# Homogeneous Linear Differential Equations

A **homogeneous linear differential equation** is a differential equation in which every term is of the form $y^{(n)}p(x)$ i.e. a derivative of $y$ times a function of $x$. In general, these are very difficult to work with, but in the case where all the constants are coefficients, they can be solved exactly. In this case, the differential equation looks like $a_{n} \dfrac{d^ny}{dx^n} + a_{n-1} \dfrac{d^{n-1}y}{dx^{n-1}} + \cdots + a_{1} \dfrac{dy}{dx} + a_{0} y = 0,$ with $a_{0}, a_{1} ,a_{n-1} ,\ldots , a_{n}$ being real constants, and almost resembles a polynomial. In fact, looking at the roots of this associated polynomial gives solutions to the differential equation.

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## Characteristic Equation

A reasonable guess for a solution to $a_ny^{(n)}+\cdots+a_0y$ is $y=e^{rx}$ - this is called an *ansatz*. Note that $\dfrac{d^ky}{dx^k} e^{rx} = r^k e^{rx},$ so for $e^{rx}$ to be a solution, it would have to satisfy $a_{n} r^n e^{rx} + a_{n-1} r^{n-1} e^{rx} + \cdots + a_{1} r e^{rx} + a_{0} e^{rx} = 0.$ Since $e ^ {rx} \neq 0$, it must be the case that $a_ {n}r^n + a_ {n-1}r^{n-1}+ \cdots + a_ {1} r + a_ {0} = 0. \qquad (3).$ Thus, $e^{rx}$ solves the differential equation when $r$ is a root of this polynomial.

For the differential equation $a_{n} \dfrac{d^ny}{dx^n} + a_{n-1} \dfrac{d^{n-1}y}{dx^{n-1}} + \cdots + a_{1} \dfrac{dy}{dx} + a_{0} y = 0,$ the associated

characteristic equationis $a_nr^n+\cdots+a_1r+a_0=0$.

Depending on the nature of the roots of the characteristic polynomial, the differential equation has slightly different solutions.

## Case of Distinct Real Roots

When the roots are real and distinct, the solutions are just the linear combinations of $e^{rx}$ for the different roots $r$.

If the $n$ roots $r_1, r_2, \ldots, r_{n}$ of the characteristic equation are real and distinct, then $y(x) = c_1 e^{r_1x} + c_2e^{r_2x} +\cdots +c_ne^{r_nx}$ is a general solution of the equation $(1),$ with $c_1, c_2, \ldots, c_n$ constants.

## Solve $$

$\begin{array}{c}&y'' + 2y' - 8y = 0, &y(0) = 5, &y'(0) = - 2.\end{array}$

We need to solve the characteristic equation $r^2 + 2r - 8 = 0.$

Observe that $r = 2, -4$ are the roots of this equation. Then since $y(x) = c_1 e^{2x} + c_2 e^{-4x}$ is the general solution, $y'(x) = 2 c_1 e^{2x} - 4 c_2 e^{-4x}$. Now, using the initial conditions, we have $y(0) = c_1 + c_2 = 5, y'(0) = 2c_1 - 4c_2 = -2 \Rightarrow c_1 = 3, c_2=2.$

Thereofore, the desired particular solution is $y(x) = 3e^{2x} + 2e^{-4x}. \ _\square$

## Case of Repeated Real Roots

When the characteristic polynomial has repeated roots, the previous theorem no longer applies.

If the characteristic equation has a repeated real root $r$ of multiplicity $k,$ then part of the general solution of the differential equation corresponding to $r$ in equation is of the form $(c_1 + c_ {2}x + c_ {3}x^2 + \cdots + c_ {k}x^{k-1}) \cdot e^{rx}.$

Find a general solution of $$ $y^{(4)} + 3 \cdot y^{(3)} + 3 \cdot y'' + y' = 0.$

The characteristic equation of the differential equation is $r^4 + 3 \cdot r^3 + 3 \cdot r^2 + r= r \cdot (r+1)^3=0$.

It has the single root $r_1 = 0,$ which gives the solution $y_1 = c_1$ to the general solution, and the triple root $(k=3)$ $r_2 = -1,$ which gives $y_2 = (c_2 + c_ {3}x + c_ {4}x^2) \cdot e^{-x}.$ Thus, the general solution of the differential equation is $y(x) = c_1 + (c_2 + c_ {3}x + c_ {4}x^2) \cdot e^{-x}. \ _\square$

## Case of Complex Roots

Because the coefficients of the differential equation and its characteristic equation are real, any root complex appears in complex conjugate pair $a \pm bi,$ where $a$ and $b$ are real and i = $\sqrt{-1}.$

If the characteristic equation has a pair of complex roots not repeated $a \pm bi$, then the relevant part to them of the general solution of equation has the form $e^{ax} \cdot (c_1 \cos bx + c_2 \sin bx)$.

Solve $y'' - 4y' + 5y = 0.$

The characteristic equation is $r^2 - 4r + 5 = 0,$ whose roots are $2 \pm i.$ Thus, the general solution is $y(x) = e^{2x} \cdot (c_1 \cos x + c_2 \sin x).$

When there are repeated complex roots, they can be accounted for in the same way as with repeated real roots.

## Find a general solution of $y^{(4)} + 4 \cdot y^{(3)} + 12 \cdot y'' + 16y' + 16y = 0.$

The characteristic equation is $(r^2 + 2r + 4)^2 = 0,$ whose roots are $-1 \pm i \cdot \sqrt{3}$ of multiplicity $2.$ Therefore, the general solution is $y(x) =e^{-x} \cdot \left(c_1 \cos x\sqrt{3} + d_1 \sin x\sqrt{3}\right) + xe^{-x} \cdot \left(c_2 \cos x\sqrt{3} + d_2 \sin x\sqrt{3}\right).$

**Cite as:**Homogeneous Linear Differential Equations.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/homogeneous-linear-differential-equations/