Differential Equations

Second Order Equations

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Second Order Linear Nonhomogeneous Differential Equations with Constant Coefficients

Solved Problems

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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

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