Stochastic Integral form

Question

I have a silly question.

I know that the stochastic differential equation in derivative form is : $$d X_{t} = aX_{t}dt +b X_{t}dB_{t}$$ can be written is the integral form as $$X_{t}-X_{0} = a \int_{0}^{t}X_{s}ds +b\int_{0}^{t}X_{s}dB_{s}$$. Now I have in derivative form the : $$d X_{t} = -(a+b^2X_{t})(1-X_{t}^2)dt +b(1- X_{t}^2)dB_{t}$$. How can I write it in the integral form ? Any help?

Maybe :

$$-(a+b^2X_{t})(1-X_{t}^2) = -a +aX_{t}^2 - b^2X_{t}+b^2X_{t}^3$$

So .. ? $$X_{t}-X_{0} = ? +b\int_{0}^{t}1 - X_{s}^2dB_{s} $$

Answer 1

The form $dX=a(X)dt+b(X)dW$ stands for $$ X_t-X_0=\int_0^t a(X_s)ds+\int_0^tb(X_s)dW_s $$ as well as any $$ \int_0^tc(X_s)dX_s=\int_0^t c(X_s)a(X_s)ds+\int_0^tc(X_s)b(X_s)dW_s $$ for any sufficiently regular function $c$.

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So in your specific case you just get $$ X_T-X_0 = -\int_0^T(a+b^2X_{t})(1-X_{t}^2)dt +\int_0^T b(1- X_{t}^2)dB_{t} $$ but also, as example, $$ \int_0^T\frac{dX_t}{1-X_t^2} = -\int_0^T(a+b^2X_{t}) dt +\int_0^T b dB_{t} $$