📂Odinary Differential Equations

Solution to Second Order Homogeneous Differential Equations

Theorem¹

$$ ay^{\prime \prime} + by^\prime + cy=0 $$

Let’s say the solutions to the characteristic equation $ar^2+br+c=0$ given above are $r_{1}$ and $r_2$. Then,

$\text{1.}$ If $r_{1}$ and $r_2$ are two distinct real numbers$(b^2-4ac>0)$, the general solution is as follows: $$ y(t)=c_{1}e^{r_{1}t}+c_2e^{r_2t} $$

$\text{2.}$ If $r_{1}$ and $r_2$ are complex conjugates $\lambda \pm i \mu$$(b^2-4ac<0)$, the general solution is as follows: $$ \begin{align*} y(t) &= c_{1}e^{(\lambda + i\mu)t} + c_2e^{(\lambda – i\mu)t} \\ &= c_{3}e^{\lambda t} \cos \mu t + c_{4} e^{\lambda t} \sin \mu t \end{align*} $$

$\text{3.}$ In the case of $r_{1}=r_2=r$$(b^2-4ac=0)$, the general solution is as follows: $$ y(t)=c_{1}e^{rt}+c_2te^{rt} $$

Solution

1. If $r_{1} \ne r_2$ and $r_{1}, r_2\in \mathbb{R}$

The general solution is as follows:

$$ y(t)=c_{1}e^{r_{1}t}+c_2e^{r_2t} $$

Here, $c_{1}, c_2$ is a constant, and if we know the two initial values $y(0)=y_{0}$ and $y^\prime (0) =y^\prime_{0}$, we can determine it exactly.

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2. If $r_{1} \ne r_2$ and $r_{1}, r_2 \in \mathbb{C}$

In the case where the discriminant of the characteristic equation is $b^2-4ac<0$. Since $r_{1}$ and $r_2$ become complex conjugates, it can be expressed as follows:

$$ r_{1}=\lambda + i\mu,\quad r_2=\lambda – i\mu $$

Then, the two solutions of the differential equation are as follows:

$$ y_{1}=e^{r_{1}t}=e^{(\lambda + i\mu)t}, y_{2}=e^{(\lambda – i\mu)t} $$

Therefore, the general solution is as follows:

$$ y(t)=c_{1}e^{(\lambda + i\mu)t} + c_2e^{(\lambda – i\mu)t} $$

Up to here, it is not particularly different from 1. If we represent the general solution using the Euler’s formula, then

$$ \begin{align*} &\ c_{1}e^{(\lambda + i\mu)t} + c_2e^{(\lambda – i\mu)t} \\ =&\ c_{1} e^{\lambda t} (\cos \mu t + i\sin \mu t) + c_2 e^{\lambda t} ( \cos \mu t – i \sin \mu t) \\ =&\ c_{3}e^\lambda \cos \mu t + c_{4} e^\lambda \sin \mu t \end{align*} $$

Therefore,

$$ y(t)=c_{3}e^{\lambda t} \cos \mu t + c_{4} e^{\lambda t} \sin \mu t $$

At this time, $c_{4}$ is a complex constant including $i$. If we know the initial values, we can determine $c_{3}$ and $c_{4}$ exactly.

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3. If $r_{1}=r_2=r=-\dfrac{b}{2a}$

In the case where the discriminant of the characteristic equation is $b^2-4ac=0$. $y_{1}$ can be found by $y_{1}=e^{\frac{-b}{2a}t}$, but $y_{2}$ cannot be found. Assume $y_{2}$ as $y(t)=\nu (t) y_{1}(t)$, then

$$ \begin{align*} y^\prime &= \nu ^\prime y_{1} + \nu y_{1}^\prime \\ y^{\prime \prime}&=\nu^{\prime \prime}y_{1} +\nu ^\prime y_{1}^\prime + \nu^\prime y_{1}^\prime + \nu y_{1}^{\prime \prime}=\nu^{\prime \prime}y_{1}+2\nu ^\prime y_{1}^\prime+ \nu y_{1}^{\prime \prime} \end{align*} $$

Substituting $y^\prime$ and $y^{\prime \prime}$ into the given differential equation gives

$$ a \left( \nu^{\prime \prime}y_{1}+2\nu ^\prime y_{1}^\prime+ \nu y_{1}^{\prime \prime} \right) + b \left( \nu ^\prime y_{1} + \nu y_{1}^\prime \right) + c\nu y_{1}=0 $$

Organizing it according to $\nu$ turns out as follows:

$$ \nu \left( ay_{1}^{\prime \prime} + b y_{1}^{\prime} + cy_{1}\right) + \nu^\prime \left( 2ay_{1}^\prime+by_{1} \right) + ay_{1} \nu ^{\prime \prime}=0 $$

Here, since $y_{1}$ is a solution to the given differential equation, the first parenthesis is $0$. Also, since $y_{1}=e^{(-b/{2a})t}$ and $y_{1}^\prime = \frac{-b}{2a}e^{({-b}/{2a})t}$,

$$ \begin{align*} &&\nu^\prime \left( 2a \dfrac{-b}{2a}e^{\frac{-b}{2a}t} + b e^{\frac{-b}{2a}t} \right) + ae^{\frac{-b}{2a}t} \nu ^{\prime \prime}&=0 \\ \implies&& \nu^\prime(-b+b)+a\nu^{\prime \prime}&=0 \\ \implies && \nu^{\prime \prime} &=0 \end{align*} $$

Therefore, it turns out that $\nu (t)=c_{1}+c_2t$. Ultimately, the general solution to the given differential equation is

$$ y(t)=\nu (t) y_{1}(t)=c_{1}y_{1}(t)+c_2ty_{1}(t)=c_{1}e^{rt}+c_2te^{rt} $$

In other words, it turns out to be $y_{2}=ty_{1}$.

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Examples

1.

$$ y^{\prime \prime}+ 5y^\prime + 6y=0 \\ y(0)=2 \\ y^\prime (0)=3 $$

The characteristic equation is $r^2+5r+6=0$. That is $(r+2)(r+3)=0$, so $r_{1}=-2$, $r_2=-3$. Therefore, the general solution is

$$ y(t)=c_{1}e^{-2t}+c_2e^{-3t} $$

Differentiating the general solution gives

$$ y^\prime(t)=-2c_{1}e^{-2t} -3c_2e^{-3t} $$

Substituting the initial values gives

$$ \begin{cases} c_{1}+c_2=2 \\ -2c_{1}-3c_2=3 \end{cases} $$

Solving them together gives

$$ c_{1}=9,\quad c_2=7 $$

Therefore, the solution for the given initial values is

$$ y(t)=9e^{-2t}+7e^{-3t} $$

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2.

$$ y^{\prime \prime}+4y^\prime + 8y=0 $$

The characteristic equation is

$$ r^2+4r+8=0 $$

The solutions of the characteristic equation are

$$ r_{1,2}= \dfrac{-4\pm \sqrt{16-32}}{2}=-2 \pm 2i $$

So $\lambda=-2$, $\mu=2$. Therefore, the general solution for the given differential equation is

$$ y(t)=c_{1}e^{-2t}\cos 2t + c_2 e^{-2t} \sin 2t $$

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3.

$$ y^{\prime \prime} + 4y^\prime + 4y=0 $$

The characteristic equation is

$$ r^2+4r+4=(r+2)^2=0 $$

The solutions of the characteristic equation are

$$ r=-2 $$

Therefore, it turns out $y_{1}(t)=e^{-2t}$ and the general solution is

$$ y(t)=c_{1}e^{-2t} + c_2te^{-2t} $$

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