# Differential Equations

## Higher Order Equations # Higher Order Linear Homogeneous Differential Equations with Constant Coefficients

The linear homogeneous differential equation of the $$n$$th order with constant coefficients can be written as

${ {y^{\left( n \right)}}\left( x \right) + {a_1}{y^{\left( {n – 1} \right)}}\left( x \right) + \cdots } + {{a_{n – 1}}y’\left( x \right) + {a_n}y\left( x \right) }={ 0,}$

where $${a_1},{a_2}, \ldots ,{a_n}$$ are constants which may be real or complex.

Using the linear differential operator $$L\left( D \right),$$ this equation can be represented as

$L\left( D \right)y\left( x \right) = 0,$

where

${L\left( D \right) }={ {D^n} + {a_1}{D^{n – 1}} + \cdots }+{ {a_{n – 1}}D + {a_n}.}$

For each differential operator with constant coefficients, we can introduce the characteristic polynomial

${L\left( \lambda \right) }={ {\lambda ^n} + {a_1}{\lambda ^{n – 1}} + \cdots }+{ {a_{n – 1}}\lambda + {a_n}.}$

The algebraic equation

${L\left( \lambda \right) }={ {\lambda ^n} + {a_1}{\lambda ^{n – 1}} + \cdots }+{ {a_{n – 1}}\lambda }+{ {a_n} = 0}$

is called the characteristic equation of the differential equation.

According to the fundamental theorem of algebra, a polynomial of degree $$n$$ has exactly $$n$$ roots, counting multiplicity. In this case the roots can be both real and complex (even if all the coefficients of $${a_1},{a_2}, \ldots ,{a_n}$$ are real).

Let us consider in more detail the different cases of the roots of the characteristic equation and the corresponding formulas for the general solution of differential equations.

### Case $$1.$$ All Roots of the Characteristic Equation are Real and Distinct

We assume that the characteristic equation $$L\left( \lambda \right) = 0$$ has $$n$$ roots $${\lambda _1},{\lambda _2}, \ldots ,{\lambda _n}.$$ In this case the general solution of the differential equation is written in a simple form:

${y\left( x \right) }={ {C_1}{e^{{\lambda _1}x}} + {C_2}{e^{{\lambda _2}x}} + \cdots }+{ {C_n}{e^{{\lambda _n}x}},}$

where $${C_1},{C_2}, \ldots ,{C_n}$$ are constants depending on initial conditions.

### Case $$2.$$ The Roots of the Characteristic Equation are Real and Multiple

Let the characteristic equation $$L\left( \lambda \right) = 0$$ of degree $$n$$ have $$m$$ roots $${\lambda _1},{\lambda _2}, \ldots ,{\lambda _m},$$ the multiplicity of which, respectively, is equal to $${k_1},{k_2}, \ldots ,{k_m}.$$ It is clear that the following condition holds:

${k_1} + {k_2} + \cdots + {k_m} = n.$

Then the general solution of the homogeneous differential equations with constant coefficients has the form

${y\left( x \right) }={ {C_1}{e^{{\lambda _1}x}} + {C_2}x{e^{{\lambda _1}x}} + \cdots } + {{C_{{k_1}}}{x^{{k_1} – 1}}{e^{{\lambda _1}x}} + \cdots } + {{C_{n – {k_m} + 1}}{e^{{\lambda _m}x}} }+{ {C_{n – {k_m} + 2}}x{e^{{\lambda _m}x}} + \cdots } + {{C_n}{x^{{k_m} – 1}}{e^{{\lambda _m}x}}.}$

It is seen that the formula of the general solution has exactly $${k_i}$$ terms corresponding to each root $${\lambda _i}$$ of multiplicity $${k_i}.$$ These terms are formed by multiplying $$x$$ to a certain degree by the exponential function $${e^{{\lambda _i}x}}.$$ The degree of $$x$$ varies in the range from $$0$$ to $${k_i} – 1,$$ where $${k_i}$$ is the multiplicity of the root $${\lambda _i}.$$

### Case $$3.$$ The Roots of the Characteristic Equation are Complex and Distinct

If the coefficients of the differential equation are real numbers, the complex roots of the characteristic equation will be presented in the form of conjugate pairs of complex numbers:

${{\lambda _{1,2}} = \alpha \pm i\beta ,\;\;}\kern-0.3pt{{\lambda _{3,4}} = \gamma \pm i\delta ,\; \ldots }$

In this case the general solution is written as

${y\left( x \right) }={ {e^{\alpha x}}\left( {{C_1}\cos \beta x + {C_2}\sin \beta x} \right) } + {{e^{\gamma x}}\left( {{C_3}\cos \delta x + {C_4}\sin \delta x} \right) + \cdots }$

### Case $$4.$$ The Roots of the Characteristic Equation are Complex and Multiple

Here, each pair of complex conjugate roots $$\alpha \pm i\beta$$ of multiplicity $$k$$ produces $$2k$$ particular solutions:

${{e^{\alpha x}}\cos \beta x,\;}\kern-0.3pt{{e^{\alpha x}}\sin\beta x,\;}\kern-0.3pt {{e^{\alpha x}}x\cos \beta x,\;}\kern-0.3pt{{e^{\alpha x}}x\sin \beta x, \ldots ,\;}\kern-0.3pt {{e^{\alpha x}}{x^{k – 1}}\cos \beta x,\;}\kern-0.3pt{{e^{\alpha x}}{x^{k – 1}}\sin\beta x.}$

Then the part of the general solution of the differential equation corresponding to a given pair of complex conjugate roots is constructed as follows:

${y\left( x \right) }={ {e^{\alpha x}}\left( {{C_1}\cos \beta x + {C_2}\sin \beta x} \right) } + {{x{e^{\alpha x}}\left( {{C_3}\cos \beta x }\right.}+{\left.{ {C_4}\sin \beta x} \right) + \cdots }} + {{{x^{k – 1}}{e^{\alpha x}}\left( {{C_{2k – 1}}\cos \beta x }\right.}+{\left.{ {C_{2k}}\sin \beta x} \right).}}$

In general, when the characteristic equation has both real and complex roots of arbitrary multiplicity, the general solution is constructed as the sum of the above solutions of the form $$1-4.$$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Solve the differential equation $$y^{\prime\prime\prime} + 2y^{\prime\prime} -$$ $$y’ – 2y = 0.$$

### Example 2

Solve the equation $${y^{\prime\prime\prime} – 7y^{\prime\prime} }+{ 11y’ – 5y = 0.}$$

### Example 3

Solve the equation $${y^{IV}} – y^{\prime\prime\prime} + 2y’ = 0.$$

### Example 4

Solve the equation $${y^V} + 18y^{\prime\prime\prime} + 81y’ = 0.$$

### Example 5

Solve the differential equation $${y^{IV}} – 4y^{\prime\prime\prime} + 5y^{\prime\prime}$$ $$-\; 4y’ + 4y = 0.$$

### Example 1.

Solve the differential equation $$y^{\prime\prime\prime} + 2y^{\prime\prime} -$$ $$y’ – 2y = 0.$$

Solution.

Write the corresponding characteristic equation:

${\lambda ^3} + 2{\lambda ^2} – \lambda – 2 = 0.$

Solving it, we find the roots:

${{\lambda ^2}\left( {\lambda + 2} \right) – \left( {\lambda + 2} \right) = 0,\;\;}\Rightarrow {\left( {\lambda + 2} \right)\left( {{\lambda ^2} – 1} \right) = 0,\;\;}\Rightarrow {{\left( {\lambda + 2} \right)\left( {\lambda – 1} \right)\left( {\lambda + 1} \right) = 0,\;\;}}\Rightarrow {{\lambda _1} = – 2,\;\;}\kern-0.3pt{{\lambda _2} = 1,\;\;}\kern-0.3pt{{\lambda _3} = – 1.}$

It is seen that all three roots are real. Therefore, the general solution of the differential equations can be written as

${y\left( x \right) }={ {C_1}{e^{ – 2x}} + {C_2}{e^x} }+{ {C_3}{e^{ – x}},}$

where $${C_1},$$ $${C_2},$$ $${C_3}$$ are arbitrary constants.

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Problem 1
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Problems 2-5