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

Let `A(x,y)` be any point on the curve `y=f(x)`.
Then the slope of the tangent to the curve at the point A is `(dy)/(dx)`
According to the given condition,
`x+y=(dy)/(dx)+5`
`:.(dy)/(dx)-y=x-5` ... (1)
This is the linear differential equation of the form
`(dy)/(dx)+Py = Q ` where `P=-1 and Q = x-5`.
`:.` I.F. `=e^(int P*dx) = e^(int-1*dx)=e^(-x)`
Multiplying both sides of equation (1) by I.F. `=e^(-x)`, we get
`e^(-x)((dy)/(dx)-y)=(x-5)*e^(-x)`
`:.e^(-x)*(dy)/(dx)-y* e^(-x)=(x-5)*e^(-x)`
`:.(d)/(dx)(e^(-x)*y)=(x-5)*e^(-x)`
Integrating both sides w.r.t. 'x',
`e^(-x)*y=int(x-5)e^(-x)*dx+c`
`:.e^(-x)*y=(x-5)int e^(-x)*dx-int [(d)/(dx)(x-5)int e^(-x)*dx]*dx + c`
`:.e^(-x)*y=(x-5)(e^(-x))/(-1)-int 1.(e^(-x))/(-1).dx+c`
`:.e^(-x)*y=(x-5)(e^(-x))/(-1)+c`
`:.y=-(x-5)-1+c*e^(x)`
`:.y=-x+5-1+c*e^(x)`
`:.y=4-x+c*e^(x)` ...(2)
This is a general equation of the curve.
`:'` The required curve is passing through the point (0,2).
On putting `x=0` and `y=2` in equation (2), we get
`2=4-0+c*e^(0)`
`:.c=-2`
`:.` Equation (2) becomes,
`y=4-x-2*e^(x)`.
This is the required equation of the curve.