Question

Find the particular solution of the differential equation (1 + e^2x) dy + (1 + y²) e^x dx = 0, given that y = 1 when x = 0.

Answer 1

The given equation is

(1 + e^2x)dy + (1 + y²) e^x dx = 0

⇒ `dy/(1 + y^2) + e^x/(1 + e^(2x))  dx = 0`

Integrating, `int dy/(1 + y^2) + int e^x/(1 + e^(2x))  dx = C_1`

⇒ `tan^-1 y + I = C_1`                     ....(1)

Where `I = int e^x/(1 + e^(2x))  dx`

Put e^x = t

⇒ e^x  dx = dt

∴ `I = int dt/(1 + t^2) = tan^-1 t + C_2`

= tan^-1 e^x + C₂

From (1), tan^-1 y + tan^-1 e^x + C₂ = C₁

or tan^-1 y + tan^-1 e^x = C                .....(2)

When x = 0, y = 1,

∴ tan^-1 (1) + tan^-1 (e⁰) = C

⇒ tan^-1 (1) + tan^-1 (1) = C

⇒ `pi/4 + pi/4 = C`

⇒ `C = pi/2`

Putting in (2) `tan^-1 y + tan^-1 e^x = pi/2,`

Which is the required solution.