Let P(x,y) be the point on the curve y = f(x) such that tangent at Pcuts the coordinate axes at A and B.

It cuts the axes at A and B so, equation of tangent at P(x,y)

Y – y = (X – x)

Now, putting Y = 0

0 – y = (X – x)

⇒ X = x – y

So, B(x – y,0)

Given, intercept on x – axis = y

⇒ x – y = y

⇒ – y = y – x

⇒

⇒ ……(1)

We can see that it is a linear differential equation.

Comparing it with

P = , Q = – 1

I.F = e^∫Pdy

= e^dy

= e ^{– logy} = y

Solution of the given equation is given by

x × I.F = ∫Q × I.F dy + c

⇒ x × = ∫ – 1 × dy + c

⇒ = – logy + c……(1)

As the equation passing through (1,1)

0 = – 1 + c

⇒ c = 1

Putting the value of c in equation (1)

∴ = – logy + 1

⇒ x = y – ylogy