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How can I solve the following SDE:

\begin{cases} dX_{t}=-\sin X_{t}\cdot\cos^{3}X_{t}dt+\cos^{2}X_{t}dB_{t}\\ X_{0}=x_{0} \end{cases}?

This SDE has the form of

$$dX_{t}=\frac{1}{2}\sigma'\left(X_{t}\right)\sigma\left(X_{t}\right)dt+\sigma\left(X_{t}\right)dB_{t},$$ and in this case, as far as I know there is a trick to use the Itô-formula for the $h\left(X_{t}\right)$ function where $h\left(x\right)=\int\frac{1}{\sigma\left(x\right)}dx.$ So in this example $h\left(x\right)=\tan x$. Using the Itô-formula for $h\left(X_{t}\right)$ we get the following: \begin{align*} h\left(X_{T}\right)-h\left(X_{0}\right)= & \tan X_{T}-\tan X_{0}\\ = & \int_{0}^{T}\frac{1}{\cos^{2}X_{t}}\left(-1\right)\sin X_{t}\cdot\cos^{3}X_{t}dt+\int_{0}^{T}\frac{1}{\cos^{2}X_{t}}\cos^{2}X_{t}dB_{t}+\int_{0}^{T}\tan X_{t}\cdot\cos^{4}X_{t}dt=\\ = & \int_{0}^{T}-\sin X_{t}\cos X_{t}+\sin X_{t}\cos^{3}X_{t}dt+B_{T}=\\ = & -\int_{0}^{T}\sin^{3}X_{t}\cos X_{t}dt+B_{T}. \end{align*} Can I get somehow a closed formula for the solution? Did I go wrong somewhere?

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1 Answer 1

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So if $Y=h(X)$, then $$ dY_t=h'(X_t)dX_t+\frac12h''(X_t)d\langle X\rangle_t \\ =\frac12h'(X_t)σ'(X_t)σ(X_t)\,dt+h'σ(X_t)\,dB_t+\frac12h''(X_t)σ(X_t)^2\,dt $$ which suggests to take $h'(X_t)σ'(X_t)+h''(X_t)σ(X_t)=0$ leading to the given form of $h$.


When applying the Ito formula, your last term does not have the required form $\frac12h''(X_t)σ(X_t)^2\,dt$, the first factor should be $-\frac{\sin x}{\cos^3x}$. There is no minus sign in the first term of the same expression.

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  • $\begingroup$ You are right...I just couldn't divide two functions -.- ... so I made a mistake calculating the second derivate of h. $\endgroup$
    – Kapes Mate
    Jul 14, 2021 at 20:17

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