Transform second order IVP to first order

Question

IVP is given by: $$y''(t)= \dfrac{1}{1+t} \\ y(0)= y'(0)=0 $$

Need to transform this IVP into $$x'(t)=Ax(t)+b(t) , x(0)=0 \tag 1 $$

I am having some issues with this specific question. I am able to transform other higher ODE's into first order systems. If anyone could explain how I can make a start in getting it into the form (1) - that would be great!

Answer 1

You could take $x(t) = y'(t)$, so the equation becomes $x' = 1/(1+t)$. This is $Ax + b(t)$ where $A = 0$. Then to get $y$ you integrate: $y(t) = \int_0^t x(s)\; ds$.

Answer 2

$$ \pmatrix {y\\ y'}'=\pmatrix { 0 & 1\\ 0 & 0 } \pmatrix {y \\ y'}+ \pmatrix {0 \\ \dfrac 1 {t+1}}$$ Substitute $u=y$ and $v=y'$:

$$ \pmatrix {u \\ v}'=\pmatrix { 0 & 1\\ 0 & 0 } \pmatrix {u \\ v}+ \pmatrix {0 \\ \dfrac 1 {t+1}}$$ $$X'=AX+B$$ With initial condition: $$X(0)=\pmatrix {u(0) \\ v(0)}=\pmatrix {y(0) \\ y'(0)}=\pmatrix {0 \\ 0}=0$$