Q22 of 462 Page 22

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.

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 – 3xdx + 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)


∴


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