)dx)=e^(-x)`
The general equation of the curve is given by the relation.
`y(I.F) = int(Q xx I.F.)dx+C`
or `y.e^(-x)=int(x-5)e^(-x)dx+C`.................(1)
Now, `int(x-5)e^(-x)dx`
`=(x-5)int^(-x)dx-int[d/(dx)(x-5).inte^(-x)dx]dx`
`=(x-5)(-e^(-x))-int(-e^(-x))dx`
`=(4-x)e^(-x)`
Therefore, equation (1) becomes
`ye^(-x)=(4-x)e^(-x)+C`
or `y=4-x+Ce^(x)`
`x+y-4=Ce^(x)`.................(2)
The curve passes through point (0,2).
Therefore, equation (2) becomes
`0+2-4=Ce^(0)`
or `C=-2`
Substituting `C=-2` in equation (2), we get
`x+y-4=-2e^(x)`
or `y=4-x-2e^(x)`