Let (left(B_{t} ight)_{t geqslant 0}) be a (mathrm{BM}^{1}). Use Lemma 21.10 to find the solution of the

Let \(\left(B_{t}\right)_{t \geqslant 0}\) be a \(\mathrm{BM}^{1}\). Use Lemma 21.10 to find the solution of the following SDE:

\[d X_{t}=\left(\sqrt{1+X_{t}^{2}}+\frac{1}{2} X_{t}\right) d t+\sqrt{1+X_{t}^{2}} d B_{t}\]

Data From 21.10 Lemma

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21.10 Lemma. Let (Bt)to be a BM. The SDE dX+ = b(X+) dt +(X+) dB, with coefficients b,: R R can be transformed into a linear SDE dz+= (a + Z+) dt + (y + 8Z+) dBt, a, b, y, 8 = R, if, and only if, d((x'(x)o(x)) dx K'(x) b(x) =0 where (x) = -(x). (21.15) () Set d(x)= dy/(y) and 8 = - [(x'(x)o(x))] / x'(x). The transformation Z = f(xt) is given by =- ed(x), if 60, f(x) = yd(x), if 8=0. Proof. Since the coefficients do not depend on t, it is clear that the transformation Z = f(x+) should not depend on t. Using the general transformation scheme from above we see that the following relations are necessary: b(x)f'(x) + (x)f" (x) = a + Bf(x), (x)f'(x) = y +8f(x), where x = g(z), z = f(x), l.e. g(z) = f(z). Case 1. If 8 +0, we set d(x) = dy/(y); then f(x) = ce6d(x) - Y (21.16a) (21.16b)