"t" is the **independent** variable, not a third dependent variable. Your equation would be, as I said, \(\displaystyle u'= du/dt= v\), and \(\displaystyle v'= dv/dt= u^3v- 3u+ t^2\) so that \(\displaystyle du= vdt\) and \(\displaystyle dv= (u^3v- 3u+ t^2)dt\). To do an Euler numerical integration, one would start with u(0)= 1, v(0)= 2. Then taking, say, dt= 0.1, \(\displaystyle du= 2(0.1)= 0.2\) and \(\displaystyle dv= (1(2)- 3(1)- 0)(0.1)= -0.1\) so u(0.1)= u(0)+ du= 1+ 0.2= 1.2 and v(0.1)= v(0)+ dv= 2- 0.1= 1.99. Then \(\displaystyle du= 1.99(0.1)= 0.199\) and \(\displaystyle dv= ((1.2)^3(1.99)- 3(1.2)+ 0.1^2)(0.1)= -0.015128\) so that \(\displaystyle u(0.2)= 1.2+ 0.199= 1.399\) and \(\displaystyle v(0.2)= 1.99- 0.015129= 1.974871. etc. (You will have to decide what accuracy makes sense.\)