Solving a DAE (differential algebraic equation)

Question

I'm interested in differential algebraic equations. [https://it.mathworks.com/help/matlab/math/solve-differential-algebraic-equations-daes.html]

I have this one:

$$y_1'(t)=-y_2(t)$$

$$0=y_1(t)-(\frac{y_2^3(t)}{3}-y_2(t))$$

I've called $y_1=y(t)$, and $y(2)=y'(t)$, and then I differentiated both the equations, and got

$$y_1'=y_2^2 y_2' - y_2'$$

$$y_2'=-y_2''-y_2y_2''-2y_2y_2'^2$$

Now, if $y_3=y_2'$, then $y_3'=y_2''$ and the system becomes:

$$y_1'=y_2^2 y_3 - y_3$$

$$y_2'=y_3$$

$$y_3'=\frac{-2y_2y_3 - y_3}{1+y_2}$$

Is my resolution right? And most of all, is this the correct way to proceed?

Answer 1

You only need to differentiate the second equation once and then isolate $y_2'$ from both equations to get $$ y_2'=\frac{y_2}{1-y_2^2}. $$ This is now an isolated scalar ODE that can be solved by separation.

Answer 2

You first equation says $y_1(t) = -y_2(t)$. This implies for the second equation: $$ 0 = y_1(t) -(y_2(t)^3/3 - y_2(t)) = y_1(t)+y_2(t) -y_2(t)^3/3 = y_2(t)^3/3. $$ From this, it follows that $y_2(t) = 0$ and therefore also $y_1(t) = 0$.