# 2nd-order ODE Solution for when Discriminant = 0

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

• The discriminant of your differential equation is not zero $\Delta =1-4=-3$ Feb 26, 2020 at 0:17
• 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 $y''-2y'+y=0$ Feb 26, 2020 at 0:27
• For your differential equation the characteristic polynomial is $r^2+r+1=0$ the discriminant is $\Delta = 1-4=-3$ Feb 26, 2020 at 0:33
• No problem Ebehr...did you calculate the derivative of the solution and plug it in your equation ? Feb 26, 2020 at 0:38
• I did the calculation and get the answer 's book $v''=0$ Feb 26, 2020 at 0:47

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 $$\Delta=b^2-4ac=0 \implies v''=0$$