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Question
Find the equation of the 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.
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Solution
We know that the slope of the tangent to the curve is `dy/dx`
We are given that
`dy/dx = x + y`
⇒ `dy/dx - y = x` ....(1)
Which is a linear equation of the type
`dy/dx + Py = Q`
Here P = -1 and Q = x
∴ `I.F. = e^(intP dx) = e^(int -dx) = e^-x`
∴ The solution is `y. (I.F.) = int Q. (I.F.) dx + C`
`y.e^-x = int x.e^-x dx + C`
`= x * (e^-x)/-1 - int (1) (e^-x)/-1 dx + C` ...[Integrating by parts]
⇒ `ye^-x = -xe^-x + int e^-x dx + C`
`= -xe^-x + (e^-x)/-1 + C`
⇒ y = -x - 1 + Cex ....(2)
Since the curve passes through the origin (0, 0)
∴ 0 = - 0 - 1 + Ce0
⇒ C = 1
Putting in (2), we get y = -x - 1 + ex
⇒ x + y + 1 = ex
Which is the required equation of the curve.
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