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

Question

Question

Answer 1

Concept:

\(\rm \int e^{x}dx = e^{x}\)

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

\(\rm -\frac{1}{2} \) e^-2y = e^x + C .... (i)

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

\(\rm -\frac{1}{2} \) = 1 + C

⇒ C = \(\rm -\frac{3}{2} \) .

Putting C = \(\rm -\frac{3}{2} \) in (i) , we get

\(\rm -\frac{1}{2} \) e^-2y = e^x \(\rm -\frac{3}{2} \)

⇒ e^-2y = - 2e^x + 3

⇒ e^2y = \(\rm \frac{1}{3- 2\ e^{x}} \)

⇒ 2y = ln \(\rm \left ( \frac{1}{3- 2\ e^{x}} \right )\)

⇒ y = \(\rm \frac{1}{2}\ln\left ( \frac{1}{3- 2\ e^{x}} \right )\) .

The correct option is 3.

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