Question #76309

Sep 20, 2016

For homogenous linear differential equation with constant coefficients, the general solution has the form

$y \left(t\right) = C {e}^{\lambda t}$. Substituing this solution into the homogeneous equation, we have

$\left({a}_{n} {\lambda}^{n} + {a}_{n - 1} {\lambda}^{n - 1} + \cdots + {a}_{1} \lambda + {a}_{0}\right) C {e}^{\lambda t} = 0$

${e}^{\lambda t} \ne 0$ so the equation is verified for $\lambda ' s$ which are the roots of

$p \left(\lambda\right) = {a}_{n} {\lambda}^{n} + {a}_{n - 1} {\lambda}^{n - 1} + \cdots + {a}_{1} \lambda + {a}_{0}$

the so called characteristic polynomial.

If ${\lambda}_{k} , k = 1 , 2 , \cdots , {\lambda}_{n}$ are the roots, supposing that they are distinct, them the general solution will be posed as

$y \left(t\right) = {\sum}_{k} {C}_{k} {e}^{{\lambda}_{k} t}$

In case of repeated roots, appear associated polinomials in $t$ with their degree equal to the repetition level.

The constants ${C}_{k}$ are determined according with the initial conditions.

https://en.wikipedia.org/wiki/Linear_differential_equation

Sep 20, 2016

See explanation.

Explanation:

It is found that the differential equation can be converted to a

polynomial equation, with the same coefficients, by the substitution

$y = {e}^{r x}$, using ${n}^{t h}$ derivative of ${e}^{r x} = {r}^{k} y , k = 1 , 2 , 3 , \ldots , n$.

With D for the differentiation operator $\frac{d}{\mathrm{dx}}$, the conversion

gives.

$\sum {a}_{k} {D}^{k} y = \left(\sum {a}_{k} {r}^{k}\right) {e}^{r x} = 0$

As ${e}^{r x} > 0 , \sum {a}_{k} {r}^{k} = 0$

It follows that , for every root ${r}_{k}$ of this polynomial equation,

$y = {e}^{{r}_{k} x}$ is a solution for the the differential equation.

Also, an arbitrary scalar ${A}_{k} \times {e}^{{r}_{k} x}$ is a solution..

Theoretically, reduction of the order of the differential equation by

every integration produces one constant of integration. So,

successive integration n times to produce the general solution

would deposit n constants of integration.

Now, the linear sum

$y = \sum {A}_{k} {e}^{{r}_{k} x}$ substituted in the differential equation leads to

$\sum \left({A}_{k}\right) \left(0\right) = 0$.

And so, we are justified in stating that

$y = \sum {A}_{k} {e}^{{r}_{k} x}$ is the general solution.

There are particular cases like ${D}^{3} y = 0$, wherein this method

Here, $y = a {x}^{2} + b x + c$ is a quadratic in x

Here, 0 is a thrice repeated root. of the characteristic equation.

For $\left({D}^{3} + {D}^{2}\right) y = {D}^{2} \left(D + 1\right) y = 0$, it is a combination

$y = a x + b + c {e}^{- x}$, against the three roots 0, 0 and -1.

The part ax + b comes from direct integration, twice in succession,

for removing D^2. The other operator D+1 gives the part $c {e}^{- x}$

Substitute separately both in the differential equation and see

how it works.

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