Given the following ode system


and $2$ solutions of the system $$u_{(1)}=\begin{pmatrix}(a+b)t^2\\at^2\\bt^2\end{pmatrix},\quad u_{(2)}=\begin{pmatrix}2\ln t+1\\ \ln t+1\\\ln t+1\end{pmatrix}$$

What is the system's general solution ?


I think it is quite clear that some sort of parameters variation is required due to given solutions. The only problem is that I am not sure how to approach this solution.

I tried to develop some sort of a matrix $$\psi(t)=\begin{pmatrix}(a+b)t^2&2\ln t+1\\at^2&\ln t+1\\bt^2&\ln t+1\end{pmatrix}$$ such that for some $v=\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}$ fulfills $\psi(t)v'(t)=\begin{pmatrix}t^2e^t\\t^3\\t^2e^t\end{pmatrix}$;

the same method we solve linear ode systems, but clearly this matrix multipication is undefined.

How should I approach this? Thank you very much!

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    $\begingroup$ These are not solutions of the system, the first is part of the homogeneous solution, the second vector seems not to have any relation to the system. $\endgroup$ – LutzL May 31 '18 at 12:11

The system matrix $$A=\pmatrix{0&2&2\\-1&3&1\\-1&1&3}$$ has the characteristic polynomial $$ \det(A-\lambda I) %=\det\pmatrix{-λ&2&2\\-1&3-λ&1\\-1&1&3-λ} =-λ(3-λ)^2-4+λ+4(3-λ) \\ =-λ^3+6λ^2-12λ+8=(-λ+2)^3 $$ so that substituting $u_k=x_k/t^2$, $u_k'=(tx_k'-2x_k)/t^3$, results in the system $$ \left\{\begin{aligned} tu_1'&=-2u_1+2u_2+2u_3+te^t\\ tu_2'&=-u_1+u_2+u_3+t^2\\ tu_3'&=-u_1+u_2+u_3+te^t \end{aligned}\right. $$ As one can see, the homogeneous part is always a multiple of $v=-u_1+u_2+u_3$ with $$ tv'=t^2\implies v=\frac12t^2+c $$ and consequently \begin{align} tu_1'&=2v+te^t&\implies u_1 &= \frac12t^2+2c\ln t+e^t+d_1\\ tu_2'&=v+t^2&\implies u_2&=\frac34t^2+c\ln t+d_2\\ tu_3'&=v+te^t&\implies u_3&=\frac14t^2+c\ln t+e^t+d_3 \end{align} with $c=-d_1+d_2+d_3$. With setting $d_2=a+c$, $d_3=b+c$, $d_1=d_2+d_3-c=a+b+c$ one gets expressions close to those you cited, but with extra terms.

  • $\begingroup$ Nice solution! But is it not $x_k = t^2 u_k$? $\endgroup$ – Stefan May 31 '18 at 13:15
  • $\begingroup$ @Stefan: Yes, you are right. $\endgroup$ – LutzL May 31 '18 at 13:30
  • $\begingroup$ It seems like the substitution you used follows from the characteristic polynomial. Could you elaborate more on this? $\endgroup$ – Chee Han May 31 '18 at 17:56
  • $\begingroup$ It is a shift in the spectrum so that the eigenvalues of the resulting matrix all become zero. $tx'-2x=(A-2I)x+b(t)$ and the integrating factor for the left side is $t^{-3}$ to get $(t^{-2}x(t))'$. $\endgroup$ – LutzL May 31 '18 at 18:53

*Another approach *

$$\cases{tx_1'=2x_2+2x_3+t^3e^t\\tx_2'=-x_1+3x_2+x_3+t^4\\tx_3'=-x_1+x_2+3x_3+t^3e^t}$$ We can observe that $r_2-r_3$ gives :

$$t(x'_2-x'_3)=2(x_2-x_3)+t^4-t^3e^t$$ Thats easy to integrate (substitute $z=x_2-x_3)$ $$ \implies tz'=2z+t^4-t^3e^t$$ $$ \implies t^2z'-2tz=t^5-t^4e^t \implies (\frac z {t^2})'=t-e^t $$ $$ \boxed{\implies x_2=x_3+\frac {t^4}2-e^tt^2+Kt^2}$$ You have $x_2$ as a function of $x_3$ and t

We also have $r_1-2r_3$ which gives another easy integrable equation $$t(x'_1-2x'_3)=2(x_1-2x_3)-t^3e^t$$ $$\implies tv'=2v-t^3e^t$$ Solve that equation with the same method as for the first one $$ \boxed{\implies x_1=2x_3-e^tt^2+Ct^2}$$ You get $x_1$ in function of $x_3$ and t

Then take any equation and substitute these values ..and integrate $$tx_3'=-x_1+x_2+3x_3+t^3e^t$$ $$tx_3'-2x_3=t^3e^t+\frac{t^4}2+Kt^2-Ct^2$$ $$(\frac {x_3}{t^2})'=e^t+\frac t 2+\frac 1t(K-C)$$ $$\boxed{\implies x_3=t^2e^t+\frac {t^4}4+(K-C)t^2\ln|t|+At^2 }$$

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    $\begingroup$ I like this approach, but what are $z,v$ role in the general solution ? are they combined in a matrix in some way ? $\endgroup$ – Yariv Levy May 31 '18 at 14:02
  • $\begingroup$ @YarivLevy the first integral gives $x_2$ in funstion of t and $x_3$ same for the second equation...once you have these relations you take any equation of the system and substitute the value of the unknown functions $\endgroup$ – Isham May 31 '18 at 14:06
  • $\begingroup$ @YarivLevy I added some lines I hope it's more clear now $\endgroup$ – Isham May 31 '18 at 14:42
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    $\begingroup$ It is, thank you ! $\endgroup$ – Yariv Levy May 31 '18 at 14:50
  • $\begingroup$ yw @YarivLevy ...... $\endgroup$ – Isham May 31 '18 at 14:51

A quite general solution to these kinds questions is tried here,

The system can be written as $$ x'(t) = A(t)x(t)+b(t). $$, with $A(t)$ commuting for all $t$. This has a general solution written as

$$ x(t) = \exp(A(t))\left (x(0) + \int_0^t\exp(-A\tau)b(\tau)\,d\tau \right ) $$

We can solve for $x(0)$ and get $$ x(0) = \exp(-A(t))x(t) - \int_0^t\exp(-A\tau)b(\tau)\,d\tau $$

Let $c(t) = \int_0^t\exp(-A\tau)b(\tau)\,d\tau$, hence $$ x(0) = \exp(-A(t)) x(t) - c(t) $$ If you have two solutions $u_1,u_2$, then you get two solutions $x_1(0),x_2(0)$. Assume $x_1(0),x_2(0)$ linearly independent, then you can take $x_3(0) = x_1(0)\times x_2(0)$ and write the general solution as $$ a_1 x_1(0)+a_2 x_2(0) + a_3 x_3(0) = \exp(-A(t)) x_{a}(t) - c(t) $$ L.H.S is the same as $$ a_1 (\exp(-A(t))x_1(t)-c(t)) + a_2 (\exp(-A(t))x_2(t)-c(t)) +a_3(\exp(-A(t))x_1(t)-c(t))\times (\exp(-A(t))x_2(t)-c(t))=L_a(t) $$ And we get,

$$ x_a(t) = \exp(A(t))(L_a(t) + c(t)) $$ And you can express all solutions with the solutions $x_1(t)$ and $x_2(t)$

  • $\begingroup$ This is semi-wrong. Here you get $x'=\frac1tAx+b(t)$ so that the values of $A(t)$ actually commute which is necessary for $\exp(\int A(s)ds)$ to have anything to do with the solution. Then $\int_1^t A(s)ds=A\ln(t)$, so that now you have to compute a matrix exponential $e^{A\ln(t)}\overset?=t^A$, which is done, surprise, using the eigen-decomposition of $A$. $\endgroup$ – LutzL May 31 '18 at 15:09
  • $\begingroup$ Thanks LutzL, I agree with you that the comuting must be ensured, I forgot about that. I'll add it $\endgroup$ – Stefan May 31 '18 at 17:12
  • $\begingroup$ The exponent is an anti-derivative of $A(t)$. Then you have to consider that $\exp(\int A(t)\,dt)$ is a matrix, so you must be careful in the order of factors. $\endgroup$ – LutzL May 31 '18 at 17:27
  • $\begingroup$ @LutzL Yeah, got the order backwards. I think I fixed it now $\endgroup$ – Stefan May 31 '18 at 17:46

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