Question

For the differential equation, find the general solution:

e^x tan y dx + (1 – e^x) sec² y dy = 0

Answer 1

We have, e^x tan y dx + (1 - e^x) sec² y dy = 0 or e^x tan y dx = -(1 - e^x) sec² y dy

⇒ `e^x/(1 - e^x) dx = (-sec^2y)/(tany)  dy`            ...(1)

Integrating (1) both sides, we get

`int(e^x)/(1 - e^x)  dx = - int(sec^2 y)/(tan y) dy`

Let `I_1 = int e^x/(1 - e^x) dx`

and `I_2 = int(sec^2y)/(tan y) dy`

Now, `I_1 = int e^x/(1 - e^x)  dx`

Putting 1- e^x = t

⇒ -e^x dx = dt

`I_1 = int (-dt)/t = -log |t| - log C_1`

`= - log (t C_1) = -log ((1 - e^x)C_1)`               ....(2)

Now, `I_2 = int (sec^2 y)/(tan y)  dy`

Putting tan y = t

⇒ sec² y dy = dt

`I_2 = int dt/t = log |t| + log  C_2`

`= log |tany| + log C_2`

= log (C₂ tan y)

Also, I₁ = -I₂

⇒ - log (C₁ (1 - e^x))

= - log (C₂ tan y)

⇒ C₁ (1 - e^x) = C₂ tan y

⇒ tan y = C (1 - e^x)

Which is the required solution