Solve the initial value problem, dy = e^{x + 2y}dx , y (0) = 0.

Question

Question

Answer 1

Concept:

ln e^y = y .

Calculation:

We have ,

dy = ex + 2y dx

⇒ dy = e^x . e^2y dx

⇒ e^-2y dy = e^x dx

On Integrating both sides, We get

e^-2y = e^x + C .... (i)

It is given that y (0) = 0 , i.e. at x = 0 , y = 0 , putting this in (i) ,

= 1 + C

⇒ C = .

Putting C = in (i) , we get

e^-2y = e^x

⇒ e^-2y = - 2e^x + 3

⇒ e^2y =

⇒ 2y = ln

⇒ y = .

The correct option is 3.

Download Solution PDF