Second Order Linear Homogeneous Differential Equations with Variable Coefficients
Solved Problems
Example 3.
Write a homogeneous linear differential equation, if its fundamental system of solutions is known and has the form \(x,{e^x}.\)
Solution.
We find the derivatives of the given functions:
\[{y'_1} = x' = 1,\;\; {y^{\prime\prime}_1} = 1' = 0;\]
\[{y'_2} = {\left( {{e^x}} \right)^\prime } = {e^x},\;\; y^{\prime\prime}_2 = \left( {{e^x}} \right)^\prime = e^x.\]
The desired differential equation is expressed in terms of the determinant:
\[\left| {\begin{array}{*{20}{c}}
{{y_1}}&{{y_2}}&y\\
{{y'_1}}&{{y'_2}}&y'\\
{{y^{\prime\prime}_1}}&{{y^{\prime\prime}_2}}&y^{\prime\prime}
\end{array}} \right| = 0,\;\; \Rightarrow
\left| {\begin{array}{*{20}{c}}
x&{{e^x}}&y\\
1&{{e^x}}&y'\\
0&{{e^x}}&y^{\prime\prime}
\end{array}} \right| = 0.\]
Expand the determinant along the \(1\)st column:
\[x\left( {{e^x}y^{\prime\prime} - {e^x}y'} \right) - 1 \cdot \left( {{e^x}y^{\prime\prime} - {e^x}y} \right) = 0,\;\; \Rightarrow
{e^x}\left[ {\left( {xy^{\prime\prime} - xy'} \right) - \left( {y^{\prime\prime} - y} \right)} \right] = 0.\]
Given that \({e^x} \ne 0,\) we obtain after simplification
\[xy^{\prime\prime} - xy' - y^{\prime\prime} + y = 0,\;\; \Rightarrow
\left( {x - 1} \right)y^{\prime\prime} - xy' + y = 0.\]
Example 4.
Find the general solution of the equation \[{x^2}y^{\prime\prime} - 2xy' + 2y = 0,\] given the particular solution \({y_1} = x.\)
Solution.
We make the substitution \(y = {y_1}z \) \(= xz.\) Then
\[y' = {\left( {xz} \right)^\prime } = z + xz',\;\; y^{\prime\prime} = {\left( {z + xz'} \right)^\prime }
= z' + z' + xz^{\prime\prime} = 2z' + xz^{\prime\prime}.\]
After substituting the original equation becomes:
\[{x^2}\left( {2z' + xz^{\prime\prime}} \right) - 2x\left( {z + xz'} \right) + 2xz = 0,\;\; \Rightarrow
\cancel{\color{blue}{2{x^2}z'}} + {x^3}z^{\prime\prime} - \cancel{\color{red}{2xz}} - \cancel{\color{blue}{2{x^2}z'}} + \cancel{\color{red}{2xz}} = 0,\;\; \Rightarrow {x^3}z^{\prime\prime} = 0.\]
By replacing \(z' = p,\) we obtain a compact equation \({x^3}p' = 0,\) whose solution is the function \(p = {C_1}.\)
From here we find the function \(z:\)
\[p = {C_1},\;\; \Rightarrow
z' = {C_1},\;\; \Rightarrow
z = {C_1}x + {C_2}.\]
Now we can write the general solution of the equation:
\[y\left( x \right) = xz
= x\left( {{C_1}x + {C_2}} \right)
= {C_1}{x^2} + {C_2}x.\]
Example 5.
Find the general solution of the equation \[\left( {{x^2} + 1} \right)y^{\prime\prime} - 2y = 0.\]
Solution.
First, choose a particular solution for the given equation. Based on the structure of the equation, we can try to find a particular solution in the form of a quadratic function:
\[{y_1} = A{x^2} + Bx + C.\]
Its derivatives will be equal to
\[{y'_1} = 2Ax + B,\;\; {y^{\prime\prime}_1} = 2A.\]
Substituting this into the differential equation, we determine the coefficients \(A, B, C:\)
\[\left( {{x^2} + 1} \right) \cdot 2A - 2\left( {A{x^2} + Bx + C} \right) \equiv 0,\;\; \Rightarrow
\cancel{2A{x^2}} + 2A - \cancel{2A{x^2}} - 2Bx - 2C \equiv 0.\]
It is seen that the coefficients must satisfy the conditions
\[\left\{ \begin{array}{l}
- 2B = 0\\
2A - 2C = 0
\end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l}
B = 0\\
A = C
\end{array} \right..\]
A function of the form \({y_1} = C\left( {{x^2} + 1} \right)\) corresponds to these conditions. Thus, we can take the function \({y_1} = {{x^2} + 1}\) as a particular solution.
We make the following substitution to reduce the order of the equation:
\[y = {y_1}z = \left( {{x^2} + 1} \right)z,\;\; \Rightarrow
y' = 2xz + \left( {{x^2} + 1} \right)z',\;\; \Rightarrow
y^{\prime\prime} = 2z + 2xz' + 2xz' + \left( {{x^2} + 1} \right)z^{\prime\prime}
= 2z + 4xz' + \left( {{x^2} + 1} \right)z^{\prime\prime}.\]
Then the equation becomes:
\[\left( {{x^2} + 1} \right) \left[ {2z + 4xz' + \left( {{x^2} + 1} \right)z^{\prime\prime}} \right] - 2\left( {{x^2} + 1} \right)z = 0,\;\; \Rightarrow
\left( {{x^2} + 1} \right) \left[ {\cancel{2z} + 4xz' + \left( {{x^2} + 1} \right)z^{\prime\prime} - \cancel{2z}} \right] = 0,\;\; \Rightarrow
\left( {{x^2} + 1} \right)z^{\prime\prime} + 4xz' = 0.\]
With another change \(z' = p\left( x \right),\) we obtain
\[\left( {{x^2} + 1} \right)p' + 4xp = 0.\]
Now we have a linear equation of the first order, which can be solved by separation of variables:
\[\left( {{x^2} + 1} \right)\frac{{dp}}{{dx}} = - 4xp,\;\; \Rightarrow
\frac{{dp}}{p} = - \frac{{4xdx}}{{{x^2} + 1}},\;\; \Rightarrow
\int {\frac{{dp}}{p}} = - 2\int {\frac{{d\left( {{x^2} + 1} \right)}}{{{x^2} + 1}}} ,\;\; \Rightarrow
\ln\left| p \right| = - 2\ln \left( {{x^2} + 1} \right) + \ln {C_1},\;\; \Rightarrow
\ln \left| p \right| = \ln \frac{{{C_1}}}{{{{\left( {{x^2} + 1} \right)}^2}}},\;\; \Rightarrow
p = \frac{{{C_1}}}{{{{\left( {{x^2} + 1} \right)}^2}}}.\]
We solve one more equation for \(z:\)
\[p = z' = \frac{{{C_1}}}{{{{\left( {{x^2} + 1} \right)}^2}}},\;\; \Rightarrow
z = \int {\frac{{dx}}{{{{\left( {{x^2} + 1} \right)}^2}}}} .\]
This integral is calculated using the reduction formula:
\[\int {\frac{{dx}}{{{{\left( {{x^2} + 1} \right)}^2}}}}
= \frac{x}{{2\left( {{x^2} + 1} \right)}} + \frac{1}{2}\int {\frac{{dx}}{{{x^2} + 1}}}
= \frac{x}{{2\left( {{x^2} + 1} \right)}} + \frac{1}{2}\arctan x + {C_2}.\]
As a result, we get the following expression for \(z:\)
\[z = {C_1}\left[ {\frac{x}{{2\left( {{x^2} + 1} \right)}} + \frac{1}{2}\arctan x + {C_2}} \right]
= \frac{{{C_1}x}}{{{x^2} + 1}} + {C_1}\arctan x + {C_2},\]
where the constants \({C_1}\) and \({C_2}\) are renormalized to simplify the expression.
Thus, the general solution has the form:
\[y = \left( {{x^2} + 1} \right)z
= {C_1}x + {C_1}\left( {{x^2} + 1} \right)\arctan x + {C_2}\left( {{x^2} + 1} \right) = {C_1}\left[ {x + \left( {{x^2} + 1} \right)\arctan x} \right] + {C_2}\left( {{x^2} + 1} \right).\]
Example 6.
Find the general solution of the equation \[{x^2}y^{\prime\prime} - 4xy' + 6y = 0\] using the Liouville formula. A particular solution of the equation is known and has the form \({y_1} = {x^2}.\)
Solution.
Let \({y_1}\) and \({y_2}\) be linearly independent particular solutions of the given equation (the solution \({y_1}\) is known). Then the Liouville formula is written as
\[W\left( x \right) = {W_{{y_1},{y_2}}}\left( x \right)
= \left| {\begin{array}{*{20}{c}}
{{y_1}}&{{y_2}}\\
{{y'_1}}&{{y'_2}}
\end{array}} \right|
= {W_0}\left( x \right)\exp \left( { - \int\limits_{{x_0}}^x {\frac{{{a_1}\left( t \right)}}{{{a_0}\left( t \right)}}dt} } \right).\]
The integral in this expression is equal to
\[\int\limits_{{x_0}}^x {\frac{{{a_1}\left( t \right)}}{{{a_0}\left( t \right)}}dt}
= \int\limits_{{x_0}}^x {\left( {\frac{{ - 4t}}{{{t^2}}}} \right)dt}
= \left. {\left( { - 4\ln t} \right)} \right|_{{x_0}}^x
= - 4\ln x + 4\ln {x_0}
= - \ln {x^4} + \ln x_0^4
= - \ln \frac{{{x^4}}}{{x_0^4}}.\]
Substituting this, we obtain
\[\left| {\begin{array}{*{20}{c}}
{{y_1}}&{{y_2}}\\
{{y'_1}}&{{y'_2}}
\end{array}} \right|
= {W_0}\left( x \right)\exp \left( {\ln \frac{{{x^4}}}{{x_0^4}}} \right)
= \frac{{{W_0}\left( x \right)}}{{x_0^4}}{x^4}
= {C_1}{x^4},\]
where \({C_1}\) is an arbitrary constant.
As the particular solution \({y_1}\) is known, we get a first order differential equation for the determination of the other particular solution \({y_2}.\) After dividing by \(y_1^2\) we have:
\[\left. {{y'_2}{y_1} - {y_2}{y'_1} = {C_1}{x^4}} \right|:y_1^2,\;\; \Rightarrow
\frac{{{y'_2}{y_1} - {y_2}{y'_1}}}{{y_1^2}} = \frac{{{C_1}{x^4}}}{{y_1^2}},\;\; \Rightarrow
\left( {\frac{{{y_2}}}{{{y_1}}}} \right)^\prime = \frac{{{C_1}{x^4}}}{{y_1^2}}
= \frac{{{C_1}\cancel{x^4}}}{{\cancel{x^4}}} = {C_1}.\]
From this we find the function \({y_2}:\)
\[\frac{{{y_2}}}{{{y_1}}} = {C_1}x + {C_2},\;\; \Rightarrow
{y_2} = {y_1}\left( {{C_1}x + {C_2}} \right)
= {x^2}\left( {{C_1}x + {C_2}} \right)
= {C_1}{x^3} + {C_2}{x^2},\]
where \({C_1},\) \({C_2}\) are constants of integration. It is clear that \({y_2}\) is actually the general solution of the equation.
Example 7.
Find the general solution of the equation \[{x^2}y^{\prime\prime} + xy' - y = 0\] (for \(x \ne 0\) by the Liouville formula, if a particular solution is known: \({y_1} = x.\)
Solution.
Using the Liouville formula we get the following equation for the determination of the second particular solution \({y_2}:\)
\[W\left( x \right) = {W_{{y_1},{y_2}}}\left( x \right)
= \left| {\begin{array}{*{20}{c}}
{{y_1}}&{{y_2}}\\
{{y'_1}}&{{y'_2}}
\end{array}} \right|
= {C_1}\exp \left( { - \int {\frac{{{a_1}\left( x \right)}}{{{a_0}\left( x \right)}}dx} } \right).\]
The integral in this formula is
\[\int {\frac{{{a_1}\left( x \right)}}{{{a_0}\left( x \right)}}dx}
= \int {\frac{x}{{{x^2}}}dx}
= \int {\frac{{dx}}{x}}
= \ln \left| x \right|.\]
Then we can write
\[{y'_2}{y_1} - {y_2}{y'_1} = {C_1}{e^{ - \ln \left| x \right|}},\;\; \Rightarrow
{y'_2}{y_1} - {y_2}{y'_1} = {C_1}{e^{\ln \frac{1}{{\left| x \right|}}}},\;\; \Rightarrow
{y'_2}{y_1} - {y_2}{y'_1} = \frac{{{C_1}}}{x},\]
where \({C_1}\) is an arbitrary constant.
Divide both sides of the equation by \(y_1^2 = {x^2}\) (assuming that \(x \ne 0\)).
\[\frac{{{y'_2}{y_1} - {y_2}{y'_1}}}{{y_1^2}} = \frac{{{C_1}}}{{xy_1^2}},\;\; \Rightarrow
\left( {\frac{{{y_2}}}{{{y_1}}}} \right)^\prime = \frac{{{C_1}}}{{x \cdot {x^2}}} = \frac{{{C_1}}}{{{x^3}}},\;\; \Rightarrow
\frac{{{y_2}}}{{{y_1}}} = \int {\frac{{{C_1}}}{{{x^3}}}dx}
= - \frac{{{C_1}}}{{2{x^2}}} + {C_2} = \frac{{{C_1}}}{{{x^2}}} + {C_2}.\]
Here, we redefined: \( - {\frac{{{C_1}}}{2}} \to {C_1}.\) The final answer is
\[{y_2} = {y_1}\left( {\frac{{{C_1}}}{{{x^2}}} + {C_2}} \right)
= x\left( {\frac{{{C_1}}}{{{x^2}}} + {C_2}} \right)
= \frac{{{C_1}}}{x} + {C_2}x.\]
