Q.Using the differential equation of linear S.H.M., obtain the expression for

[!TLDR] From $\frac{d^2x}{dt^2}+\omega^2x=0$: integrating once (with $v=0$ at $x=A$) gives $v=\pm\omega\sqrt{A^2-x^2}$; integrating again gives $x=A\sin(\omega t+\phi)$, from which $a=-\omega^2x$. [!ANSWER] $v=\omega\sqrt{A^2-x^2}$, $a=-\omega^2x$.

1. Velocity: starting from the S.H.M. differential equation $\frac{d^2x}{dt^2}+\omega^2x=0$, write $\frac{d^2x}{dt^2}=v\frac{dv}{dx}$, so $v\,dv=-\omega^2x\,dx$. Integrating both sides, $\frac{v^2}{2}=-\frac{\omega^2x^2}{2}+C$. Using the boundary condition that $v=0$ at the extreme position $x=A$ to fix $C=\frac{\omega^2A^2}{2}$, we get $\frac{v^2}{2}=\frac{\omega^2}{2}(A^2-x^2)$, i.e. $v=\pm\omega\sqrt{A^2-x^2}$.
2. Acceleration: integrating the velocity-displacement relation a second time (separating variables and integrating $dx/\sqrt{A^2-x^2}$) gives $x=A\sin(\omega t+\phi)$. Differentiating twice, $\frac{d^2x}{dt^2}=-A\omega^2\sin(\omega t+\phi)=-\omega^2x$, so $a=-\omega^2x$ -- directly from the original differential equation itself, since it can be rearranged as $\frac{d^2x}{dt^2}=-\omega^2x$. [!ANSWER] $v=\pm\omega\sqrt{A^2-x^2}$ and $a=-\omega^2x$.

Integrate the S.H.M. differential equation once (with the v=0-at-extreme boundary condition) for velocity, and read acceleration directly off the rearranged differential equation (or the second derivative of x(t)).

Forgetting the boundary condition needed to fix the constant of integration C when deriving the velocity expression, or dropping the negative sign in $a=-\omega^2x$.