# Higher Order Linear Nonhomogeneous Differential Equations with Constant Coefficients

These equations have the form

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

where $${a_1},{a_2}, \ldots ,{a_n}$$ are real or complex numbers, and the right-hand side $$f\left( x \right)$$ is a continuous function on some interval $$\left[ {a,b} \right].$$

Using the linear differential operator $$L\left( D \right)$$ equal to

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

the nonhomogeneous differential equation can be written as

$L\left( D \right)y\left( x \right) = f\left( x \right).$

The general solution $$y\left( x \right)$$ of the nonhomogeneous equation is the sum of the general solution $${y_0}\left( x \right)$$ of the corresponding homogeneous equation and a particular solution $${y_1}\left( x \right)$$ of the nonhomogeneous equation:

$y\left( x \right) = {y_0}\left( x \right) + {y_1}\left( x \right).$

For an arbitrary right side $$f\left( x \right)$$, the general solution of the nonhomogeneous equation can be found using the method of variation of parameters. If the right-hand side is the product of a polynomial and exponential functions, it is more convenient to seek a particular solution by the method of undetermined coefficients.

### Method of Variation of Parameters

We assume that the general solution of the homogeneous differential equation of the $$n$$th order is known and given by

${{y_0}\left( x \right) }={ {C_1}{Y_1}\left( x \right) }+{ {C_2}{Y_2}\left( x \right) + \cdots } + {{C_n}{Y_n}\left( x \right).}$

According to the method of variation of constants (or Lagrange method), we consider the functions $${C_1}\left( x \right),$$ $${C_2}\left( x \right), \ldots ,$$ $${C_n}\left( x \right)$$ instead of the regular numbers $${C_1},$$ $${C_2}, \ldots ,$$ $${C_n}.$$ These functions are chosen so that the solution

${y = {C_1}\left( x \right){Y_1}\left( x \right) }+{ {C_2}\left( x \right){Y_2}\left( x \right) + \cdots } + {{C_n}\left( x \right){Y_n}\left( x \right)}$

satisfies the original nonhomogeneous equation.

The derivatives of $$n$$ unknown functions $${C_1}\left( x \right),$$ $${C_2}\left( x \right), \ldots ,$$ $${C_n}\left( x \right)$$ are determined from the system of $$n$$ equations:

$\left\{ \begin{array}{l} {{C’_1}\left( x \right){Y_1}\left( x \right) }+{ {C’_2}\left( x \right){Y_2}\left( x \right) + \cdots }+{ {C’_n}\left( x \right){Y_n}\left( x \right) = 0}\\ {{C’_1}\left( x \right){Y’_1}\left( x \right) }+{ {C’_2}\left( x \right){Y’_2}\left( x \right) + \cdots }+{ {C’_n}\left( x \right){Y’_n}\left( x \right) = 0}\\ \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\ {{C’_1}\left( x \right)Y_1^{\left( {n – 1} \right)}\left( x \right) }+{ {C’_2}\left( x \right)Y_2^{\left( {n – 1} \right)}\left( x \right) + \cdots }+{ {C’_n}\left( x \right)Y_n^{\left( {n – 1} \right)}\left( x \right) }={ f\left( x \right)} \end{array} \right.$

The determinant of this system is the Wronskian of $${Y_1},$$ $${Y_2}, \ldots ,$$ $${Y_n}$$ forming a fundamental system of solutions. By the linear independence of these functions, the determinant is not zero and the system is uniquely solvable. The final expressions for the functions $${C_1}\left( x \right),$$ $${C_2}\left( x \right), \ldots ,$$ $${C_n}\left( x \right)$$ can be found by integration.

### Method of Undetermined Coefficients

If the right-hand side $$f\left( x \right)$$ of the differential equation is a function of the form

${{P_n}\left( x \right){e^{\alpha x}}\;\;\text{or}\;\;}\kern-0.3pt {\left[ {{P_n}\left( x \right)\cos \beta x }\right.}+{\left.{ {Q_m}\left( x \right)\sin\beta x} \right]{e^{\alpha x}},}$

where $${P_n}\left( x \right),$$ $${Q_m}\left( x \right)$$ are polynomials of degree $$n$$ and $$m,$$ respectively, then the method of undetermined coefficients may be used to find a particular solution.

In this case, we seek a particular solution in the form corresponding to the structure of the right-hand side of the equation. For example, if the function has the form

$f\left( x \right) = {P_n}\left( x \right){e^{\alpha x}},$

the particular solution is given by

${y_1}\left( x \right) = {x^s}{A_n}\left( x \right){e^{\alpha x}},$

where $${A_n}\left( x \right)$$ is a polynomial of the same degree $$n$$ as $${P_n}\left( x \right).$$ The coefficients of the polynomial $${A_n}\left( x \right)$$ are determined by direct substitution of the trial solution $${y_1}\left( x \right)$$ in the nonhomogeneous differential equation.

In the so-called resonance case, when the number of $$\alpha$$ in the exponential function coincides with a root of the characteristic equation, an additional factor $${x^s},$$ where s is the multiplicity of the root, appears in the particular solution. In the non-resonance case, we set $$s = 0.$$

The same algorithm is used when the right-hand side of the equation is given in the form

${f\left( x \right) }={ \left[ {{P_n}\left( x \right)\cos \beta x }\right.}+{\left.{ {Q_m}\left( x \right)\sin\beta x} \right]{e^{\alpha x}}.}$

Here the particular solution has a similar structure and can be written as

${{y_1}\left( x \right) }={ {x^s}\left[ {{A_n}\left( x \right)\cos \beta x }\right.}+{\left.{ {B_n}\left( x \right)\sin\beta x} \right]{e^{\alpha x}},}$

where $${{A_n}\left( x \right)},$$ $${{B_n}\left( x \right)}$$ are polynomials of degree $$n$$ (for $$n \ge m$$), and the degree $$s$$ in the additional factor $${x^s}$$ is equal to the multiplicity of the complex root $$\alpha \pm \beta i$$ in the resonance case (i.e. when the numbers $$\alpha$$ and $$\beta$$ coincide with the complex root of the characteristic equation), and accordingly, $$s = 0$$ in the non-resonance case.

### Superposition Principle

The superposition principle is stated as follows. Let the right-hand side $$f\left( x \right)$$ be the sum of two functions:

$f\left( x \right) = {f_1}\left( x \right) + {f_2}\left( x \right).$

Suppose that $${y_1}\left( x \right)$$ is a solution of the equation

$L\left( D \right)y\left( x \right) = {f_1}\left( x \right),$

and the function $${y_2}\left( x \right)$$ is, accordingly, a solution of the second equation

$L\left( D \right)y\left( x \right) = {f_2}\left( x \right).$

Then the sum of the functions

$y\left( x \right) = {y_1}\left( x \right) + {y_2}\left( x \right)$

will be a solution of the linear nonhomogeneous equation

${L\left( D \right)y\left( x \right) = f\left( x \right) } = {{f_1}\left( x \right) + {f_2}\left( x \right).}$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Find the general solution of the differential equation $${y^{\prime\prime\prime} + 3y^{\prime\prime} – 10y’ }={ x – 3.}$$

### Example 2

Solve the differential equation $$y^{\prime\prime\prime} – y’ = \sin 3x.$$

### Example 3

Solve the differential equation $${y^{IV}} – y = 2\cos x.$$

### Example 4

Solve the equation $${{y^{IV}} + y^{\prime\prime\prime} – 3y^{\prime\prime} }-{ 5y’ – 2y }={ {e^{2x}} – {e^{ – x}}.}$$

### Example 5

Find the general solution of the equation $$y^{\prime\prime\prime} + y’ = {\large\frac{1}{{\cos x}}\normalsize}$$ using the method of variation of constants.
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Concept
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Problems 1-5