Higher Order Linear Homogeneous Differential Equations with Variable Coefficients
Solved Problems
Example 1.
Show that the functions \(x,\) \(\sin x,\) \(\cos x\) are linearly independent.
Solution.
We find the Wronskian matrix \(W\left( x \right)\) for this system of functions:
\[W\left( x \right) = \left| {\begin{array}{*{20}{c}}
x&{\sin x}&{\cos x}\\
1&{\cos x}&{ - \sin x}\\
0&{ - \sin x}&{ - \cos x}
\end{array}} \right|
= x\left| {\begin{array}{*{20}{c}}
{\cos x}&{ - \sin x}\\
{ - \sin x}&{ - \cos x}
\end{array}} \right|
- 1 \cdot \left| {\begin{array}{*{20}{c}}
{\sin x}&{\cos x}\\
{ - \sin x}&{ - \cos x}
\end{array}} \right|
= x\left( { - {{\cos }^2}x - {{\sin }^2}x} \right)
- \left( { - \cancel{\sin x\cos x} + \cancel{\sin x\cos x}} \right)
= - x \ne 0.\]
Since the Wronskian is not identically zero, it follows that the given system of functions is linearly independent.
Example 2.
Show that the functions \(x,{x^2},{x^3},{x^4}\) form a linearly independent system.
Solution.
We compute the corresponding Wronskian:
\[W\left( x \right) = \left| {\begin{array}{*{20}{c}}
x&{{x^2}}&{{x^3}}&{{x^4}}\\
1&{2x}&{3{x^2}}&{4{x^3}}\\
0&2&{6x}&{12{x^2}}\\
0&0&6&{24x}
\end{array}} \right|\begin{array}{*{20}{c}}
{}\\
{\small{{R_1} - x{R_2}}\normalsize}\\
{}\\
{}
\end{array}
= \left( { - \frac{1}{x}} \right) \cdot \left| {\begin{array}{*{20}{c}}
x&{{x^2}}&{{x^3}}&{{x^4}}\\
0&{ - {x^2}}&{ - 2{x^3}}&{ - 3{x^4}}\\
0&2&{6x}&{12{x^2}}\\
0&0&6&{24x}
\end{array}} \right|
= \left( { - \frac{1}{x}} \right) \cdot x \cdot \left( { - 1} \right) \cdot \left| {\begin{array}{*{20}{c}}
{{x^2}}&{2{x^3}}&{3{x^4}}\\
2&{6x}&{12{x^2}}\\
0&6&{24x}
\end{array}} \right|
= {{x^2}\left| {\begin{array}{*{20}{c}}
{6x}&{12{x^2}}\\
6&{24x}
\end{array}} \right|
- 2\left| {\begin{array}{*{20}{c}}
{2{x^3}}&{3{x^4}}\\
6&{24x}
\end{array}} \right|}
= {x^2}\left( {144{x^2} - 72{x^2}} \right)
- 2\left( {48{x^4} - 18{x^4}} \right)
= 12{x^4} \ne 0.\]
As the determinant is not identically equal to zero, these functions are linearly independent.
Example 3.
Make a differential equation, which is determined by the fundamental system of functions \(1,{x^2},{e^x}.\)
Solution.
This equation is written in terms of the determinant as follows:
\[\left| {\begin{array}{*{20}{c}}
1&{{x^2}}&{{e^x}}&y\\
0&{2x}&{{e^x}}&{y'}\\
0&2&{{e^x}}&{y^{\prime\prime}}\\
0&0&{{e^x}}&{y^{\prime\prime\prime}}
\end{array}} \right| = 0,\;\; \Rightarrow
1 \cdot \left| {\begin{array}{*{20}{c}}
{2x}&{{e^x}}&{y'}\\
2&{{e^x}}&{y^{\prime\prime}}\\
0&{{e^x}}&{y^{\prime\prime\prime}}
\end{array}} \right| = 0,\;\; \Rightarrow
2x\left( {{e^x}y^{\prime\prime\prime} - {e^x}y^{\prime\prime}} \right) - 2\left( {{e^x}y^{\prime\prime\prime} - {e^x}y'} \right) = 0,\;\; \Rightarrow
2x{e^x}y^{\prime\prime\prime} - 2x{e^x}y^{\prime\prime\prime} - 2{e^x}y^{\prime\prime\prime} + 2{e^x}y' = 0,\;\; \Rightarrow
2{e^x}\left( {xy^{\prime\prime\prime} - xy^{\prime\prime} - y^{\prime\prime\prime} + y'} \right) = 0,\;\; \Rightarrow
\left( {x - 1} \right)y^{\prime\prime\prime} - xy^{\prime\prime} + y' = 0.\]
Example 4.
Find the general solution of the equation
\[\left( {2x - 3} \right)y^{\prime\prime\prime} - \left( {6x - 7} \right)y^{\prime\prime} + 4xy' - 4y = 0,\]
if the particular solutions \({y_1} = {e^x},\) \({y_2} = {e^{2x}}\) are known.
Solution.
We make the substitution: \(y = {y_1}z = {e^x}z.\) The derivatives will be
\[y' = {\left( {{e^x}z} \right)^\prime } = {e^x}z + {e^x}z' = {e^x}\left( {z + z'} \right),\]
\[y^{\prime\prime} = {\left[ {{e^x}\left( {z + z'} \right)} \right]^\prime } = {e^x}\left( {z + z'} \right) + {e^x}\left( {z' + z^{\prime\prime}} \right) = {e^x}\left( {z + 2z' + z^{\prime\prime}} \right),\]
\[y^{\prime\prime\prime} = {\left[ {{e^x}\left( {z + 2z' + z^{\prime\prime}} \right)} \right]^\prime } = {e^x}\left( {z + 2z' + z^{\prime\prime}} \right) + {e^x}\left( {z' + 2z^{\prime\prime} + z^{\prime\prime\prime}} \right) = {e^x}\left( {z + 3z' + 3z^{\prime\prime} + z^{\prime\prime\prime}} \right).\]
Note that the derivative of the \(n\)th order of the product of two functions \({y_1}z\) can be immediately calculated from the Leibniz formula:
\[{y^{\left( n \right)}}\left( x \right) = {\left( {{y_1}z} \right)^{\left( n \right)}} = \sum\limits_{i = 0}^n {\left[ {C_n^iy_1^{\left( i \right)}{z^{\left( {n - i} \right)}}} \right]} .\]
Substituting the derivatives into the equation and dividing by \({e^x},\) we have
\[{\left( {2x - 3} \right) \left( {z + 3z' + 3z^{\prime\prime} + z^{\prime\prime\prime}} \right) } - \left( {6x - 7} \right)\left( {z + 2z' + z^{\prime\prime}} \right) + 4x\left( {z + z'} \right) - 4z = 0.\]
After simple algebra, the equation becomes:
\[\cancel{\left( {2x - 3} \right)z} + \left( {6x - 9} \right)z'
+ \left( {6x - 9} \right)z^{\prime\prime} + \left( {2x - 3} \right)z^{\prime\prime\prime}
- \cancel{\left( {6x - 7} \right)z} - \left( {12x - 14} \right)z'
- \left( {6x - 7} \right)z^{\prime\prime}
+ \cancel{4xz} + 4xz' - \cancel{4z} = 0,\;\; \Rightarrow
\left( {2x - 3} \right)z^{\prime\prime\prime} - 2z^{\prime\prime} - \left( {2x - 5} \right)z' = 0.\]
By setting \(z' = u,\) we obtain a homogeneous linear second-order equation:
\[\left( {2x - 3} \right)u^{\prime\prime} - 2u' - \left( {2x - 5} \right)u = 0.\]
Its order can be reduced again by one using the known second particular solution \({y_2} = {e^{2x}}.\) The function \({z_2}\) corresponds to this solution, so that we can write:
\[{y_2} = {y_1}{z_2},\;\; \Rightarrow {z_2} = \frac{{{y_2}}}{{{y_1}}} = \frac{{{e^{2x}}}}{{{e^x}}} = {e^x}.\]
From this we obtain a particular solution \({u_1}:\)
\[{u_1} = {z'_2} = {\left( {{e^x}} \right)^\prime } = {e^x}.\]
Further we act in the same manner. We make the following change:
\[u = {u_1}v = {e^x}v,\;\; \Rightarrow u' = {e^x}\left( {v + v'} \right),\;\; \Rightarrow u^{\prime\prime} = {e^x}\left( {v + 2v' + v^{\prime\prime}} \right).\]
We obtain the differential equation for the new variable \(v:\)
\[\left( {2x - 3} \right)\left( {v + 2v' + v^{\prime\prime}} \right) - 2\left( {v + v'} \right) - \left( {2x - 5} \right)v = 0,\;\; \Rightarrow
\cancel{\left( {2x - 3} \right)v} + \left( {4x - 6} \right)v'
+ \left( {2x - 3} \right)v^{\prime\prime} - \cancel{2v} - 2v' - \cancel{\left( {2x - 5} \right)v} = 0,\;\; \Rightarrow
\left( {2x - 3} \right)v^{\prime\prime} + \left( {4x - 8} \right)v' = 0.\]
Denote \(v' = w.\) Then we can write:
\[\left( {2x - 3} \right)w' + \left( {4x - 8} \right)w = 0.\]
The last equation is a first-order equation with separable variables. We find its general solution:
\[\left( {2x - 3} \right)\frac{{dw}}{{dx}} = - \left( {4x - 8} \right)w,\;\; \Rightarrow \frac{{dw}}{w} = - \frac{{4x - 8}}{{2x - 3}}dx,\;\; \Rightarrow \int {\frac{{dw}}{w}} = - \int {\frac{{4x - 8}}{{2x - 3}}dx} ,\;\; \Rightarrow \int {\frac{{dw}}{w}} = - \int {\left( {2 - \frac{2}{{2x - 3}}} \right)dx} ,\;\; \Rightarrow \ln \left| w \right| = - 2x + \ln \left| {2x - 3} \right| + \ln {C_1},\;\; \Rightarrow \ln \left| w \right| = \ln {e^{ - 2x}} + \ln \left| {2x - 3} \right| + \ln {C_1},\;\; \Rightarrow \ln \left| w \right| = \ln \left( {{C_1}\left| {2x - 3} \right|{e^{ - 2x}}} \right),\;\; \Rightarrow w = {C_1}\left( {2x - 3} \right){e^{ - 2x}}.\]
Now we can restore the function \(v\) by integrating the resulting expression for \(w:\)
\[v = \int {wdx} = {C_1}\int {\left( {2x - 3} \right){e^{ - 2x}}dx} .\]
This integral is calculated by parts:
\[v = {C_1}\int {\left( {2x - 3} \right){e^{ - 2x}}dx} = {C_1}\left( {2x - 3} \right)\left( { - \frac{1}{2}} \right){e^{ - 2x}} - \int {2\left( { - \frac{1}{2}} \right){e^{ - 2x}}dx} = {C_1}\left[ {\left( { - x + \frac{3}{2}} \right){e^{ - 2x}} + \int {{e^{ - 2x}}dx} } \right] = {C_1}\left[ {\left( { - x + \frac{3}{2}} \right){e^{ - 2x}} - \frac{1}{2}{e^{ - 2x}}} \right] + {C_2} = - {C_1}{e^{ - 2x}}\left( {x - \frac{3}{2} + \frac{1}{2}} \right) + {C_2} = - {C_1}\left( {x - 1} \right){e^{ - 2x}} + {C_2}.\]
Next we find the function \(u:\)
\[u = {u_1}v = {e^x}v = {e^x}\left[ { - {C_1}\left( {x - 1} \right){e^{ - 2x}} + {C_2}} \right] = -{C_1}\left( {x - 1} \right){e^{ - x}} + {C_2}{e^x}.\]
Performing another integration, we find the function \(z:\)
\[z = \int {udx} = \int {\left[ { - {C_1}\left( {x - 1} \right){e^{ - x}} + {C_2}{e^x}} \right]dx} = - {C_1}\int {\left( {x - 1} \right){e^{ - x}}dx} + {C_2}\int {{e^x}dx} = - {C_1}\Big[ { - \left( {x - 1} \right){e^{ - x}} - \int {\left( { - {e^{ - x}}} \right)dx} } \Big] + {C_2}\int {{e^x}dx} = - {C_1}\left[ { - \left( {x - 1} \right){e^{ - x}} - {e^{ - x}}} \right] + {C_2}{e^x} + {C_3} = {C_1}x{e^{ - x}} + {C_2}{e^x} + {C_3}.\]
Finally, we find the general solution \(y\left( x \right):\)
\[y = {e^x}z = {e^x}\left( {{C_1}x{e^{ - x}} + {C_2}{e^x} + {C_3}} \right) = {C_1}x + {C_2}{e^{2x}} + {C_3}{e^x},\]
where \({C_1},{C_2},{C_3}\) are arbitrary numbers.
