$\mathbf{Attempt}:$
Part 1: A second order ODE can be written as a combination two $1$st order ODEs.
Justification: A $2$nd order ODE can be written as $\frac{d^2y}{dx^2}=F(x,y,\frac{dy}{dx})$. We put $u=y$ and $v=y'$.
Now, we have $\frac{dv}{dx}=F(x,u,v) \ \ \ ...(i)$ and $\frac{du}{dx}=v \ \ \ ... (ii)$.
$(i)$ and $(ii)$, when combined gives the solution of the given 2nd order ODE.
Part 2: Combination of two 1st order ODEs gives a 2nd order ODE.
Justification: Let $\frac{dp}{dx}=G(x,p,q)$ and $\frac{dq}{dx}=H(x,p,q)$be two general first order coupled ODEs.
We write $p=I(x,q,\frac{dq}{dx})$ $ \ \ \ $[from $\frac{dq}{dx}=H]$.
Now, $\displaystyle\frac{d^2q}{dx^2}=J(x,p,\frac{dp}{dx},\frac{dq}{dx},q)=J(x,I(x,q,\frac{dq}{dx}),G(x,I(x,q,\frac{dq}{dx}),q),\frac{dq}{dx},q))=K(x,q,\frac{dq}{dx}) \implies \displaystyle\frac{d^2q}{dx^2}=K(x,q,\frac{dq}{dx})$, which is the general form of a 2nd order ODE.
Hereby, our justification is complete.
Is this correct?
Kindly $\mathbf{VERIFY}$.