# Second Order Linear Nonhomogeneous Differential Equations with Constant Coefficients

## Solved Problems

Click or tap a problem to see the solution.

### 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 3.

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

Solution.

First we solve the related homogeneous equation $$y^{\prime\prime} - 5y' + 4y = 0.$$ The roots of the characteristic equation are

${k^2} - 5k + 4 = 0,\;\; \Rightarrow D = 25 - 4 \cdot 4 = 9,\;\; \Rightarrow {k_{1,2}} = \frac{{5 \pm \sqrt 9 }}{2} = \frac{{5 \pm 3}}{2} = 4,1.$

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

${y_0}\left( x \right) = {C_1}{e^{4x}} + {C_2}{e^x},$

where $${C_1},$$ $${C_2}$$ are constant numbers.

Find a particular solution of the nonhomogeneous differential equation. Notice that the power of the exponential function on the right coincides with the root $${k_1} = 4$$ of the auxiliary characteristic equation. Therefore we will look for a particular solution of the form

${y_1} = Ax{e^{4x}}.$

The derivatives are given by

${y'_1} = {\left( {Ax{e^{4x}}} \right)^\prime } = A{e^{4x}} + 4Ax{e^{4x}} = \left( {A + 4Ax} \right){e^{4x}};$
$y^{\prime\prime}_1 = \left[ {\left( {A + 4Ax} \right){e^{4x}}} \right]^\prime = 4A{e^{4x}} + \left( {4A + 16Ax} \right){e^{4x}} = \left( {8A + 16Ax} \right){e^{4x}}.$

Substituting the function $${y_1}$$ and its derivatives in the differential equation yields:

$\left( {8A + 16Ax} \right){e^{4x}} - 5\left( {A + 4Ax} \right){e^{4x}} + 4Ax{e^{4x}} = {e^{4x}},\;\; \Rightarrow 8A + \cancel{16Ax} - 5A - \cancel{20Ax} + \cancel{4Ax} = 1,\;\; \Rightarrow 3A = 1,\;\; \Rightarrow A = \frac{1}{3}.$

Thus, the particular solution to the differential equation can be written in the form:

${y_1} = \frac{x}{3}{e^{4x}}.$

Now we can write the full solution of the nonhomogeneous equation:

$y = {y_0} + {y_1} = {C_1}{e^{4x}} + {C_2}{e^x} + \frac{x}{3}{e^{4x}}.$

### Example 4.

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

Solution.

First we determine the general solution of the related homogeneous equation. Solve the auxiliary characteristic equation:

${k^2} + 9 = 0,\;\; \Rightarrow {k^2} = - 9,\;\; \Rightarrow {k_{1,2}} = \pm 3i.$

The solution is written in the form:

${y_0}\left( x \right) = {C_1}\cos 3x + {C_2}\sin 3x.$

Now we construct a particular solution. The right-hand side of the given equation is a quadratic function. So we can guess on a particular solution of the same form:

${y_1} = A{x^2} + Bx + C,$

where the numbers $$A, B, C$$ can be determined by the method of undetermined coefficients. Hence, we can write:

${y'_1} = 2Ax,\;\; {y^{\prime\prime}_1} = 2A.$

Substituting this into the original nonhomogeneous differential equation, we have

$2A + 9\left( {A{x^2} + Bx + C} \right) = 2{x^2} - 5,\;\; \Rightarrow 2A + 9A{x^2} + 9Bx + 9C = 2{x^2} - 5.$

By equating the coefficients of like powers of $$x,$$ we obtain:

$\left\{ \begin{array}{l} 9A = 2\\ 9B = 0\\ 2A + 9C = - 5 \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} A = \frac{2}{9}\\ B = 0\\ C = - \frac{{49}}{{81}} \end{array} \right..$

Thus, the particular solution is given by

${y_1} = \frac{2}{9}{x^2} - \frac{{49}}{{81}}.$

Then the general solution of the original nonhomegeneous differential equation is expressed by the formula

$y = {y_0} + {y_1} = {C_1}\cos 3x + {C_2}\sin 3x + \frac{2}{9}{x^2} - \frac{{49}}{{81}}.$

### Example 5.

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

Solution.

First of all we solve the related homogeneous equation. The characteristic equation has roots:

${k^2} + 16 = 0,\;\; \Rightarrow {k^2} = - 16,\;\; \Rightarrow {k_{1,2}} = \pm 4i,$

so the general solution has the form:

${y_0}\left( x \right) = {C_1}\cos 4x + {C_2}\sin 4x.$

Now we find a particular solution for the nonhomogeneous equation. Rewrite the right-hand side as

$2{\cos ^2}x = \cos 2x + 1.$

It follows from here that the particular solution is defined by the function

${y_1} = A\cos 2x + B\sin 2x + C,$

where the numbers $$A, B,$$ and $$C$$ can be calculated using the method of undetermined coefficients. The first and second derivatives of the function $${y_1}$$ are

${y'_1} = - 2A\sin 2x + 2B\cos 2x,$
$y^{\prime\prime}_1 = - 4A\cos 2x - 4B\sin 2x.$

Substituting this back into the differential equation produces:

$- 4A\cos 2x - 4B\sin 2x + 16\left( {A\cos 2x + B\sin 2x + C} \right) = \cos 2x + 1,$
$- 4A\cos 2x - 4B\sin 2x + 16A\cos 2x + 16B\sin 2x + 16C = \cos 2x + 1,$
$12A\cos 2x + 12B\sin 2x + 16C = \cos 2x + 1.$

The latter expression is identical. Therefore we can write the following system of equations to determine the coefficients $$A, B, C:$$

$\left\{ \begin{array}{l} 12A = 1\\ 12B = 0\\ 16C = 1 \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} A = \frac{1}{{12}}\\ B = 0\\ C = \frac{1}{{16}} \end{array} \right..$

Thus, the particular solution has the form:

${y_1} = \frac{1}{{12}}\cos 2x + \frac{1}{{16}}.$

Respectively, the general solution of the original nonhomogeneous equation is written as

$y = {y_0} + {y_1} = {C_1}\cos 4x + {C_2}\sin 4x + \frac{1}{{12}}\cos 2x + \frac{1}{{16}}.$

### Example 6.

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

Solution.

Find the solution for the related homogeneous equation $$y^{\prime\prime} + y = 0.$$ The characteristic equation has roots:

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

Hence, the general solution of the homogeneous equation is

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

Find now the general solution of the original nonhomogeneous equation. According to the method of variation of constants we will consider the coefficients $${C_1}$$ and $${C_2}$$ as functions $${C_1}\left( x \right)$$ and $${C_2}\left( x \right).$$

The derivatives $${C'_1}\left( x \right),$$ $${C'_2}\left( x \right)$$ are defined by the following system of equations:

$\left\{ \begin{array}{l} {{C'_1}\left( x \right)\cos x} + {{C'_2}\left( x \right)\sin x} = 0\\ {{C'_1}\left( x \right){\left( {\cos x} \right)^\prime }} + {{C'_2}\left( x \right){\left( {\sin x} \right)^\prime }} = {{\sec ^2}x} \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} {{C'_1}\left( x \right)\cos x} + {{C'_2}\left( x \right)\sin x} = 0\\ { - {C'_1}\left( x \right)\sin x} + {{C'_2}\left( x \right)\cos x} = {\frac{1}{{{{\cos }^2}x}}} \end{array} \right..$

It follows from the first equation that

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

Substituting this into the second equation, we get

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

Then we have

${C'_1}\left( x \right) = - \frac{1}{{\cos x}} \cdot \frac{{\sin x}}{{\cos x}} = - \frac{{\sin x}}{{{{\cos }^2}x}}.$

Integrate these expressions to find the functions $${C_1}\left( x \right),$$ $${C_2}\left( x \right):$$

${C_1}\left( x \right) = \int {{C'_1}\left( x \right)dx} = \int {\left( { - \frac{{\sin x}}{{{{\cos }^2}x}}} \right)dx} = \int {\frac{{d\left( {\cos x} \right)}}{{{{\cos }^2}x}}} = - \frac{1}{{\cos x}} + {A_1},$
${C_2}\left( x \right) = \int {{C'_2}\left( x \right)dx} = \int {\frac{{dx}}{{\cos x}}} = \int {\frac{{\cos xdx}}{{{{\cos }^2}x}}} = \int {\frac{{d\left( {\sin x} \right)}}{{1 - {\sin^2}x}}} = \frac{1}{2}\ln \left| {\frac{{1 + \sin x}}{{1 - \sin x}}} \right| + {A_2}.$

As a result, the general solution of the nonhomogeneous equation is represented in the form:

$y\left( x \right) = {C_1}\left( x \right)\cos x + {C_2}\left( x \right)\sin x = \left( { - \frac{1}{{\cos x}} + {A_1}} \right)\cos x + \left( {\frac{1}{2}\ln \left| {\frac{{1 + \sin x}}{{1 - \sin x}}} \right| + {A_2}} \right)\sin x = {A_1}\cos x + {A_2}\sin x - 1 + \frac{{\sin x}}{2}\ln \left| {\frac{{1 + \sin x}}{{1 - \sin x}}} \right|,$

where $${A_1}, {A_2}$$ are constant numbers.

### Example 7.

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

Solution.

Consider the related homogeneous equation $$y^{\prime\prime} - 7y' + 12y = 0.$$ The roots of the auxiliary equation are

${k^2} - 7k + 12 = 0,\;\; \Rightarrow D = 49 - 4 \cdot 12 = 1,\;\; \Rightarrow {k_{1,2}} = \frac{{7 \pm 1}}{2} = 4,3.$

Hence, the general solution is given by

${y_0}\left( x \right) = {C_1}{e^{4x}} + {C_2}{e^{3x}}.$

We see that the right-hand side is the sum of two functions. According to the superposition principle, a particular solution is expressed by the formula

${y_1}\left( x \right) = {y_2}\left( x \right) + {y_3}\left( x \right),$

where $${y_2}\left( x \right)$$ is a particular solution for the differential equation $$y^{\prime\prime} - 7y' + 12y = 8\sin x,$$ and $${y_3}\left( x \right)$$ is a particular solution for the equation $$y^{\prime\prime} - 7y' + 12y = {e^{3x}}.$$

First we determine the function $${y_2}\left( x \right).$$ In this case we will be looking for a solution in the form

${y_2}\left( x \right) = A\cos x + B\sin x.$

Substitute the function $${y_2}\left( x \right)$$ and its derivatives

${y'_2}\left( x \right) = - A\sin x + B\cos x,\;\; {y^{\prime\prime}_2}\left( x \right) = - A\cos x - B\sin x$

into the corresponding differential equation:

$y^{\prime\prime} - 7y' + 12y = 8\sin x,\;\; \Rightarrow - A\cos x - B\sin x - 7\left( { - A\sin x + B\cos x} \right) + 12\left( {A\cos x + B\sin x} \right) = 8\sin x,\;\; \Rightarrow - \color{blue}{A\cos x} - \color{red}{B\sin x} + \color{red}{7A\sin x} - \color{blue}{7B\cos x} + \color{blue}{12A\cos x} + \color{red}{12B\sin x} = \color{black}{8\sin x},\;\;\Rightarrow \left( {11A - 7B} \right)\cos x + \left( {11B + 7A} \right)\sin x = 8\sin x.$

Hence,

$\left\{ \begin{array}{l} 11A - 7B = 0\\ 11B + 7A = 8 \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} A = \frac{{28}}{{85}}\\ B = \frac{{44}}{{85}} \end{array} \right..$

Then we get

${y_2}\left( x \right) = {\frac{{28}}{{85}}}\cos x + {\frac{{44}}{{85}}}\sin x.$

Similarly, we can construct a particular solution $${y_3}\left( x \right)$$ for the equation $$y^{\prime\prime} - 7y' + 12y = {e^{3x}}.$$ Notice, that the power of the exponential function coincides with the root $${k_2} = 3$$ of the characteristic equation of the related homogeneous equation. Therefore, we will look for a particular solution in the form:

${y_3}\left( x \right) = Ax{e^{3x}}.$

The derivatives are given by

${y'_3}\left( x \right) = \left( {Ax{e^{3x}}} \right)^\prime = A{e^{3x}} + 3Ax{e^{3x}},$
${y^{\prime\prime}_3}\left( x \right) = \left( {A{e^{3x}} + 3Ax{e^{3x}}} \right)^\prime = 3A{e^{3x}} + 3A{e^{3x}} + 9Ax{e^{3x}} = 6A{e^{3x}} + 9Ax{e^{3x}}.$

Substitute the function $${y_3}\left( x \right)$$ and its derivatives into the differential equation:

$6A{e^{3x}} + 9Ax{e^{3x}} - 7\left( {A{e^{3x}} + 3Ax{e^{3x}}} \right) + 12Ax{e^{3x}} = {e^{3x}},\;\; \Rightarrow \color{blue}{6A{e^{3x}}} + \cancel{\color{red}{9Ax{e^{3x}}}} - \color{blue}{7A{e^{3x}}} - \cancel{\color{red}{21Ax{e^{3x}}}} + \cancel{\color{red}{12Ax{e^{3x}}}} = \color{black}{e^{3x}},\;\; \Rightarrow - A{e^{3x}} = {e^{3x}}.$

As it can be seen, $$A = -1.$$ Hence, the particular solution $${y_3}\left( x \right)$$ has the form:

${y_3}\left( x \right) = - x{e^{3x}}.$

As a result, the general solution of the original nonhomogeneous equation is given by

$y = {y_0} + {y_2} + {y_3} = {C_1}{e^{4x}} + {C_2}{e^{3x}} + \frac{{28}}{{85}}\cos x + \frac{{44}}{{85}}\sin x - x{e^{3x}}.$