Refer to figure 1.Sticking to the convention (counterclockwise is positive) the differential equation for a physical pendulum is I*D^2(θ) =-mgdsin(θ) since the torque of the gravity force is clockwise, so it has a negative sign in front which with the small angle approx becomes D^2(θ) + (mgd/I)θ = 0 where I is the moment of inertia and d the distante to the center of mass D^2 is the second derivative with rispect to time

No problem there, now refering to figure 2 I want to find the equation using clockwise as positive I*D^2(θ) = mgdsin(θ) , now the torque has a positive sign, but something must be wrong because this lead to the equation: D^2(θ) - (mgd/I)θ = 0, whose solution won't be the same as in the previous case ,I guess I have to put a "-" in front of D^2(θ) , but I don't know how to justify it

Refer to figure $1$.

Sticking to the convention (counterclockwise is positive) the differential equation for a physical pendulum is $$I\cdot D^2(θ) =-mgd\sin(θ)$$ since the torque of the gravity force is clockwise, so it has a negative sign in front which with the small angle approx becomes $$D^2(θ) + (mgd/I)θ = 0,$$ where $I$ is the moment of inertia, $d$ is the distance to the center of mass, and $D^2$ is the second derivative with respect to time.

No problem there, now referring to figure $2$.

I want to find the equation using clockwise as positive $$I*D^2(θ) = mgd\sin(θ),$$ now the torque has the positive sign, but something must be wrong, because this lead to the equation: $$D^2(θ) - (mgd/I)θ = 0,$$ whose solution won't be the same as in the previous case.

I guess I have to put a “-” in front of $D^2(θ)$, but I don't know how to justify it.

Refer to figure 1.Sticking to the convention (counterclockwise is positive) the differential equation for a physical pendulum is I*D^2(θ) =-mgdsin(θ) since the torque of the gravity force is clockwise, so it has a negative sign in front which with the small angle approx becomes D^2(θ) + (mgd/I)θ = 0 where I is the moment of inertia and d the distante to the center of mass

No problem there, now refering to figure 2 I want to find the equation using clockwise as positive I*D^2(θ) = mgdsin(θ) , now the torque has a positive sign, but something must be wrong because this lead to the equation: D^2(θ) - (mgd/I)θ = 0, whose solution won't be the same as in the previous case ,I guess I have to put a "-" in front of D^2(θ) , but I don't know how to justify it

Refer to figure 1.Sticking to the convention (counterclockwise is positive) the differential equation for a physical pendulum is I*D^2(θ) =-mgdsin(θ) since the torque of the gravity force is clockwise, so it has a negative sign in front which with the small angle approx becomes D^2(θ) + (mgd/I)θ = 0 where I is the moment of inertia and d the distante to the center of mass D^2 is the second derivative with rispect to time

No problem there, now refering to figure 2 I want to find the equation using clockwise as positive I*D^2(θ) = mgdsin(θ) , now the torque has a positive sign, but something must be wrong because this lead to the equation: D^2(θ) - (mgd/I)θ = 0, whose solution won't be the same as in the previous case ,I guess I have to put a "-" in front of D^2(θ) , but I don't know how to justify it