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I have the following differential equation, I am not sure how to integrate this.

$$\left((\ddot y - \ddot M y)\delta_{t,\pi/2} + (\dot y - \dot MM^{-1}y)(1-\delta_{t,\pi/2}) \right) = 0$$

Now when $t\ne\pi/2$, I know that the equation comes to $$\begin{align} &(\dot y - \dot MM^{-1}y) dt = 0\\ \end{align}$$ which has a solution $y = \exp(\int\dot MM^{-1} dt)y_0$ for $y_0 = y(0)$.

However, I don't know how to evaluate when $t=\pi/2$.

Any help is appreciated.

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  • $\begingroup$ Changes on a set of measure zero, here in one point, do not change the integral in any equivalent integral equation, are you sure you did not misunderstand something here? $\endgroup$ Commented Apr 11, 2020 at 14:42

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The integrating factor is $M^{-1}$, $$\frac{d}{dt}(M^{-1}y)=M^{-1}\dot y-M^{-1}\dot MM^{-1}y=0\implies y(t)=M(t)C.$$

What you wrote makes sense for scalar functions, with the exponential of matrices however you have to be careful, as the scalar treatment only generalizes if everything commutes.

If one understands the first term as a very unusual way to write an initial condition, one concludes from $\ddot M(\frac\pi2)C=\ddot y(\frac\pi2)=\ddot M(\frac\pi2)y(\frac\pi2)$ that $C=y(\frac\pi2)$ which in turn necessitates $$ (M(\tfrac\pi2)-I)y(\tfrac\pi2)=0 $$

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  • $\begingroup$ Yes, $M$ is a matrix here, and I mean the exponential of matrices. $\endgroup$
    – exp ikx
    Commented Apr 11, 2020 at 14:48

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