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.

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 + 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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