Problem:
Solve the stochastic differential equation
$dX_t = e^{2X_t}dB_t + (e^{4X_t} − e^{2X_t})dt, X_0 = 0$
Idea:
It seems I should apply Ito's formula.
I tried using $f(t,x) = e^{-2x}$. Then,
$\frac{\partial f(t,x)}{\partial t} = 0$
$\frac{\partial f(t,x)}{\partial x} = -2e^{-2x}$
$\frac{\partial^2 f(t,x)}{\partial x^2} = 4e^{-2x}$
Then applying Ito's Formula,
$de^{-2X_t} = -2e^{-2X_t}(e^{4X_t}-e^{2X_t})dt + -2e^{-2X_t}(e^{2X_t})dB_t+\frac{1}{2}\cdot4e^{-2X_t}(e^{2X_t})^2dt$
Then I'd like to integrate with repsect to time.
However, this doesn't seem to work out correctly. Is there a better choice for $f(t,x)$? Or is it not Ito's Formula I should be using?
Best Answer
Your approach is fine. There are several terms which cancel, and so you end up with an easy equation. Using that $(e^{2X_t})^2 =e^{4X_t}$, it follows that
\begin{align*} d(e^{-2X_t}) &= (-2 e^{2X_t} + 2) \, dt - 2 dB_t + 2 e^{2X_t} \, dt \\ &= 2 \, dt - 2 \, dB_t. \end{align*}
Hence, $$e^{-2X_t} - e^{-2X_0} = 2 t - 2 B_t.$$ Rearranging the terms, gives you $X_t$.