A book I'm using to teach myself differential equations claims the following:

If $y_{1}$ and $y_{2}$ are solutions to the differential equation $y' - a(t)y = q(t)$, then $y = y_{1} - y_{2}$ will be a null solution by linearity.

I understand there exists some linear combination of $y_{1}$ and $y_{2}$ that would provide the null solution, but how can I be sure it is exactly $y = y_{1} - y_{2}$?

  • $\begingroup$ So is the book trying to prove that $y_1=y_2$? $\endgroup$ – 5xum Nov 9 '16 at 15:33

You can check this by hand:

Suppose $y_1$ and $y_2$ are solutions, so that

$y_1' - a(t) y_1 = q(t)$ and $y_2'- a(t) y_2 = q(t).$

Then, subtracting these two equations yields

$$y_1'-y_2' - a(t) y_1 +a(t) y_2 = 0.$$

A slight rearranging shows

$$(y_1-y_2)' - a(t) (y_1-y_2) = 0,$$

so indeed, $y=y_1-y_2$ is a null solution.

  • $\begingroup$ Thank you, that makes sense if you have an idea that the linear combination $y_{1} - y_{2}$ is a null solution and I didn't even think to do that. But how do I understand that this solution is a possibility to begin with? Can I view this in terms of vector spaces? $\endgroup$ – lhoernle Nov 9 '16 at 17:15
  • $\begingroup$ Think of the equation as $\mathcal {L} y=q $ where $\mathcal {L}$ is a linear operator, meaning it inputs $y $ and outputs $y'-a (t) $. Note that if I want a null solution then I need to get rid of $q $ on the right hand side by some linear combination of solutions on the left hand side. However, of course the easiest way to do this is to simply subtract two solutions so I have $q-q=0$ on the right hand side. In terms of vector spaces, indeed the solution set of the equation $\mathcal {L} y=0$ forms a vector space. $\endgroup$ – Matt Nov 9 '16 at 18:04
  • $\begingroup$ Thank you so much. Work got in the way and I was just able to see this now. $\endgroup$ – lhoernle Nov 10 '16 at 15:54

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