Question

The family passing through (0, 0) and satisfying the differential equation (y₂/y₁) = 1[where y_n = (dⁿy/(dxⁿ)] is: 

(A) y = k 

(B) y = ke^x 

(C) y = k(e^x – 1) 

(D) y = k(e^x + 1)

Answer 1

The correct option (C) y = k(e^x – 1)   

Explanation:

(y₂/y₁) = 1 

∴ (d/dx) (log y₁) = c 

∴ (logy₁) = x + c 

∴ y₁ = e^x+c 

i.e. (dy/dx) e^x+c = e^x ∙ e^c 

Integrating 

∴ y = e^x ∙ e^c + k 

when x = 0, y = 0 ⇒ k = – e^c 

y = e^xe^c – e^c = e^c(e^x – 1) 

∴ y = k(e^x – 1) ---- put e^c = k