Higher Order Linear Nonhomogeneous Differential Equations with Variable Coefficients
Solved Problems
Example 1.
Find the general solution of the differential equation
\[\left( {{x^2} - 2} \right)y^{\prime\prime\prime} - 2xy^{\prime\prime} - \left( {{x^2} - 2} \right)y' + 2xy = 2x - \frac{4}{y}.\]
Solution.
First we find the general solution of the homogeneous equation
\[\left( {{x^2} - 2} \right)y^{\prime\prime\prime} - 2xy^{\prime\prime} - \left( {{x^2} - 2} \right)y' + 2xy = 0.\]
We use the symmetry of the equation and introduce the new variable
\[v = y^{\prime\prime} - y.\]
Then the equation becomes:
\[\left( {{x^2} - 2} \right)v' - 2xv = 0.\]
The resulting equation can be easily solved by separation of variables:
\[\left( {{x^2} - 2} \right)\frac{{dv}}{{dx}} = 2xv,\;\; \Rightarrow
\frac{{dv}}{v} = \frac{{2xdx}}{{{x^2} - 2}},\;\; \Rightarrow
\int {\frac{{dv}}{v}} = \int {\frac{{2xdx}}{{{x^2} - 2}}} ,\;\; \Rightarrow
\int {\frac{{dv}}{v}} = \int {\frac{{d\left( {{x^2} - 2} \right)}}{{{x^2} - 2}}} ,\;\; \Rightarrow
\ln \left| v \right| = \ln \left| {{x^2} - 2} \right| + \ln {B_1}\;\left( {{B_1} \gt 0} \right),\;\; \Rightarrow
\ln \left| v \right| = \ln \left( {{B_1}\left| {{x^2} - 2} \right|} \right),\;\; \Rightarrow
\left| v \right| = {B_1}\left| {{x^2} - 2} \right|,\;\; \Rightarrow
v = {B_2}\left( {{x^2} - 2} \right),\]
where \({B_2}\) ia an arbitrary number.
We now find the function \(y\left( x \right):\)
\[y^{\prime\prime} - y = v,\;\; \Rightarrow y^{\prime\prime} - y = {B_2}\left( {{x^2} - 2} \right).\]
We have obtained a nonhomogeneous equation of order \(2.\) The solution of the corresponding homogeneous equation is given by
\[y^{\prime\prime} - y = 0,\;\; \Rightarrow
{\lambda ^2} - 1 = 0,\;\; \Rightarrow
{\lambda _{1,2}} = \pm 1,\;\; \Rightarrow
{y_0}\left( x \right) = {C_1}{e^x} + {C_2}{e^{ - x}}.\]
Given that the right side \({B_2}\left( {{x^2} - 2} \right)\) is a quadratic polynomial, we seek a particular solution in the form
\[{y_1} = D{x^2} + Ex + F.\]
We substitute this function and its derivatives
\[{y'_1} = 2Dx + E,\;\; {y^{\prime\prime}_1} = 2D\]
in our nonhomogeneous equation and find the coefficients \(D, E, F:\)
\[2D - \left( {D{x^2} + Ex + F} \right) = {B_2}{x^2} - 2{B_2},\;\; \Rightarrow
2D - D{x^2} - Ex - F = {B_2}{x^2} - 2{B_2}.\]
Consequently,
\[\left\{ \begin{array}{l}
- D = {B_2}\\
- E = 0\\
2D - F = - 2{B_2}
\end{array} \right.,\;\; \Rightarrow
\left\{ \begin{array}{l}
D = - {B_2}\\
E = 0\\
F = 0
\end{array} \right..\]
Thus, the particular solution \({y_1}\) is given by
\[{y_1} = - {B_2}{x^2}.\]
Replacing the arbitrary number \( - {B_2}\) with \({C_3},\) we finally obtain the general solution of the homogeneous equation:
\[{y_0}\left( x \right) = {C_1}{e^x} + {C_2}{e^{ - x}} + {C_3}{x^2}.\]
Here, the functions \({Y_1} = {e^x},\) \({Y_2} = {e^{ - x}},\) \({Y_3} = {x^2}\) form a fundamental system of solutions.
Now we find the solution of the nonhomogeneous equation using the method of variation of constants. The general solution is represented as
\[y\left( x \right) = {C_1}\left( x \right){e^x} + {C_2}\left( x \right){e^{ - x}} + {C_3}\left( x \right){x^2},\]
where the derivatives of the unknown functions \({C_1}\left( x \right),\) \({C_2}\left( x \right),\) \({C_3}\left( x \right)\) satisfy the system of equations
\[\left\{ \begin{array}{l}
{C'_1}{e^x} + {C'_2}{e^{ - x}} + {C'_3}{x^2} = 0\\
{C'_1}{e^x} - {C'_2}{e^{ - x}} + 2{C'_3}x = 0\\
{C'_1}{e^x} + {C'_2}{e^{ - x}} + 2{C'_3} = 2x - \frac{4}{x}
\end{array} \right..\]
Calculate the determinants of this system:
\[W = \left| {\begin{array}{*{20}{c}}
{{e^x}}&{{e^{ - x}}}&{{x^2}}\\
{{e^x}}&{ - {e^{ - x}}}&{2x}\\
{{e^x}}&{{e^{ - x}}}&2
\end{array}} \right|
= {e^x}{e^{ - x}}\left| {\begin{array}{*{20}{c}}
1&1&{{x^2}}\\
1&{ - 1}&{2x}\\
1&1&2
\end{array}} \right|
= {1 \cdot \left[ {1\left( { - 2 - 2x} \right) - 1\left( {2 - {x^2}} \right) + 1\left( {2x + {x^2}} \right)} \right]}
= - 2 - \cancel{2x} - 2 + {x^2} + \cancel{2x} + {x^2}
= 2{x^2} - 4;\]
\[{W_1} = \left| {\begin{array}{*{20}{c}}
0&{{e^{ - x}}}&{{x^2}}\\
0&{ - {e^{ - x}}}&{2x}\\
{2x - \frac{4}{x}}&{{e^{ - x}}}&2
\end{array}} \right|
= \left( {2x - \frac{x}{4}} \right) \left( {2x{e^{ - x}} + {x^2}{e^{ - x}}} \right)
= \left( {2{x^2} - 4} \right)\left( {x + 2} \right){e^{ - x}},\]
\[{W_2} = \left| {\begin{array}{*{20}{c}}
{{e^x}}&0&{{x^2}}\\
{{e^x}}&0&{2x}\\
{{e^x}}&{2x - \frac{4}{x}}&2
\end{array}} \right|
= - \left( {2x - \frac{x}{4}} \right) \left( {2x{e^x} - {x^2}{e^x}} \right)
= \left( {2{x^2} - 4} \right)\left( {x - 2} \right){e^x},\]
\[{W_3} = \left| {\begin{array}{*{20}{c}}
{{e^x}}&{{e^{ - x}}}&0\\
{{e^x}}&{ - {e^{ - x}}}&0\\
{{e^x}}&{{e^{ - x}}}&{2x - \frac{4}{x}}
\end{array}} \right|
= {e^{ - x}}{e^x}\left| {\begin{array}{*{20}{c}}
1&1&0\\
1&{ - 1}&0\\
1&1&{2x - \frac{4}{x}}
\end{array}} \right|
= \left( {2x - \frac{x}{4}} \right)\left( { - 1 - 1} \right)
= - \frac{2}{x}\left( {2{x^2} - 4} \right).\]
Then the derivatives \({C'_1},\) \({C'_2},\) \({C'_3}\) are
\[{C'_1} = \frac{{{W_1}}}{W}
= \frac{{\cancel{\left( {2{x^2} - 4} \right)}\left( {x + 2} \right){e^{ - x}}}}{\cancel{2{x^2} - 4}}
= \left( {x + 2} \right){e^{ - x}},\]
\[{C'_2} = \frac{{{W_2}}}{W}
= \frac{{\cancel{\left( {2{x^2} - 4} \right)}\left( {x - 2} \right){e^{x}}}}{\cancel{2{x^2} - 4}}
= \left( {x - 2} \right){e^{x}},\]
\[{C_3} = \frac{{{W_3}}}{W}
= \frac{{ - \frac{2}{x}\cancel{\left( {2{x^2} - 4} \right)}}}{\cancel{2{x^2} - 4}}
= - \frac{2}{x}.\]
Integrating, we find the functions \({C_1}\left( x \right),\) \({C_2}\left( x \right),\) \({C_3}\left( x \right):\)
\[{C_1}\left( x \right) = \int {\left( {x + 2} \right){e^{ - x}}dx}
= \left[ {\begin{array}{*{20}{l}}
{u = x + 2}\\
{v' = {e^{ - x}}}\\
{u' = 1}\\
{v = - {e^{ - x}}}
\end{array}} \right]
= - \left( {x + 2} \right){e^{ - x}} - \int {\left( { - {e^{ - x}}} \right)dx}
= - \left( {x + 2} \right){e^{ - x}} + \int {{e^{ - x}}dx}
= - \left( {x + 2} \right){e^{ - x}} - {e^{ - x}} + {A_1}
= - \left( {x + 3} \right){e^{ - x}} + {A_1},\]
\[{C_2}\left( x \right) = \int {\left( {x - 2} \right){e^x}dx}
= \left[ {\begin{array}{*{20}{l}}
{u = x - 2}\\
{v' = {e^x}}\\
{u' = 1}\\
{v = {e^x}}
\end{array}} \right]
= \left( {x - 2} \right){e^x} - \int {{e^x}dx}
= \left( {x - 2} \right){e^x} - {e^x} + {A_2}
= \left( {x - 3} \right){e^x} + {A_2},\]
\[{C_3}\left( x \right) = \int {\left( { - \frac{2}{x}} \right)dx}
= - 2\int {\frac{{dx}}{x}}
= - 2\ln \left| x \right| + {A_3}.\]
Now we can write the general solution of the nonhomogeneous equation:
\[y\left( x \right)
= {C_1}\left( x \right){Y_1}\left( x \right) + {C_2}\left( x \right){Y_2}\left( x \right) + {C_3}\left( x \right){Y_3}\left( x \right)
= \left[ { - \left( {x + 3} \right){e^{ - x}} + {A_1}} \right]{e^x}
+ \left[ {\left( {x - 3} \right){e^x} + {A_2}} \right]{e^{ - x}}
+ \left[ { - 2\ln \left| x \right| + {A_3}} \right]{x^2}
= {A_1}{e^x} + {A_2}{e^{ - x}} + {A_3}{x^2}
- \left( {x + 3} \right) + x - 3 - 2{x^2}\ln \left| x \right|
= {A_1}{e^x} + {A_2}{e^{ - x}} + {A_3}{x^2} - 2{x^2}\ln \left| x \right| - 6.\]
