Stability of non-homogeneous ODE

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I try to examine stability of non-homogeneous ODE system:
begin{cases} Dy_{1} = y_{1}+2y_{2} +frac{3}{x^4} \ Dy_{2}= 3y_{1}+4y_{2}+ frac{3}{x^4} end{cases}



I tried to find solutions of such system and then examine whether solutions are stable, but I can't find them, is there any other method to determine stability of such system?






differential-equations stability-theory







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  • Can you show us a screenshot or picture of what you tried? That would really help us help you better. Also, personally, this looks like a problem for variation of parameters. Did you try that?
    – Noble Mushtak
    2 days ago


















0














I try to examine stability of non-homogeneous ODE system:
begin{cases} Dy_{1} = y_{1}+2y_{2} +frac{3}{x^4} \ Dy_{2}= 3y_{1}+4y_{2}+ frac{3}{x^4} end{cases}



I tried to find solutions of such system and then examine whether solutions are stable, but I can't find them, is there any other method to determine stability of such system?






differential-equations stability-theory







share|cite|improve this question







New contributor




Cga1235 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • Can you show us a screenshot or picture of what you tried? That would really help us help you better. Also, personally, this looks like a problem for variation of parameters. Did you try that?
    – Noble Mushtak
    2 days ago
















0












0








0







I try to examine stability of non-homogeneous ODE system:
begin{cases} Dy_{1} = y_{1}+2y_{2} +frac{3}{x^4} \ Dy_{2}= 3y_{1}+4y_{2}+ frac{3}{x^4} end{cases}



I tried to find solutions of such system and then examine whether solutions are stable, but I can't find them, is there any other method to determine stability of such system?






differential-equations stability-theory







share|cite|improve this question







New contributor




Cga1235 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I try to examine stability of non-homogeneous ODE system:
begin{cases} Dy_{1} = y_{1}+2y_{2} +frac{3}{x^4} \ Dy_{2}= 3y_{1}+4y_{2}+ frac{3}{x^4} end{cases}



I tried to find solutions of such system and then examine whether solutions are stable, but I can't find them, is there any other method to determine stability of such system?






differential-equations stability-theory




differential-equations stability-theory






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  • Can you show us a screenshot or picture of what you tried? That would really help us help you better. Also, personally, this looks like a problem for variation of parameters. Did you try that?
    – Noble Mushtak
    2 days ago




















  • Can you show us a screenshot or picture of what you tried? That would really help us help you better. Also, personally, this looks like a problem for variation of parameters. Did you try that?
    – Noble Mushtak
    2 days ago


















Can you show us a screenshot or picture of what you tried? That would really help us help you better. Also, personally, this looks like a problem for variation of parameters. Did you try that?
– Noble Mushtak
2 days ago






Can you show us a screenshot or picture of what you tried? That would really help us help you better. Also, personally, this looks like a problem for variation of parameters. Did you try that?
– Noble Mushtak
2 days ago












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First, solve the homogeneous equation:



$$Dy_1=y_1+2y_2$$
$$Dy_2=3y_1+4y_2$$



I am assuming you already know how to do this, so I will just write the solution here:



$$left[begin{matrix} y_1 \ y_2end{matrix}right]=left[begin{matrix} frac{1}{6}(-3-sqrt{33})e^{(5-sqrt{33})/2} & frac{1}{6}(-3+sqrt{33})e^{(5+sqrt{33})/2} \ e^{(5-sqrt{33})/2} & e^{(5+sqrt{33})/2}end{matrix}right]left[begin{matrix} c_1 \ c_2end{matrix}right] text{ where } c_1,c_2inmathbb{R}$$



Since there is one negative eigenvalue $frac{5-sqrt{33}}{2}$ and one positive eigenvalue $frac{5+sqrt{33}}{2}$, the origin is a saddle point and the system is unstable. Therefore, at the very least, you now know the stability of the ODE system.



Now, to actually solve the non-homogeneous equation, use variation of parameters by changing $c_1,c_2$ to $u_1(t),u_2(t)$:



$$left[begin{matrix} y_1 \ y_2end{matrix}right]=left[begin{matrix} frac{1}{6}(-3-sqrt{33})e^{(5-sqrt{33})/2} & frac{1}{6}(-3+sqrt{33})e^{(5+sqrt{33})/2} \ e^{(5-sqrt{33})/2} & e^{(5+sqrt{33})/2}end{matrix}right]left[begin{matrix} u_1(t) \ u_2(t)end{matrix}right]$$
$$left[begin{matrix} frac{1}{6}(-3-sqrt{33})e^{(5-sqrt{33})/2} & frac{1}{6}(-3+sqrt{33})e^{(5+sqrt{33})/2} \ e^{(5-sqrt{33})/2} & e^{(5+sqrt{33})/2}end{matrix}right]left[begin{matrix} u_1'(t) \ u_2'(t)end{matrix}right]=left[begin{matrix} frac{3}{x^4} \ frac{3}{x^4}end{matrix}right]$$



Here, you can use the second equation to solve for $u_1'(t),u_2'(t)$, integrate to solve for $u_1(t),u_2(t)$, and then plug into the first equation to solve for $y_1,y_2$. Good luck!






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    1 Answer
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    1 Answer
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    First, solve the homogeneous equation:



    $$Dy_1=y_1+2y_2$$
    $$Dy_2=3y_1+4y_2$$



    I am assuming you already know how to do this, so I will just write the solution here:



    $$left[begin{matrix} y_1 \ y_2end{matrix}right]=left[begin{matrix} frac{1}{6}(-3-sqrt{33})e^{(5-sqrt{33})/2} & frac{1}{6}(-3+sqrt{33})e^{(5+sqrt{33})/2} \ e^{(5-sqrt{33})/2} & e^{(5+sqrt{33})/2}end{matrix}right]left[begin{matrix} c_1 \ c_2end{matrix}right] text{ where } c_1,c_2inmathbb{R}$$



    Since there is one negative eigenvalue $frac{5-sqrt{33}}{2}$ and one positive eigenvalue $frac{5+sqrt{33}}{2}$, the origin is a saddle point and the system is unstable. Therefore, at the very least, you now know the stability of the ODE system.



    Now, to actually solve the non-homogeneous equation, use variation of parameters by changing $c_1,c_2$ to $u_1(t),u_2(t)$:



    $$left[begin{matrix} y_1 \ y_2end{matrix}right]=left[begin{matrix} frac{1}{6}(-3-sqrt{33})e^{(5-sqrt{33})/2} & frac{1}{6}(-3+sqrt{33})e^{(5+sqrt{33})/2} \ e^{(5-sqrt{33})/2} & e^{(5+sqrt{33})/2}end{matrix}right]left[begin{matrix} u_1(t) \ u_2(t)end{matrix}right]$$
    $$left[begin{matrix} frac{1}{6}(-3-sqrt{33})e^{(5-sqrt{33})/2} & frac{1}{6}(-3+sqrt{33})e^{(5+sqrt{33})/2} \ e^{(5-sqrt{33})/2} & e^{(5+sqrt{33})/2}end{matrix}right]left[begin{matrix} u_1'(t) \ u_2'(t)end{matrix}right]=left[begin{matrix} frac{3}{x^4} \ frac{3}{x^4}end{matrix}right]$$



    Here, you can use the second equation to solve for $u_1'(t),u_2'(t)$, integrate to solve for $u_1(t),u_2(t)$, and then plug into the first equation to solve for $y_1,y_2$. Good luck!






    share|cite|improve this answer


























      0














      First, solve the homogeneous equation:



      $$Dy_1=y_1+2y_2$$
      $$Dy_2=3y_1+4y_2$$



      I am assuming you already know how to do this, so I will just write the solution here:



      $$left[begin{matrix} y_1 \ y_2end{matrix}right]=left[begin{matrix} frac{1}{6}(-3-sqrt{33})e^{(5-sqrt{33})/2} & frac{1}{6}(-3+sqrt{33})e^{(5+sqrt{33})/2} \ e^{(5-sqrt{33})/2} & e^{(5+sqrt{33})/2}end{matrix}right]left[begin{matrix} c_1 \ c_2end{matrix}right] text{ where } c_1,c_2inmathbb{R}$$



      Since there is one negative eigenvalue $frac{5-sqrt{33}}{2}$ and one positive eigenvalue $frac{5+sqrt{33}}{2}$, the origin is a saddle point and the system is unstable. Therefore, at the very least, you now know the stability of the ODE system.



      Now, to actually solve the non-homogeneous equation, use variation of parameters by changing $c_1,c_2$ to $u_1(t),u_2(t)$:



      $$left[begin{matrix} y_1 \ y_2end{matrix}right]=left[begin{matrix} frac{1}{6}(-3-sqrt{33})e^{(5-sqrt{33})/2} & frac{1}{6}(-3+sqrt{33})e^{(5+sqrt{33})/2} \ e^{(5-sqrt{33})/2} & e^{(5+sqrt{33})/2}end{matrix}right]left[begin{matrix} u_1(t) \ u_2(t)end{matrix}right]$$
      $$left[begin{matrix} frac{1}{6}(-3-sqrt{33})e^{(5-sqrt{33})/2} & frac{1}{6}(-3+sqrt{33})e^{(5+sqrt{33})/2} \ e^{(5-sqrt{33})/2} & e^{(5+sqrt{33})/2}end{matrix}right]left[begin{matrix} u_1'(t) \ u_2'(t)end{matrix}right]=left[begin{matrix} frac{3}{x^4} \ frac{3}{x^4}end{matrix}right]$$



      Here, you can use the second equation to solve for $u_1'(t),u_2'(t)$, integrate to solve for $u_1(t),u_2(t)$, and then plug into the first equation to solve for $y_1,y_2$. Good luck!






      share|cite|improve this answer
























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        First, solve the homogeneous equation:



        $$Dy_1=y_1+2y_2$$
        $$Dy_2=3y_1+4y_2$$



        I am assuming you already know how to do this, so I will just write the solution here:



        $$left[begin{matrix} y_1 \ y_2end{matrix}right]=left[begin{matrix} frac{1}{6}(-3-sqrt{33})e^{(5-sqrt{33})/2} & frac{1}{6}(-3+sqrt{33})e^{(5+sqrt{33})/2} \ e^{(5-sqrt{33})/2} & e^{(5+sqrt{33})/2}end{matrix}right]left[begin{matrix} c_1 \ c_2end{matrix}right] text{ where } c_1,c_2inmathbb{R}$$



        Since there is one negative eigenvalue $frac{5-sqrt{33}}{2}$ and one positive eigenvalue $frac{5+sqrt{33}}{2}$, the origin is a saddle point and the system is unstable. Therefore, at the very least, you now know the stability of the ODE system.



        Now, to actually solve the non-homogeneous equation, use variation of parameters by changing $c_1,c_2$ to $u_1(t),u_2(t)$:



        $$left[begin{matrix} y_1 \ y_2end{matrix}right]=left[begin{matrix} frac{1}{6}(-3-sqrt{33})e^{(5-sqrt{33})/2} & frac{1}{6}(-3+sqrt{33})e^{(5+sqrt{33})/2} \ e^{(5-sqrt{33})/2} & e^{(5+sqrt{33})/2}end{matrix}right]left[begin{matrix} u_1(t) \ u_2(t)end{matrix}right]$$
        $$left[begin{matrix} frac{1}{6}(-3-sqrt{33})e^{(5-sqrt{33})/2} & frac{1}{6}(-3+sqrt{33})e^{(5+sqrt{33})/2} \ e^{(5-sqrt{33})/2} & e^{(5+sqrt{33})/2}end{matrix}right]left[begin{matrix} u_1'(t) \ u_2'(t)end{matrix}right]=left[begin{matrix} frac{3}{x^4} \ frac{3}{x^4}end{matrix}right]$$



        Here, you can use the second equation to solve for $u_1'(t),u_2'(t)$, integrate to solve for $u_1(t),u_2(t)$, and then plug into the first equation to solve for $y_1,y_2$. Good luck!






        share|cite|improve this answer












        First, solve the homogeneous equation:



        $$Dy_1=y_1+2y_2$$
        $$Dy_2=3y_1+4y_2$$



        I am assuming you already know how to do this, so I will just write the solution here:



        $$left[begin{matrix} y_1 \ y_2end{matrix}right]=left[begin{matrix} frac{1}{6}(-3-sqrt{33})e^{(5-sqrt{33})/2} & frac{1}{6}(-3+sqrt{33})e^{(5+sqrt{33})/2} \ e^{(5-sqrt{33})/2} & e^{(5+sqrt{33})/2}end{matrix}right]left[begin{matrix} c_1 \ c_2end{matrix}right] text{ where } c_1,c_2inmathbb{R}$$



        Since there is one negative eigenvalue $frac{5-sqrt{33}}{2}$ and one positive eigenvalue $frac{5+sqrt{33}}{2}$, the origin is a saddle point and the system is unstable. Therefore, at the very least, you now know the stability of the ODE system.



        Now, to actually solve the non-homogeneous equation, use variation of parameters by changing $c_1,c_2$ to $u_1(t),u_2(t)$:



        $$left[begin{matrix} y_1 \ y_2end{matrix}right]=left[begin{matrix} frac{1}{6}(-3-sqrt{33})e^{(5-sqrt{33})/2} & frac{1}{6}(-3+sqrt{33})e^{(5+sqrt{33})/2} \ e^{(5-sqrt{33})/2} & e^{(5+sqrt{33})/2}end{matrix}right]left[begin{matrix} u_1(t) \ u_2(t)end{matrix}right]$$
        $$left[begin{matrix} frac{1}{6}(-3-sqrt{33})e^{(5-sqrt{33})/2} & frac{1}{6}(-3+sqrt{33})e^{(5+sqrt{33})/2} \ e^{(5-sqrt{33})/2} & e^{(5+sqrt{33})/2}end{matrix}right]left[begin{matrix} u_1'(t) \ u_2'(t)end{matrix}right]=left[begin{matrix} frac{3}{x^4} \ frac{3}{x^4}end{matrix}right]$$



        Here, you can use the second equation to solve for $u_1'(t),u_2'(t)$, integrate to solve for $u_1(t),u_2(t)$, and then plug into the first equation to solve for $y_1,y_2$. Good luck!







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        answered 2 days ago









        Noble MushtakNoble Mushtak

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