The slope of the tangent at each point of a curve is equal to the sum of the coordinates of the point. Find the curve that passes through the origin.

Question

Philoid · Accepted Answer

Given slope at any point = sum of coordinates = x + y

⇒

We can see that it is a linear differential equation.

Comparing it with

P = – 1, Q = x

I.F = e^∫Pdx

= e ^{– dx}

= e ^{– x}

Solution of the given equation is given by

y × I.F = ∫Q × I.F dx + c

⇒ y × e ^{– x} = ∫ x × e ^{– x} dx + c

⇒ ye ^{– x} = ∫ x × e ^{– x} dx + c (Using integration by parts)

⇒ ye ^{– x} = – x e ^{– x –} e ^{– x} + c

⇒y = – x – 1 + ce^x……(1)

As the equation passing through origin,

0 = 0 – 1 + c× 1

⇒ c = 1

Putting the value of c in equation (1)

∴ y = – x – 1 + e^x

⇒ x + y + 1 = e^x