# Differential Equations - Homogeneous Equations with Constant Coefficients

A real linear homogeneous differential equation of \(n^\text{th}\) order with constant real coefficients is an equation in the form

\[a_{n} \dfrac{d^ny}{dx^n} + a_{n-1} \dfrac{d^{n-1}y}{dx^{n-1}} + \cdots + a_{1} \dfrac{dy}{dx} + a_{0} y = 0 \qquad (1) \]

with \(a_{0}, a_{1} ,a_{n-1} ,\ldots , a_{n}\) real constants.

#### Contents

## Characteristic Equation

Observation:

\[ \dfrac{d^ky}{dx^k} e^{rx} = r^k e^{rx}. \qquad (2)\]

Now, substituting \( y = e^{rx}\) in \((1)\) and using \((2),\) we obtain

\[ a_{n} r^n e^{rx} + a_{n-1} r^{n-1} e^{rx} + \cdots + a_{1} r e^{rx} + a_{0} e^{rx} = 0.\]

Since \( e ^ {rx} \neq 0 \), we can cancel out this term to obtain the polynomial

\[ a_ {n}r^n + a_ {n-1}r^{n-1}+ \cdots + a_ {1} r + a_ {0} = 0. \qquad (3) \]

We see that \(y = e^{rx}\) will be the solution of equation \((1)\) precisely when \(r\) is a root of this polynomial.

This equation is called the **characteristic equation** or auxiliary equation of the differential equation \((1).\)

## Case of Distinct Real Roots

## Theorem 1 \[\]

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. \(_\square\)

## Solve \[\]

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

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

If there are repeated real roots of the characteristic equation \((3)\), then we cannot generate \(n\) linearly independent solutions to equation \((1)\) by the method of **Theorem 1**. Let us now consider the case that characteristic equation \[ a_ {n}r^n + a_ {n-1}r^{n-1}+ \cdots + a_ {1} r + a_ {0} = 0 \qquad (3) \] has repeated real roots.

## Theorem 2 \[\]

If the characteristic equation \((3)\) has a repeated real root \(r\) of multiplicity \(k,\) then part of the general solution of the differential equation corresponding to \(r\) in equation \((1)\) is of the form

\[(c_1 + c_ {2}x + c_ {3}x^2 + \cdots + c_ {k}x^{k-1}) \cdot e^{rx}. \ _\square\]

## 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 = \(\sqrt{-1}.\)

## Theorem 3: (No repeated complex roots) \[\]

If the characteristic equation \((3)\) has a pair of complex roots not repeated \(a \pm bi\), then the relevant part to them of the general solution of equation \((1)\) has the form \(e^{ax} \cdot (c_1 \cos bx + c_2 \sin bx). \ _\square\)

## 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 \(y(x) = e^{2x} \cdot (c_1 \cos x + c_2 \sin x). \ _\square\)

**Theorem 2** holds for the case of repeated complex roots. If the conjugate pair \(a \pm bi\) have multiplicity \(k,\) then the relevant part to them of the general solution has the form

\[\begin{align} &\left(A_1 + A_ {2}x + A_ {3}x^2 + \cdots + A_ {k}x^{k-1}\right) \cdot e^{(a+bi)x} + \left(B_1 + B_ {2}x + B_ {3}x^2 + \cdots + B_ {k}x^{k-1}\right) \cdot e^{(a-bi)x}\\ &= \sum_{p=0}^{k-1}x^{p} \cdot e^{ax} \cdot (c_{p} \cos bx + d_{p} \sin bx). \qquad (4) \end{align}\]

## 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 equation \((4)\) gives the general solution

\[y(x) =e^{-x} \cdot \left(c_1 \cos x\sqrt{3} + d_1 \sin x\sqrt{3}\right) + xe^{-x} \cdot \left(c_2 \cos x\sqrt{3} + d_2 \sin x\sqrt{3}\right). \ _\square\]

\[\] DIFFERENTIAL EQUATIONS - NO HOMOGENEOUS EQUATIONS WITH CONSTANT COEFFICIENTS

Let \(y_p\) be a particular solution of the no homogeneous equation \[a_{n}(x) \dfrac{d^ny}{dx^n} + a_{n-1}(x) \dfrac{d^{n-1}y}{dx^{n-1}} + \cdots + a_{1}(x) \dfrac{dy}{dx} + a_{0}(x) y = f(x) \qquad (1) \] in an open interval \(I\) in which the functions \(f(x)\) and \(a_{i}(x)\) are continuous. Let \(y_1, y_2, \ldots , y_n\) be linearly independent solutions of the associated differential homogeneous equation \[a_{n}(x) \dfrac{d^ny}{dx^n} + a_{n-1}(x) \dfrac{d^{n-1}y}{dx^{n-1}} + \cdots + a_{1}(x) \dfrac{dy}{dx} + a_{0}(x) y = 0. \qquad (*) \] If \(Y\) is a solution of equation \((1)\) in \(I,\) then there exists constants \(c_1, c_2, \ldots , c_n\) such that \[ Y(x) = c_1 y_1(x) + c_2 y_2(x) + \cdots + c_n y_n(x) + y_p(x)\] for all \(x\) in \(I.\) \(_\square\)

Sketch of Proof:

\(Y\) solution of \((1)\)

\(\Rightarrow\) \( Y - y_p\) solution of the associated differential equation \((*)\)

\(\Rightarrow\) there exist \(c_1, c_2, \ldots , c_n\) such that \[ Y(x) = c_1 y_1(x) + c_2 y_2(x) + \ldots + c_n y_n(x) + y_p(x)\] for all \(x\) in \(I.\) \(_\square\)

Let now the (???) no homogeneous differential equation of \(n^\text{th}\) order with constant coefficients \[a_{n} \dfrac{d^ny}{dx^n} + a_{n-1} \dfrac{d^{n-1}y}{dx^{n-1}} + \cdots + a_{1} \dfrac{dy}{dx} + a_{0} y = f(x), \qquad (1) \] then the general solution of \((1)\) is \[ Y = y_c + y_p, \] where \(y_c\) is a general solution of the associated homogeneous equation \[a_{n} \dfrac{d^ny}{dx^n} + a_{n-1} \dfrac{d^{n-1}y}{dx^{n-1}} + \cdots + a_{1} \dfrac{dy}{dx} + a_{0} y = 0, \qquad (2) \] and \(y_p\) is a simple particular solution of \((1).\)

\[\] UNDETERMINED COEFFICIENTS METHOD

The method of undetermined coefficients is applied when \(f(x)\) according to equation \((1)\) is a finite linear combination of products of functions of three types:

- A polynomial in \(x.\)
- An exponential function.
- \(\cos kx\) or \(\sin kx.\)

Let's see some examples:

## Find a particular solution of \[\]

\[ y'' + 3y' + 4y = 3x +2.\]

Here \(f(x) = 3x +2.\) Let's seek \(y_p(x) = Ax + B.\) Then \(y'_p(x) = A\) and \(y''_p(x) = 0\), so \(y_p\) will satisfy the differential equation \(3A + 4(Ax + B) = 3x +2\) if and only if \(4A = 3\) and \(3A +4B = 2,\) i.e. \(A = \frac{3}{4}\) and \(B = \frac {-1}{16}.\) Therefore, \[y_p(x) = \frac {3}{4}x - \frac {1}{16}. \ _\square \]

## Find a particular solution of \[\]

\[ y'' + 4y = 3x^3.\]

The complementary function of this equation is \(y_c(x) = c_1 \cdot \cos 2x + c_2 \cdot \sin 2x\). Here \(f(x) = 3x^3\). Let's seek

\[\begin{align} y_p(x) &= Ax^3 + Bx^2 + Cx + D \\ y'_p(x) &= 3Ax^2+ 2Bx + C\\ y''_p(x) &= 6Ax + 2B. \end{align}\]

Substituting, we have

\[\begin{align} y''_p + 4y_p &= (6Ax + 2B) + 4\left(Ax^3 + Bx^2 + Cx + D\right) \\ &= 4Ax^3 + 4Bx^2 + (6A +4C)x + (2B + D) \\ &= 3x^3 \\ \\ \Rightarrow 4A &= 3, ~4B = 0 , ~6A + 4C = 0, ~2B + D = 0 \\ A &= \frac {3}{4}, ~B = 0, ~C = \frac{-9}{8}, ~D = 0\\ \\ \Rightarrow y_p(x) &= \frac{3}{4}x^3 - \frac{9}{8}x. \ _\square \end{align}\]

## Solve \[\]

\[\begin{array} &y'' - 3y' + 8y = 3e^{-x} - 10 \cos 3x, &y(0) = 1, &y'(0) = 2.\end{array}\]

The characteristic equation \(r^2 - 3r + 2 = 0\) has the roots \(r=1\) and \(r=2.\) Then the complementary function is \[y_c = c_1 e^{x} + c_2 e^{2x}.\] Let's seek a particular solution \[\begin{align} y_p &= Ae^{-x} + B\cos 3x + C\sin 3x\\ y'_p &= -Ae^{-x} - 3B\sin 3x + 3C\cos 3x\\ y''_p &= Ae^{-x} - 9B\cos 3x - 9C\sin 3x. \end{align}\] Substituing, we have \[\begin{align} y''_p - 3y'_p + 2y_p &=6Ae^{-x} + (-7B - 9C)\cos 3x + (9B - 7C)\sin 3x\\ &= 3e^{-x} - 10\cos 3x\\ \\ \Rightarrow 6A &= 3, ~-7B - 9C = -10, ~9B - 7C = 0 \\ A &= \frac{1}{2}, ~B = \frac{7}{13} , ~C = \frac{9}{13}\\ \\ \Rightarrow y_p(x) &= \frac{1}{2}e^{-x} + \frac{7}{13}\cos 3x + \frac{9}{13}\sin 3x. \end{align}\] Then the general solution is \[y(x) = y_c(x) + y_p(x) = c_1e^{x} + c_2e^{2x} + \frac{1}{2}e^{-x} + \frac{7}{13}\cos 3x + \frac{9}{13}\sin 3x.\] The initial conditions lead to \[\begin{align} y(0) &= c_1 + c_2 + \frac{1}{2} + \frac{7}{12} = 1\\ y'(0) &= c_1 + 2c_2 - \frac{1}{2} + \frac{27}{13} = 2 \\ \Rightarrow c_1 &= - \frac{1}{2}, ~c_2 = \frac{6}{13}. \end{align}\] Therefore, the solution sought is \[y(x) = - \frac{1}{2}e^{x} + \frac{6}{13}e^{2x} + \frac{1}{2}e^{-x} + \frac{7}{13}\cos 3x + \frac{9}{13}\sin 3x. \ _\square\]

References

Editorial: PHH *Ecuaciones diferenciales elementales y problemas con condiciones en la frontera* authors: C.H Edwards, David Penney

**Cite as:**Differential Equations - Homogeneous Equations with Constant Coefficients.

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