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My textbook says the following:

Knowing $y(t)=ce^{rt}$ is a solution for $ay''+by'+cy=0$, we find for the case of $b^2-4ac=0$ that $r=\frac{-b}{2a}$. Therefore one particular solution is $y(t) = \exp\left(\frac{-b}{2a}t \right)$.

However, another linearly independent solution exists and that is $y(t)=t\exp\left(\frac{-b}{2a}t \right)$.

The derivation to find the second equation is that we can say $y(t)=ce^{rt}=v(t)\exp\left(\frac{-b}{2a}t \right)$ with variation of parameters, and then plug that $v(t)\exp\left(\frac{-b}{2a}t \right)$ into the equation (1)

$$ay''+by'+c=0 \tag{1}$$

we find that $v''(t) = 0$ and therefore v(t) is a linear function chosen to be $v(t)=t$. Hence the 2nd solution is $y(t)=te^{rt}$ and solution space is $y(t)=c_1e^{rt}+c_2te^{rt}$.

I have plugged $v(t)\exp\left(\frac{-b}{2a}t \right)$ many times into equation (1) yet can never seem to get the conclusion that $v''(t) = 0$. What does my textbook mean by this?

CITATION: page 213, Differential Equations & Linear Algebra, Second Edition, by Farlow, Hall, McDill, West

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    $\begingroup$ The discriminant of your differential equation is not zero $\Delta =1-4=-3$ $\endgroup$
    – user577215664
    Feb 26, 2020 at 0:17
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    $\begingroup$ The polynomial characteristic of your differential equation is $r^2+r+1=0$ and the discriminant is not zero. Did you calculate it ? You should choose a DE with discriminant equal to zero for example $y''-2y'+y=0$ $\endgroup$
    – user577215664
    Feb 26, 2020 at 0:27
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    $\begingroup$ For your differential equation the characteristic polynomial is $r^2+r+1=0$ the discriminant is $ \Delta = 1-4=-3$ $\endgroup$
    – user577215664
    Feb 26, 2020 at 0:33
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    $\begingroup$ No problem Ebehr...did you calculate the derivative of the solution and plug it in your equation ? $\endgroup$
    – user577215664
    Feb 26, 2020 at 0:38
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    $\begingroup$ I did the calculation and get the answer 's book $v''=0$ $\endgroup$
    – user577215664
    Feb 26, 2020 at 0:47

1 Answer 1

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Calculate first the derivatives $$y=ve^{-bt/2a}$$ $$y'=e^{-bt/2a}(v'-\dfrac {bv}{2a})$$ $$y''=e^{-bt/2a}(v''-\frac b a v'+\dfrac {b^2v}{4a^2})$$ Your equation becomes: $$ay''+by'+c=0$$ $$av''+v(-\dfrac {b^2}{4a}+c)=0$$ Since $$\Delta=b^2-4ac=0 \implies v''=0$$

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