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I am new to stochastic differential equations. I would like to solve something like this:

$dx_t = (a) x_t dt + (b) dZ_t$

The solution is:

$x_t = e^{at} x_0 + b e^{at} \int_0^t e^{-au} dZ_u$

I would like to understand the steps to find the solution. All examples I have managed to find are of the type:

$dx_t = (a) x_t dt + (b) x_t dZ_t$

Could someone help me understand the steps or direct me towards a useful resource?

NC520
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  • You have a sign error somewhere in front of $a$. // The solution is here via integrating factor, consider $y_t=e^{at}x_t$ and apply the Ito formula (which is trivial, does not have extra terms over the product formula, as $y$ is linear in $x$). – Lutz Lehmann May 30 '22 at 10:31
  • Thank you. I will solve it and add the solution below. – NC520 May 30 '22 at 10:33

1 Answers1

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To solve the SDE, consider $W_t = e^{- a t} x_t$. By Ito's lemma: \begin{align} dW_t &= - a e^{- a t} x_t dt + e^{- a t} dx_t \\ dW_t &= - a e^{- a t} x_t dt + e^{- a t} (a x_t dt + b dZ_t) \nonumber \\ dW_t &= (- a e^{- a t} x_t + a e^{- a t} x_t) dt + b e^{- a t} dZ_t \nonumber \\ dW_t &= b e^{- a t} dZ_t \nonumber \end{align} Integrating both sides: \begin{align} \int_0^t d(e^{- a u} x_u) &= b \int_0^t e^{- a u} dZ_u \\ e^{- a t} x_t &= x_0 + b \int_0^t e^{- a u} dZ_u \nonumber \\ x_t &= e^{a t} x_0 + e^{a t} b \int_0^t e^{- a u} dZ_u \nonumber \end{align}

NC520
  • 229