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16. Homogeneous Linear Equations with Constant Coefficients

Dated: 28-11-2024

We know that

\[\frac{dy}{dx}+my=0\]

where \(m\) being a constant, has the following solution on \((-\infty, \infty)\)

\[y = c_1e^{-mx}\]

Recall

The equation

\[a_{n}(x)\frac{d^{n}y}{dx^{n}}+a_{n-1}(x)\frac{d^{n-1}y}{dx^{n-1}}+…+a_{1}(x)\frac{dy}{dx}+a_{0}(x)y=g(x)\]

for \(n = 2\) becomes

\[a_{2}\frac{d^{2}y}{dx^{2}}+a_{1}\frac{dy}{dx}+a_{0}y=0\]

Which can be re-written as

\[a\frac{d^{2}y}{dx^{2}}+b\frac{dy}{dx}+cy=0\]

Now we try out a solution of exponential form

\[y = e^{mx}\]

Then

\[y^\prime = me^{mx}\]
\[y^{\prime\prime} = m^2e^{mx}\]

Substituting into original equation, we have

\[e^{mx}(am^{2}+bm+c)=0\]
\[\because e^{mx}\neq 0, \forall x\in (-\infty,\infty)\]
\[\therefore am^{2}+bm+c=0\]

This algebraic equation is known as auxiliary equation.

Case 1: Distinct Real Roots

If the auxiliary equation has distinct roots, \(m_1\) and \(m_2\) then the solutions are

\[y_1 = e^{m_1x}\]
\[y_2 = e^{m_2x}\]

These solutions are linearly independent1

\[W(y_1, y_2) = \begin{vmatrix} y_1 & y_2\\ y_1^\prime & y_2^\prime \end{vmatrix} \]
\[= (m_2 - m_1) e^{(m_1 + m_2)x}\]

Since \(m_1 \ne m_2\) and \(e^{(m_1 + m_2)} \ne 0\)
Therefore,

\[W(y_1, y_2) \ne 0, \forall x \in (- \infty, \infty)\]
\[y = c_1e^{m_1x} + c_2e^{m_2x}\]

Case 2: Repeated Roots

if we have real roots such that \(m_1 = m_2\) then we have only one solution

\[y = c_1 e^{mx}\]

We can re-write the equation to the following form

\[y''+\frac{b}{a}y'+\frac{c}{a}y=0\]
\[y''+Py'+Qy=0\]
\[P=\frac{b}{a}\]

Therefore, the second solution is

\[y_2=y_1\int\frac{e^{-\int Pdx}}{(y_1)^2}dx=e^{mx}\int\frac{e^{-\frac{b}{a}x}}{e^{2mx}}dx\]
\[\because \text{Disc} = b^2 - 4ac = 0\]
\[m=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\]
\[2m=-\frac{b}{a}\]

Therefore

\[y_2=e^{mx}\int\frac{e^{2mx}}{e^{2mx}}dx=xe^{mx}\]

Hence the general solution2 is

\[y=c_1e^{mx}+c_2xe^{mx}=(c_1+c_2x)e^{mx}\]

Case 3: Complex Roots

If the auxiliary equation has the complex roots \(\alpha \pm \iota \beta\) then

\[m_1 = \alpha + \iota \beta\]
\[m_2 = \alpha - \iota \beta\]

Where \(\alpha, \beta > 0\) and are real.

\[y=c_{1}e^{(\alpha+i\beta)x}+c_{2}e^{(\alpha-i\beta)x}\]

Choosing the following pairs for \((c_1, c_2)\)

\[(1, 1)\]
\[(1, -1)\]

Thus we have

\[y_{1}=e^{(\alpha+i\beta)x}+e^{(\alpha-i\beta)x}\]
\[y_{2}=e^{(\alpha+i\beta)x}-e^{(\alpha-i\beta)x}\]

By Euler's Formula

\[e^{\iota \theta} = \cos (\theta) + \iota \sin(\theta), \theta \in \mathbb R\]
\[y_1=e^{\alpha x}(e^{i\beta x}+e^{-i\beta x})=2e^{\alpha x}\cos \beta x\]
\[y_2=e^{\alpha x}(e^{i\beta x}-e^{-i\beta x})=2ie^{\alpha x}\sin \beta x\]

We can drop the constants to write

\[y_1 = e^{\alpha x} \cos (\beta x)\]
\[y_2 = e^{\alpha x} \sin (\beta x)\]

The Wronskian

\[W(e^{\alpha x}\cos \beta x, e^{\alpha x}\sin \beta x)=\beta e^{2\alpha x}\neq 0 \quad \forall x\]

Therefore the following form a fundamental solution2 of the differential equation \((- \infty, \infty)\).

\[e^{\alpha x}\cos \beta x, e^{\alpha x}\sin \beta x\]

Therefore, the general solution2 is

\[y=c_1e^{\alpha x}\cos \beta x+c_2e^{\alpha x}\sin \beta x\]
\[y=e^{\alpha x}(c_1\cos \beta x+c_2\sin \beta x)\]

Example

\[y^{\prime\prime} - 10y^{\prime}+25y=0\]

Solution

We put

\[y=e^{mx}\]
\[y^{\prime}=me^{mx}, y^{\prime\prime}=m^2e^{mx}\]

Substituting in the given differential equation, we have

\[(m^2 - 10m + 25)e^{mx} = 0\]
\[\because e^{mx} \ne 0, \forall x\]
\[m^2-10m+25=0\]
\[(m-5)^2=0\Rightarrow m=5,5\]

Hence,

\[y=c_1e^{5x}+c_2xe^{5x}\]
\[y=(c_1+c_2x)e^{5x}\]

Higher order Equations

If we consider the following \(nth\) order homogeneous3 linear differential equation.4

\[a_{n}\frac{d^{n}y}{dx^{n}}+a_{n-1}\frac{d^{n-1}y}{dx^{n-1}}+\ldots+a_{1}\frac{dy}{dx}+a_{0}y=0\]

then the auxiliary equation is

\[a_{n}m^{n}+a_{n-1}m^{n-1}+\ldots+a_{1}m+a_{0}=0\]

Case 1: Real Distinct Roots

\[y=c_{1}e^{m_{1}x}+c_{2}e^{m_{2}x}+\ldots+c_{n}e^{m_{n}x}\]

Case 2: Real and Repeated Roots

\[y=c_{1}e^{m_{1}x}+c_{2}xe^{m_{1}x}+\dots+c_{n}x^{n-1}e^{m_{1}x}\]

where

\[e^{m_1x}, xe^{m_1x}, \ldots, x^{n-1}e^{m_1x}\]

are linearly independent.4

Case 3: Complex Roots

  • General solution when all roots are complex \(n = 6\):
\[\begin{align} y &= e^{\alpha_1 x}(c_1 \cos \beta_1 x + c_2 \sin \beta_1 x)\\ &+ e^{\alpha_2 x}(c_3 \cos \beta_2 x + c_4 \sin \beta_2 x)\\ &+ e^{\alpha_3 x}(c_5 \cos \beta_3 x + c_6 \sin \beta_3 x) \end{align} \]
  • General solution when 4 roots are complex and 2 are real \(n = 6\):
\[ \begin{align} y &= e^{\alpha_1 x}(c_1 \cos \beta_1 x + c_2 \sin \beta_1 x)\\ &+ e^{\alpha_2 x}(c_3 \cos \beta_2 x + c_4 \sin \beta_2 x)\\ &+ c_5 e^{m x} + c_6 x e^{m x} \end{align} \]
  • Complex root of multiplicity \(k\):
    If \(m_1 = \alpha + \iota\beta\) is a root of multiplicity \(k\), its conjugate \(m_2 = \alpha - \iota\beta\) is also of multiplicity \(k\).
  • General solution involves:

\(\(e^{(\alpha + i\beta)x}, \, x e^{(\alpha + i\beta)x}, \, x^2 e^{(\alpha + i\beta)x}, \, \dots, x^{k-1} e^{(\alpha + i\beta)x}\)\)

\[e^{(\alpha - i\beta)x}, \, x e^{(\alpha - i\beta)x}, \, x^2 e^{(\alpha - i\beta)x}, \, \dots, x^{k-1} e^{(\alpha - i\beta)x}\]
  • General solution (using Euler's formula):
\[y = e^{\alpha x} \left[ \sum_{n=0}^{k-1} c_n x^n \cos \beta x + \sum_{n=0}^{k-1} d_n x^n \sin \beta x \right]\]
  • Example for \(k = 3\):
\[y = e^{\alpha x} \left[ (c_1 + c_2 x + c_3 x^2) \cos \beta x + (d_1 + d_2 x + d_3 x^2) \sin \beta x \right]\]

Solving the Auxiliary Equation

Solving the Auxiliary Equation

  1. The auxiliary equation is:

$$ P_n(m) = 0, \quad n > 2 $$

  1. To solve, guess a root \(m_1\). Then, \(m - m_1\) is a factor of \(P_n(m)\).

  2. By synthetic or regular division, factorize:

$$ P_n(m) = (m - m_1) Q(m) $$

  1. Solve the quotient polynomial equation:

$$ Q(m) = 0 $$

  1. If \(m_1 = \frac{p}{q}\) is a rational root of \(P_n(m)\), then:
  2. \(p\) is a factor of \(a_0\) (constant term).

  3. \(q\) is a factor of \(a_n\) (leading coefficient).

  4. Use this fact to construct a list of all possible rational roots and test each using synthetic division.

Example

\[y^{\prime\prime\prime} + 3y^{\prime\prime} - 4y = 0\]

Solution

\[y = e^{mx}\]

Then,

\[y' = me^{mx}, y'' = m^2e^{mx}, \text{ and } y''' = m^3e^{mx}\]

Substituting this in the given differential equation, we have

\[(m^3 + 3m^2 - 4)e^{mx} = 0\]

Since \(e^{mx} \neq 0\), therefore

\[m^3 + 3m^2 - 4 = 0\]

So that the auxiliary equation is

\[m^3 + 3m^2 - 4 = 0\]

Solution of the auxiliary equation:
If we take \(m = 1\), then we see that

\[m^3 + 3m^2 - 4 = 1 + 3 - 4 = 0\]

Therefore \(m = 1\) satisfies the auxiliary equation so that \(m - 1\) is a factor of the polynomial \(m^3 + 3m^2 - 4\).

By synthetic division, we can write

\[m^3 + 3m^2 - 4 = (m - 1)(m^2 + 4m + 4)\]

or

\[m^3 + 3m^2 - 4 = (m - 1)(m + 2)^2\]

Therefore

\[m^3 + 3m^2 - 4 = 0\]
\[\Rightarrow (m - 1)(m + 2)^2 = 0\]
\[\Rightarrow m = 1, -2, -2\]

Hence solution of the differential equation is

\[y = c_1e^x + c_2e^{-2x} + c_3xe^{-2x}\]

References

Read more about notations and symbols.

  1. Read more about linear dependence. ↩

  2. Read more about solutions to higher order differential equations. ↩↩↩

  3. Read more about homogeneous differential equations. ↩

  4. Read more about linear differential equation. ↩↩