Question

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.

Answer 1

Let the co-ordinates be x and y, then

`x + y = dy/dx + 5`

⇒ `dy/dx - y = x - 5`                 ....(1)

Which is a linear differential equation of the type `dy/dx + Py = Q`

Here P = -1 and Q = x - 5

∴ `I.F. = e^(int Pdx) = e^(int -1  dx) = e^-x`

∴ the solution is `y. (I.F.) = int Q. (I.F.)  dx + C`

`y.e^-x = int (x - 5) e^-x  dx + C`

`= int xe^-x  dx - 5 int e^-x  dx + C`

`= x (e^-x/-1) - int (1)  (e^-x)/-1  dx - 5  (e^-x)/-1 + C`      ....[Intergrating by parts]

⇒ `ye^-x = -xe^-x + e^-x/-1 + 5e^-x + C`

`= -xe^-x + 4e^-x + C`

⇒ `y = -x + 4 + Ce^x`              ....(2)

Since the curve passes through (0, 2), we get

2 = -0 + 4 + C

⇒ C = -2

Putting C = -2 in (2), we get

y = -x + 4 - 2e^x

⇒ y = 4 - x - 2e^x

Which is the required equation of the curve.