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For a homework problem, we are provided:

$\frac{dx}{dt}=-y + t$

$\frac{dy}{dt}=x-t$

Putting these into differential operator notation and separating the dependent variables from the independent:

$Dx-y=t$

$Dy-x=-t$

My first inclination is to apply the D operator to the second equation to eliminate Dx and get:

$D^2y+y=t-1$

I solve the homogenous part and end up with $y_c=C_1\cos(t) + C_2\sin(t).$

Using annihilator approach and method of undetermined coefficients, I determine that $y_p=t-1$.

General solution for $y(t) = C_1\cos(t)+C_2\sin(t)+t-1$.

After plugging $y$ into the second equation, I get $x(t)=-C_1\sin(t)+C_2\cos(t)+1+t$

Checking my answer against the back of the book, they show: $x(t) = C_1\cos(t)+C_2\sin(t)+t+1$ and $y(t)=C_1\sin(t)-C_2\cos(t)+t-1$

I can't seem to find what I did wrong. Chegg solutions shows to eliminate y instead of x, and got the book's solution. Does the variable chosen for elimination matter? Halp!

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  • $\begingroup$ Oops, you're correct. I interchanged the x(t) and y(t) from the back of the book. Will fix now. $\endgroup$ – Irongrave Mar 16 '15 at 0:37
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    $\begingroup$ Both your solution and the book solution are correct and coincide up to renaming resp. replacing the constants. $\endgroup$ – Dr. Lutz Lehmann Mar 16 '15 at 3:16
  • $\begingroup$ Look here. $\endgroup$ – Kw08 Jan 22 '17 at 1:32
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Are you sure the original system is written as it is in the book? (Problem was updated to correct $x(t), y(t))$.

I get $$x'' +x = t + 1 \implies x(t) = c_1 \cos t + c_2 \sin t + t + 1$$ and $$y'' + y = t - 1 \implies y(t) = c_1 \sin t + c_2 \cos t + t - 1.$$

Note: this can be written as $y(t) = c_1 \sin t - c_2 \cos t + t - 1$ because $c_2$ is an arbitrary constant. Showing the negative removes the confusion when plugging back in so the authors decided to show it as part of the solution.

You can easily verify this solution by plugging it back into the original system.

Update Lets do the first in more detail. We have:

$$x'' +x = t + 1$$

To solve the homogeneous, we have $m^2 + 1 = 0 \implies m_{1,2} = \pm ~ i$, yielding:

$$x_h(t) = c_1 \cos t + c_2 \sin t$$

For the particular, we can choose $x_p = a + b t$, and substituting back into the DEQ, yields:

$$x'' + x = a + bt = 1 + t \implies a = b = 1$$

This produces:

$$x(t) = x_h(t) + x_p(t) = c_1 \cos t + c_2 \sin t + t + 1$$

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  • $\begingroup$ Thanks, I just fixed that. I am still having difficulty understanding where I went wrong. $\endgroup$ – Irongrave Mar 16 '15 at 0:39
  • $\begingroup$ I will add details on the $x(t)$ calculation. $\endgroup$ – Amzoti Mar 16 '15 at 0:41
  • $\begingroup$ Thanks. I get that that works if you choose to eliminate y, but why doesn't eliminating x work the same way? $\endgroup$ – Irongrave Mar 16 '15 at 0:46
  • $\begingroup$ It does, in my solution I absorb the negative they show into the constant. Clear? $\endgroup$ – Amzoti Mar 16 '15 at 0:47
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    $\begingroup$ @Irongrave: Of course do the signs and the order of the constants matter, since the two solution components are connected by them. However, it is quite permissible to do a linear transformation of the constants by replacing $C_1=D_2$ and $C_2=-D_1$ to go from the book solution with constants $(C_1,C_2)$ to your solution form with renamed constants $(D_1,D_2)$. $\endgroup$ – Dr. Lutz Lehmann Mar 16 '15 at 3:14
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You could of course also see the complex differential equation $$ \dot z=i·z+(1-i)·t $$ in this system.

Its homogeneous solution is $z=c·e^{it}$. The inhomogeneous solution can be constructed as usual per linear ansatz $z_p=a+b·t$ leading to $$b=i·a+i·b·t+(1-i)·t$$ and thus $b=1+i$ and $a=1-i$ and the full solution $$ z=c·e^{it}+(1-i)·(1+i·t) $$ Separating into real and imaginary part gives the solution of the original system.

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