Find the equation of the curve which passes through the origin and has the slope x + 3y – 1 at any point (x, y) on it.

Question

Philoid · Accepted Answer

Given Slope of the equation at any point (x,y) = x + 3y – 1

⇒

We can see that it is a linear differential equation.

Comparing it with

P = – 3, Q = x – 1

I.F = e^∫Pdx

= e ^{– 3dx}

= e ^{– 3x}

Solution of the given equation is given by

y × I.F = ∫Q × I.F dx + c

⇒ y × e ^{– 3x} = ∫ (x – 1) × e ^{– 3x}dx + c

⇒ y × e ^{– 3x} = (x – 1) × e ^{– 3x} – ∫(1)e ^{– 3x} dx + c

⇒ y × e ^{– 3x} = (x – 1) × e ^{– 3x +} (e ^{– 3x} ) ⁺ c

⇒

⇒ ……(1)

As the equation passing through origin(0,0)

0 = 0 + + c

⇒ c =

Putting the value of c in equation (1)

∴