Q. 305.0( 1 Vote )

# 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.

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 + cex……(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 + ex

x + y + 1 = ex

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