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I have an equation system

$$y'(t) = M(t)y(t)+h(t)$$

where $[M(t)]_{2\times2}$ square matrix and $[h(t)]_{2 \times1}$ is the nonhomogeneous part of the system. I can solve numerically homogeneous systems as $y'(t)=M(t)y(t)$ with my algorithm which is in my topic(Is it true for solving differential equations by getting constant coefficient matrix with magnus expansion) but for nonhomogeneous one I am not sure how can i do it.

In this paper (http://personales.upv.es/serblaza/2012APNUMdoi.pdf) equation(17-18), I found some useful informations about numerical solutions of nonhomogeneous systems but, i'm still suspicious solving nonhomogeneous systems with my algorithm.

Is there any way converting nonhomogeneous systems to homogeneous systems for solving numerically as above type equations ?

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3 Answers 3

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You can transform the system into a homogeneous one by adding an additional constant component $y_{n+1}\equiv 1$ so that $$ \pmatrix{\dot y_{1:n}(t)\\\dot y_{n+1}(t)} = \pmatrix{M(t)&h(t)\\0&0} \pmatrix{y_{1:n}(t)\\y_{n+1}(t)} $$ with $y_{n+1}(t_0)=1$.

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Find an annihilator, I.e., an operator which reduces the RHS to zero. For example, if your ODE has a polynomial of degree $d$ as forcing function, you differentiate $d$ times, a term of the form $\mathrm{e}^{\alpha t}$ makes you apply $\mathrm{D} -\alpha$, I.e., differentiate the equation and subtract it multiplied by $\alpha$.

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  • $\begingroup$ thanks for your solution but i can not understand explicitly your method. Can you give me an example how to do?. $\endgroup$
    – drxy
    Apr 6, 2014 at 12:36
  • $\begingroup$ I don't want to solve analytically my system so how can i convert nonhomogeneous system to homogeneous ? Can anyone help ? $\endgroup$
    – drxy
    Apr 9, 2014 at 14:10
  • $\begingroup$ What my method does is to get rid of the forcing function. What you do with the resulting function is up to you. $\endgroup$
    – vonbrand
    Apr 9, 2014 at 16:44
  • $\begingroup$ Thank you for your method can you give me an example how can i convert my nonhomogeneous problem to homogeneous to solve numerically ? $\endgroup$
    – drxy
    Apr 12, 2014 at 20:52
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To convert the non-homogeneous differential equation to a homogeneous differential equation, simply remove the "non-homogeneous part"! That is, to convert the non-homogeneous differential equation y'(t)= M(t)y(t)+ h(t) just write it as y'(t)= M(t)y(t).

After finding the general solution to that equation, you can get the general solution to the original equation by adding any single solution to the entire equation.

To give a simple one-dimensional example, suppose the problem is to find the general solution to y'(t)= 2y(t)+ 4t. The associated homogeneous equation is y'(t)= 2y(t) which we can write as dy/y= dt/2 and, integrating, ln(y)= ln(t)/2+ C which is equivalent to $y(t)= C't^{1/2}$. To find a single solution to the entire equation since the 'non-homogeneous' part is linear, we try a linear solution, y(t)= at+ b, y'(t)= a, so the equation becomes $a= 2(at+b)+ 4t$. That is the same as $-2at+ (a+ 2b)= 4t$ so we have -2a= 4, a+ 2b= 0. From -2a= 4, a= -2. Then a+ 2b= -2+ 2b= 0 gives b= 1.

The general solution to the original equation, y'(t)= 2y(t)+ 4t, is $y(t)= C't^{1/2}- 2t+ 1$.

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  • $\begingroup$ I think the user wants to solve the ode system with numerical algorithms. $\endgroup$
    – MrYouMath
    Mar 11, 2018 at 18:49

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