# 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

$1.\;f\left( x \right) = {P_n}\left( x \right){e^{\alpha x}};$
$2.\;f\left( x \right) = \left[ {{P_n}\left( x \right)\cos\left( {\beta x} \right) + {Q_m}\left( x \right)\sin\left( {\beta x} \right)} \right]{e^{\alpha x}},$

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

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

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 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:

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

### Example 2.

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

Solution.

We will use the method of undetermined coefficients. The right side of the given equation is a linear function $$f\left( x \right) = ax + b.$$ Therefore, we will look for a particular solution in the form

${y_1} = Ax + B.$

Then the derivatives are

${y'_1} = A,\;\;{y^{\prime\prime}_1} = 0.$

Substituting this in the differential equation gives:

$0 + A - 6\left( {Ax + B} \right) = 36x,\;\; \Rightarrow A - 6Ax - 6B = 36x.$

The last equation must be valid for all values of $$x,$$ so the coefficients with the like powers of $$x$$ in the right and left sides must be identical:

$\left\{ \begin{array}{l} - 6A = 36\\ A - 6B = 0 \end{array} \right..$

We find from this system that $$A = -6,$$ $$B = -1.$$ As a result, the particular solution is written as

${y_1} = - 6x - 1.$

Now we find the general solution of the homogeneous differential equation. Calculate the roots of the auxiliary characteristic equation:

${k^2} + k - 6 = 0,\;\; \Rightarrow D = 1 - 4 \cdot \left( { - 6} \right) = 25,\;\; \Rightarrow {k_{1,2}} = \frac{{ - 1 \pm \sqrt {25} }}{2} = \frac{{ - 1 \pm 5}}{2} = - 3,2.$

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

${y_0}\left( x \right) = {C_1}{e^{ - 3x}} + {C_2}{e^{2x}}.$

Thus, the general solution of the initial nonhomogeneous equation is expressed by the formula

$y = {y_0} + {y_1} = {C_1}{e^{ - 3x}} + {C_2}{e^{2x}} - 6x - 1.$

See more problems on Page 2.