Let $y(t)$ be a continuous semimartingale and define $P(t)$ as follows. $P(t)=\exp(-y(t)(T-t))$ (t $\in[0,T]$)
Is following calculation using Ito's lemma correct? And how should I write its differential form?
$P(t,y(t))=P(0,y(0))+\int^t_0y(s)\exp(-y(s)(T-t))ds+\int^t_0{-(T-t)\exp(-y(s)(T-t))}dy(s)+1/2\int^t_0(T-t)^2\exp(-y(s)(T-t))d<y>(s)$
Let $P(Y_t,t)=\exp(-Y_t(T-t))$, then $$\frac{\partial P(y,t)}{\partial t}=yP(y,t), \ \ \ \ \ \frac{\partial P(y,t)}{\partial y}=-(T-t)P(y,t), \ \ \ \ \ \frac{\partial^2 P(y,t)}{\partial y^2}=(T-t)^2P(y,t)$$ Therefore, by Ito: $$dP=\frac{\partial P(Y_t,t)}{\partial t}dt+\frac{\partial P(Y_t,t)}{\partial y}dY_t+\frac{1}{2}\frac{\partial^2 P(Y_t,t)}{\partial y^2}d[Y]_t$$ and making the integral form more explicit $$P(Y_t,t)-P(Y_0,0)=\int_0^tY_sP(Y_s,s)ds-\int_0^t(T-s)P(Y_s,s)dY_s+ \\ +\frac{1}{2}\int_0^t(T-s)^2P(Y_s,s)d[Y]_s$$
Now let $P(Y_t,t)=\exp(-Y_t)$, then $$\frac{\partial P(y,t)}{\partial t}=0, \ \ \ \ \ \frac{\partial P(y,t)}{\partial y}=-P(y,t), \ \ \ \ \ \frac{\partial^2 P(y,t)}{\partial y^2}=P(y,t)$$ So the differential and the integral are respectively $$dP=\frac{\partial P(Y_t,t)}{\partial y}dY_t+\frac{1}{2}\frac{\partial^2 P(Y_t,t)}{\partial y^2}d[Y]_t=-P(Y_t,t)dY_t+\frac{1}{2}P(Y_t,t)d[Y]_t$$ $$P(Y_t,t)-P(Y_0,0)=-\int_0^tP(Y_s,s)dY_s +\frac{1}{2}\int_0^tP(Y_s,s)d[Y]_s$$