# 7.1 Second-order linear equations  (Page 5/15)

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$y\left(x\right)={c}_{1}{e}^{-2x}+{c}_{2}{e}^{-7x}.$

## Single repeated real root

Things are a little more complicated if the characteristic equation has a repeated real root, $\lambda \text{.}$ In this case, we know ${e}^{\lambda x}$ is a solution to [link] , but it is only one solution and we need two linearly independent solutions to determine the general solution. We might be tempted to try a function of the form $k{e}^{\lambda x},$ where k is some constant, but it would not be linearly independent of ${e}^{\lambda x}.$ Therefore, let’s try $x{e}^{\lambda x}$ as the second solution. First, note that by the quadratic formula,

$\lambda =\frac{\text{−}b±\sqrt{{b}^{2}-4ac}}{2a}.$

But, $\lambda$ is a repeated root, so ${b}^{2}-4ac=0$ and $\lambda =\frac{\text{−}b}{2a}.$ Thus, if $y=x{e}^{\lambda x},$ we have

${y}^{\prime }={e}^{\lambda x}+\lambda x{e}^{\lambda x}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}y\text{″}=2\lambda {e}^{\lambda x}+{\lambda }^{2}x{e}^{\lambda x}.$

Substituting these expressions into [link] , we see that

$\begin{array}{cc}\hfill ay\text{″}+b{y}^{\prime }+cy& =a\left(2\lambda {e}^{\lambda x}+{\lambda }^{2}x{e}^{\lambda x}\right)+b\left({e}^{\lambda x}+\lambda x{e}^{\lambda x}\right)+cx{e}^{\lambda x}\hfill \\ & =x{e}^{\lambda x}\left(a{\lambda }^{2}+b\lambda +c\right)+{e}^{\lambda x}\left(2a\lambda +b\right)\hfill \\ & =x{e}^{\lambda x}\left(0\right)+{e}^{\lambda x}\left(2a\left(\frac{\text{−}b}{2a}\right)+b\right)\hfill \\ & =0+{e}^{\lambda x}\left(0\right)\hfill \\ & =0.\hfill \end{array}$

This shows that $x{e}^{\lambda x}$ is a solution to [link] . Since ${e}^{\lambda x}$ and $x{e}^{\lambda x}$ are linearly independent, when the characteristic equation has a repeated root $\lambda ,$ the general solution to [link] is given by

$y\left(x\right)={c}_{1}{e}^{\lambda x}+{c}_{2}x{e}^{\lambda x},$

where ${c}_{1}$ and ${c}_{2}$ are constants.

For example, the differential equation $y\text{″}+12{y}^{\prime }+36y=0$ has the associated characteristic equation ${\lambda }^{2}+12\lambda +36=0.$ This factors into ${\left(\lambda +6\right)}^{2}=0,$ which has a repeated root $\lambda =-6.$ Therefore, the general solution to this differential equation is

$y\left(x\right)={c}_{1}{e}^{-6x}+{c}_{2}x{e}^{-6x}.$

## Complex conjugate roots

The third case we must consider is when ${b}^{2}-4ac<0.$ In this case, when we apply the quadratic formula, we are taking the square root of a negative number. We must use the imaginary number $i=\sqrt{-1}$ to find the roots, which take the form ${\lambda }_{1}=\alpha +\beta i$ and ${\lambda }_{2}=\alpha -\beta i\text{.}$ The complex number $\alpha +\beta i$ is called the conjugate of $\alpha -\beta i\text{.}$ Thus, we see that when ${b}^{2}-4ac<0,$ the roots of our characteristic equation are always complex conjugates .

This creates a little bit of a problem for us. If we follow the same process we used for distinct real roots—using the roots of the characteristic equation as the coefficients in the exponents of exponential functions—we get the functions ${e}^{\left(\alpha +\beta i\right)x}$ and ${e}^{\left(\alpha -\beta i\right)x}$ as our solutions. However, there are problems with this approach. First, these functions take on complex (imaginary) values, and a complete discussion of such functions is beyond the scope of this text. Second, even if we were comfortable with complex-value functions, in this course we do not address the idea of a derivative for such functions. So, if possible, we’d like to find two linearly independent real-value solutions to the differential equation. For purposes of this development, we are going to manipulate and differentiate the functions ${e}^{\left(\alpha +\beta i\right)x}$ and ${e}^{\left(\alpha -\beta i\right)x}$ as if they were real-value functions. For these particular functions, this approach is valid mathematically, but be aware that there are other instances when complex-value functions do not follow the same rules as real-value functions. Those of you interested in a more in-depth discussion of complex-value functions should consult a complex analysis text.

Based on the roots $\alpha ±\beta i$ of the characteristic equation, the functions ${e}^{\left(\alpha +\beta i\right)x}$ and ${e}^{\left(\alpha -\beta i\right)x}$ are linearly independent solutions to the differential equation. and the general solution is given by

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nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
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learn
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Bhagvanji
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Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
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narayan
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Damian
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Professor
I think
Professor
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Alexandre
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Rafiq
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