Solution to Linear Time-invariant Matrix ODEdPdt=AP+B

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potomakavkl · Accepted Answer

Step 1Guessing the solution is a time-honored way of solving differential equations.Prove that your ansatz is correct by calculating its time derivativeAP=AeAtP0+eAtB-B⇒AP+B=AeAtP0+eAtBAP˙=A2eAtP0+AeAtB=A(AP+B)&amp;nbsp;(Eq#1)P˙=AP+B&amp;nbsp;(QED)If A is singular, try substituting the pseudoinverse&amp;nbsp;A+P=eAtP0+A+(eAt−I)B&amp;nbsp;(ansatz)AP=AeAtP0+(eAt−I)B&amp;nbsp;(recall&amp;nbsp;:AA+A=A)AP˙=A2eAtP0+AeAtB&amp;nbsp;(time&amp;nbsp;derivative)=A(AP+B)&amp;nbsp;(simplify)So this assumption yielded a solution of (Eq 1)AP˙=A(AP+B)if not the original ODE.

ophelialee4xn · Accepted Answer

Step 1Assuming A is constant (i.e. independent of t), you can use&amp;nbsp;exp⁡(−At)&amp;nbsp;as an integrating factor; left-multiplying by it givesddt(e−AtP(t))=e−AtB(t), so thate−AtP(t)−P(0)=∫0te−AsB(s),ds,and henceP(t)=eAtP(0)+∫0teA(t−s)B(s),ds,which holds in general, and reduces to your formula if A is invertible and B is constant.

Solution to Linear Time-invariant Matrix ODE \frac{d\mathbf{P}}{dt} = \mathbf{A}\mathbf{P} + \mathbf{B}

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