# Second Order Linear Nonhomogeneous Differential Equations with Constant Coefficients

### Structure of the General Solution

The nonhomogeneous differential equation of this type has the form

${y^{\prime\prime} + py’ + qy }={ f\left( x \right),}$

where $$p, q$$ are constant numbers (that can be both as real as complex numbers). For each equation we can write the related homogeneous or complementary equation:

${y^{\prime\prime} + py’ + qy }={ 0.}$

#### Theorem.

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

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

Below we consider two methods of constructing the general solution of a nonhomogeneous differential equation.

### Method of Variation of Constants

If the general solution $${y_0}$$ of the associated homogeneous equation is known, then the general solution for the nonhomogeneous equation can be found by using the method of variation of constants.

Let the general solution of a second order homogeneous differential equation be

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

Instead of the constants $${C_1}$$ and $${C_2}$$ we will consider arbitrary functions $${C_1}\left( x \right)$$ and $${C_2}\left( x \right).$$ We will find these functions such that the solution

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

satisfies the nonhomogeneous equation with the right side $$f\left( x \right).$$

The unknown functions $${C_1}\left( x \right)$$ and $${C_2}\left( x \right)$$ can be determined from the system of two equations:

$\left\{ \begin{array}{l} {C’_1}\left( x \right){Y_1}\left( x \right) + {C’_2} \left( x \right){Y_2}\left( x \right) = 0\\ {C’_1} \left( x \right){Y’_1} \left( x \right) + {C’_2} \left( x \right){Y’_2} \left( x \right) = f\left( x \right) \end{array} \right.$

### Method of Undetermined Coefficients

The right side $$f\left( x \right)$$ of a nonhomogeneous differential equation is often an exponential, polynomial or trigonometric function or a combination of these functions. In this case, it’s more convenient to look for a solution of such an equation using the method of undetermined coefficients.

The given method works only for a restricted class of functions in the right side, such as

1. $$f\left( x \right)$$ $$= {P_n}\left( x \right){e^{\alpha x}};$$
2. $$f\left( x \right)$$ $$= \left[ {{P_n}\left( x \right)\cos\left( {\beta x} \right) }\right.$$ $$\left.{+ {Q_m}\left( x \right)\sin\left( {\beta x} \right)} \right]{e^{\alpha x}},$$
3. where $${{P_n}\left( x \right)}$$ and $${{Q_m}\left( x \right)}$$ are polynomials of degree $$n$$ and $$m,$$ respectively.

In both cases, a choice for the particular solution should match the structure of the right side of the nonhomogeneous equation.

In case $$1,$$ if the power $$\alpha$$ of the exponential function coincides with a root of the auxiliary characteristic equation, the particular solution will contain the additional factor $${x^s},$$ where $$s$$ is the order of the root $$\alpha$$ in the characteristic equation.

In case $$2,$$ if the number $$\alpha + \beta i$$ coincides with a root of the characteristic equation, the expected expression for the particular solution should be multiplied by the additional factor $$x.$$

The unknown coefficients can be determined by substitution of the expected type of the particular solution into the original nonhomogeneous differential equation.

### Superposition Principle

If the right side of a nonhomogeneous equation is the sum of several functions of kind

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

then a particular solution of the differential equation is also the sum of particular solutions constructed separately for each term in the right side.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Solve the differential equation $$y^{\prime\prime} + y = \sin \left( {2x} \right).$$

### Example 2

Find the general solution of the equation $$y^{\prime\prime} + y’ – 6y$$ $$= 36x.$$

### Example 3

Solve the differential equation $$y^{\prime\prime} – 5y’ + 4y$$ $$= {e^{4x}}.$$

### Example 4

Find the general solution of the equation $$y^{\prime\prime} + 9y$$ $$= 2{x^2} – 5.$$

### Example 5

Solve the differential equation $$y^{\prime\prime} + 16y$$ $$= 2{\cos ^2}x.$$

### Example 6

Solve the equation $$y^{\prime\prime} + y = {\sec ^2}x$$ using the method of variation of constants.

### Example 7

Find the solution of the differential equation $$y^{\prime\prime} – 7y’ + 12y$$ $$= 8\sin x + {e^{3x}}.$$

### Example 1.

Solve the differential equation $$y^{\prime\prime} + y = \sin \left( {2x} \right).$$

Solution.

First we solve the related homogeneous equation $$y^{\prime\prime} + y = 0.$$ The roots of the corresponding characteristic equation are purely imaginary:

${{k^2} + 1 = 0,\;\;}\Rightarrow {{k^2} = – 1,\;\;}\Rightarrow {{k_{1,2}} = \pm i.}$

Hence, the general solution of the homogeneous equation is given by

${{y_0}\left( x \right) }={ {C_1}\cos x }+{ {C_2}\sin x.}$

Let’s go back to the nonhomogeneous equation. We will seek for its solution in the form

${y\left( x \right) }={ {C_1}\left( x \right)\cos x }+{ {C_2}\left( x \right)\sin x,}$

using the method of variation of constants.

The functions $${C_1}\left( x \right)$$ and $${C_2}\left( x \right)$$ can be determined from the following system of equations:

$\left\{ \begin{array}{l} {{C’_1}\left( x \right)\cos x} + {{C’_2}\left( x \right)\sin x} = 0\\ {C’_1}\left( x \right){\left( {\cos x} \right)^\prime } + {C’_2}\left( x \right){\left( {\sin x} \right)^\prime } = {\sin 2x} \end{array} \right.$

Then

$\left\{ \begin{array}{l} {{C’_1}\left( x \right)\cos x} + {{C’_2}\left( x \right)\sin x} = 0\\ {{C’_1}\left( x \right)\left( { – \sin x} \right)} + {{C’_2}\left( x \right)\cos x }= {\sin 2x} \end{array} \right.$

We can express the derivative $${C’_1}\left( x \right)$$ from the first equation:

${C’_1}\left( x \right) = – {C’_2}\left( x \right)\frac{{\sin x}}{{\cos x}}.$

Substituting this in the second equation, we find the derivative $${C’_2}\left( x \right):$$

${{\left( { – {C’_2}\left( x \right)\frac{{\sin x}}{{\cos x}}} \right)\left( { – \sin x} \right) }+{ {C’_2}\left( x \right)\cos x = \sin 2x,\;\;}}\Rightarrow {{ {C’_2}\left( x \right)\left( {\frac{{{{\sin }^2}x}}{{\cos x}} + \cos x} \right) }={ \sin 2x,\;\;}}\Rightarrow {{ {C’_2}\left( x \right)\frac{{{{\sin }^2}x + {{\cos }^2}x}}{{\cos x}} }={ \sin 2x,\;\;}}\Rightarrow { {C’_2}\left( x \right)\frac{1}{{\cos x}} = \sin 2x,\;\;}\Rightarrow { {C’_2}\left( x \right) = \sin 2x\cos x.}$

It follows from here that

${{C’_1}\left( x \right) }={ – \sin 2x\cos x \cdot \frac{{\sin x}}{{\cos x}} } = { – \sin 2x\sin x.}$

Integrating the expressions for the derivatives $${C’_1}\left( x \right)$$ and $${C’_2}\left( x \right),$$ gives

${{C_1}\left( x \right) }={ \int {\left( { – \sin 2x\sin x} \right)dx} } = { – 2\int {{{\sin }^2}x\cos xdx} } = { – 2\int {{{\sin }^2}xd\left( {\sin x} \right)} } = { – 2 \cdot \frac{{{{\sin }^3}x}}{3} + {A_1} } = { – \frac{2}{3}{\sin ^3}x + {A_1},}$

${{C_2}\left( x \right) }={ \int {\left( {\sin 2x\cos x} \right)dx} } = {2\int {\sin x\,{{\cos }^2}xdx} } = { – 2\int {{\cos^2}xd\left( {\cos x} \right)} } = { – 2 \cdot \frac{{{\cos^3}x}}{3} + {A_2} } = { – \frac{2}{3}{\cos^3}x + {A_2}.}$

where $${A_1},$$ $${A_2}$$ are constants of integration.

Now we substitute the found functions $${C_1}\left( x \right)$$ and $${C_2}\left( x \right)$$ into the formula for $${y_1}\left( x \right)$$ and write the general solution of the nonhomogeneous equation:

${y\left( x \right) }={ {C_1}\left( x \right)\cos x + {C_2}\left( x \right)\sin x } = {{\left( { – \frac{2}{3}{\sin^3}x + {A_1}} \right)\cos x }}+{{ \left( { – \frac{2}{3}{\cos^3}x + {A_2}} \right)\sin x }} = {{{A_1}\cos x + {A_2}\sin x }-{ \frac{2}{3}{\sin ^3}x\cos x }}-{{ \frac{2}{3}{\cos^3}x\sin x }} = {{{A_1}\cos x + {A_2}\sin x }}-{{ \frac{2}{3}\sin x\cos x\left( {\underbrace {{{\sin }^2}x + {{\cos }^2}x}_1} \right) }} = {{{A_1}\cos x + {A_2}\sin x }-{ \frac{1}{3} \cdot 2\sin x\cos x }} = {{{A_1}\cos x + {A_2}\sin x }-{ \frac{1}{3}\sin 2x.}}$

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