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This is the system:

$$ y_1^{'} = 2y_1+y_2+xe^x \\ y_2^{'}=3y_1 + 4y_2 + 2e^x$$

Solution to this system is sum of solution to the corresponding homogeneous system and one particular solution of this system. So when i find solution of homogeneous system i can use variation of parameters method (and i am supposed to, since text of the problem requires it even though i'm not very smooth with this method), but the problem is, when i try to solve homogeneous system:

$$ y_1^{'} = 2y_1+y_2\\ y_2^{'}=3y_1 + 4y_2 $$

i have the following:

firstly i find characteristic polynomial and it's $$\lambda^2-6\lambda+5$$ so i have characteristic numbers $$\lambda_1=1$$ and $$\lambda_2=5$$

Now, for $\lambda_1$ i end up with the following system

$$ y_1^{'} = y_1+y_2\\ y_2^{'}=3y_1 + 3y_2 $$ and since solutions are following type of functions $$y=ae^x$$ where $a$ is constant i end up with this algebraic system of equations $$a_1=a_1+a_2 \\ a_2=3a_2+3a_1$$ and it has only trivial solution, and similar thing happens for the second characteristic number, which made me stuck here, even before what i thought that would be the hard part of the problem, and that is the method of variations of parameters. Any ideas on how to solve this?

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  • $\begingroup$ Do you know Laplace Transform? $\endgroup$
    – xpaul
    Aug 22, 2017 at 14:23
  • $\begingroup$ @cdummie: Are you sure the equations are written correctly? $\endgroup$
    – Moo
    Aug 22, 2017 at 14:25
  • $\begingroup$ @xpaul No, that's beyond this course, i am not supposed to use Laplace transform $\endgroup$
    – cdummie
    Aug 22, 2017 at 15:25
  • $\begingroup$ @Moo They are, why? $\endgroup$
    – cdummie
    Aug 22, 2017 at 15:25
  • $\begingroup$ I added the eigenvector calculations below, but as written, the non-homogeneous system does not have a solution - so I suspect there is a typo. $\endgroup$
    – Moo
    Aug 22, 2017 at 15:27

3 Answers 3

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If I am understanding, you are trying to solve the homogeneous part first.

For $\lambda_1 = 1$, the RREF of $[A - \lambda_1 I] v_1 = [A - I] v_1 = 0$ gives

$$\begin{bmatrix}1 & 1 \\0 & 0\end{bmatrix} v_1 = 0 \implies v_1 = \begin{bmatrix} - 1 \\ 1 \end{bmatrix}$$

RREF details for $\lambda_1$

$$[A - I] v_1 = \begin{bmatrix}1 & 1 \\3 & 3\end{bmatrix} v_1 = 0$$

  • Divide row $2$ by $3$:

$$\begin{bmatrix}1 & 1 \\1 & 1\end{bmatrix} v_1 = 0$$

  • Set row $2 =$ row $2 - $ row $1$:

$$\begin{bmatrix}1 & 1 \\0 & 0\end{bmatrix} v_1 = 0$$

  • We can now solve for $v_1$:

$$v_1 = \begin{bmatrix} - 1 \\ 1 \end{bmatrix}$$

Repeating this process for $\lambda_2 = 5$, the RREF of $[A - \lambda_2 I] v_2 = [A - 5I] v_2 = 0$ gives

$$\begin{bmatrix}1 & -\dfrac{1}{3} \\0 & 0\end{bmatrix} v_2 = 0 \implies v_2 = \begin{bmatrix} 1 \\ 3\end{bmatrix}$$

I will assume you can continue here to write the homogeneous solution.

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  • $\begingroup$ well, it is $$y_1^{'} = 2y_1+y_2+ t e^t\\ y_2^{'}=3y_1 + 4y_2 + 2e^t$$ ,exactly as you written it, except you used variable t instead of x, but it makes no difference and i have no idea how you got those RREF's considering the equations i have. $\endgroup$
    – cdummie
    Aug 22, 2017 at 16:51
  • $\begingroup$ well, i use it when i want to determine rank of matrix, so i suppose i do, but isn't Gauss-Jordan used for RREF? $\endgroup$
    – cdummie
    Aug 22, 2017 at 17:07
  • $\begingroup$ I added more details on the RREF calculation for one eigenvalue. You can now try the second one. $\endgroup$
    – Moo
    Aug 22, 2017 at 17:14
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    $\begingroup$ i see now, i actually overlooked the fact that, when i plug in the characteristic values i have rhs equal to zero, that was the main problem, i solved it now, thanks! $\endgroup$
    – cdummie
    Aug 23, 2017 at 10:22
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Since exp(x) appears in both forcing terms, it makes sense to factor it out:

y = g exp(x)

Under which the equations become

g1' = g1 + g2 + x

g2'= 3( g1 + g2 ) +2

Add them up, with g= g1+g2

g' = 4 g +x +2

Solution is simple... g= C exp(4x) - x/4 - 9/16

Then get g2 by integration

g2' = 3 g + 2

Or g2 = 3C/4 + 5/16 x - 3/8 x^2 + D

And g1= g-g2

If the purpose of the problem was to get you to use the general formalism of characteristic equation etc, this doesn't help much. More often than not I find that formalism gets people into trouble on text book problems.

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  • $\begingroup$ The reason you had trouble is that it's one of those special cases designed to trip you up.. one of the forcing functions is a homogeneous solution. In dynamics it's called a resonance. Usually you just multiply by x to get a particular solution. $\endgroup$
    – Mhw
    Aug 22, 2017 at 17:58
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From the second equation, one has $$ y_1=\frac13y_2^{'}-\frac43y_2-\frac23e^x. \tag{1}$$ Differentiating the second equation, one has $$ y_2''=3y_1' + 4y_2' + 2e^x.\tag{2}$$ Using (1) and the first equation, one has $$ y_2''=3(2y_1+y_2+xe^x)+4y_2' + 2e^x=6(\frac13y_2^{'}-\frac43y_2-\frac23e^x)+3y_2+3xe^x+4y_2' + 2e^x$$ or $$ y_2''-6y_2'+5y_2=(3x-2)e^x.\tag{3}$$ Note that $y_2''-6y_2'+5y_2=0$ has two independent solutions $e^x,e^{5x}$ with the Wronskian $$ W(e^x,e^{5x})=\left|\begin{matrix}e^x&e^{5x}\\e^x&5e^{5x}\end{matrix}\right|=4e^{6x}$$ and that (3) has a particular solution $$ y_p=-e^x\int\frac{e^{5x}(3x-2)e^x}{W(e^x,e^{5x})}dx+e^{5x}\int\frac{e^{x}(3x-2)e^x}{W(e^x,e^{5x})}dx$$ which gives $$ y_2=\frac{1}{64}(-24x^2+20x+5)e^x.$$ Thus the solution of (3) is $$ y_2=\frac{1}{64}(-24x^2+20x+5)e^x+C_1e^x+C_2e^{5x} $$ and hence one has from (1) $$ y_1=\frac{1}{192}(72x^2-108x-123)e^x-C_1e^x+\frac13e^{5x}.$$

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