Answer
$\theta=-t^2+\dfrac{t}{2}+\sqrt 2$
Work Step by Step
The anti-derivative for $\dfrac{d^3 \theta}{dt^3}=0$ is:
$\dfrac{d^2 \theta}{dt^2}=c$; $\dfrac{d \theta}{dt}=ct+c'$, $\theta=\dfrac{ct^2}{2}+c'x+c''$
Now, apply Initial conditions $\theta''(0)=-2 ; \theta'(0)=\dfrac{1}{2},\theta(0)=\sqrt 2$, we have
This implies $c=-2$ and $c'=\dfrac{1}{2}$ and $c''=\sqrt 2$
Hence, $\theta=-t^2+\dfrac{t}{2}+\sqrt 2$
