Second Order Euler Equation
Solved Problems
Example 1.
Find the general solution of the differential equation \[4{x^2}y^{\prime\prime} + y = 0\] assuming that \(x \gt 0.\)
Solution.
We make the substitution \(x = {e^t}.\) As
\[y^{\prime\prime} = {e^{ - 2t}}\left( {\frac{{{d^2}y}}{{d{t^2}}} - \frac{{dy}}{{dt}}} \right),\]
the equation becomes:
\[4\cancel{e^{2t}}\cancel{e^{ - 2t}}\left( {\frac{{{d^2}y}}{{d{t^2}}} - \frac{{dy}}{{dt}}} \right) + y = 0,\;\; \Rightarrow 4\frac{{{d^2}y}}{{d{t^2}}} - 4\frac{{dy}}{{dt}} + y = 0.\]
Calculate the roots of the corresponding characteristic equation:
\[4{k^2} - 4k + 1 = 0,\;\; \Rightarrow D = 16 - 16 = 0,\;\; \Rightarrow k = \frac{4}{{2 \cdot 4}} = \frac{1}{2}.\]
The equation has one root of order \(2.\) Then the general solution for the function \(y\left( t \right)\) is given by
\[y\left( t \right) = \left( {{C_1} + {C_2}t} \right){e^{\frac{t}{2}}}.\]
Now it's easy to write the solution for the original function \(y\left( x \right):\)
\[y\left( x \right) = \left( {{C_1} + {C_2}\ln x} \right){e^{\frac{{\ln x}}{2}}} = \left( {{C_1} + {C_2}\ln x} \right){x^{\frac{1}{2}}} = \left( {{C_1} + {C_2}\ln x} \right)\sqrt x ,\]
where \({C_1}\) and \({C_2}\) are arbitrary real numbers.
Example 2.
Find the general solution of the equation \[{x^2}y^{\prime\prime} - xy' - 8y = 0\] assuming that \(x \gt 0.\)
Solution.
We use the second method to solve the given equation, i.e. we will seek the solution of the form \(y = {x^k}.\) Then
\[y' = k{x^{k - 1}},\;\; y^{\prime\prime} = k\left( {k - 1} \right){x^{k - 2}}.\]
Substituting into the original differential equation gives:
\[{x^2}k\left( {k - 1} \right){x^{k - 2}} - xk{x^{k - 1}} - 8{x^k} = 0,\;\; \Rightarrow k\left( {k - 1} \right){x^k} - k{x^k} - 8{x^k} = 0,\;\; \Rightarrow \left[ {k\left( {k - 1} \right) - k - 8} \right]{x^k} = 0.\]
The corresponding characteristic equation has the roots:
\[k\left( {k - 1} \right) - k - 8 = 0,\;\; \Rightarrow {k^2} - 2k - 9 = 0,\;\; \Rightarrow D = 4 + 32 = 36,\;\; \Rightarrow {k_{1,2}} = \frac{{2 \pm 6}}{2} = 4, - 2.\]
Hence the general solution for the function \(y\left( t \right)\) is given by
\[y\left( t \right) = {C_1}{e^{4t}} + {C_2}{e^{ - 2t}}.\]
Returning to the variable \(x,\) we have the following final answer:
\[y\left( x \right) = {C_1}{e^{4\ln x}} + {C_2}{e^{ - 2\ln x}} = {C_1}{x^4} + {C_2}{x^{ - 2}} = {C_1}{x^4} + \frac{{{C_2}}}{{{x^2}}}.\]
Here \({C_1},\) \({C_2}\) are arbitrary real constants.
Example 3.
Find the general solution of the Euler equation \[{x^2}y^{\prime\prime} + xy' + y = 5{x^2}\] for \(x \gt 0.\)
Solution.
First we construct the general solution of the homogeneous equation:
\[{x^2}y^{\prime\prime} + xy' + y = 0.\]
Make the substitution:
\[x = {e^t},\;\; y' = {e^{ - t}}\frac{{dy}}{{dt}},\;\; y^{\prime\prime} = {e^{ - 2t}}\left( {\frac{{{d^2}y}}{{d{t^2}}} - \frac{{dy}}{{dt}}} \right).\]
As a result the homogeneous differential equation is written in the form:
\[\cancel{e^{2t}}\cancel{e^{ - 2t}}\left( {\frac{{{d^2}y}}{{d{t^2}}} - \frac{{dy}}{{dt}}} \right) + \cancel{e^t}\cancel{e^{ - t}}\frac{{dy}}{{dt}} + y = 0,\;\; \Rightarrow \frac{{{d^2}y}}{{d{t^2}}} - \cancel{\frac{{dy}}{{dt}}} + \cancel{\frac{{dy}}{{dt}}} + y = 0,\;\; \Rightarrow \frac{{{d^2}y}}{{d{t^2}}} + y = 0.\]
Solve the characteristic equation:
\[{k^2} + 1 = 0,\;\; \Rightarrow {k_{1,2}} = \pm i.\]
Thus, the roots of the characteristic equation are imaginary. Therefore, the general solution of the homogeneous equation is given by
\[{y_0}\left( t \right) = {C_1}\cos t + {C_2}\sin t,\]
where \({C_1}\) and \({C_2}\) are real constants.
Determine a particular solution of the non-homogeneous equation:
\[\frac{{{d^2}y}}{{d{t^2}}} + y = 5{e^{2t}}.\]
Taking into account the structure of the right side, we look for a particular solution of the form \({y_1}\left( t \right) = a{e^{2t}}\) where \(a\) is a constant coefficient. Then
\[\frac{{d{y_1}}}{{dt}} = 2a{e^{2t}},\;\; \frac{{{d^2}{y_1}}}{{d{t^2}}} = 4a{e^{2t}}.\]
Substitute the function and its derivative into the equation to find the coefficient \(a:\)
\[4a{e^{2t}} + a{e^{2t}} = 5{e^{2t}},\;\; \Rightarrow 5a{e^{2t}} = 5{e^{2t}},\;\; \Rightarrow a = 1.\]
Thus, a particular solution of the non-homogeneous equation is given by
\[{y_1}\left( t \right) = {e^{2t}}.\]
Now we can write the general solution of the non-homogeneous equation:
\[y\left( t \right) = {y_0}\left( t \right) + {y_1}\left( t \right) = {C_1}\cos t + {C_2}\sin t + {e^{2t}}.\]
Going back to the variable \(x,\) we obtain
\[y\left( x \right) = {C_1}\cos \left( {\ln x} \right) + {C_2}\sin \left( {\ln x} \right) + {e^{2\ln x}}.\]
As \({e^{2\ln x}} = {e^{\ln {x^2}}} = {x^2},\) the final answer is written in the form
\[y\left( x \right) = {C_1}\cos \left( {\ln x} \right) + {C_2}\sin \left( {\ln x} \right) + {x^2}.\]
Example 4.
Solve the non-homogeneous Euler equation \[{x^2}y^{\prime\prime} - 2xy' + 2y = 6{x^2} + 4\ln x\] assuming that \(x \gt 0.\)
Solution.
First we solve the homogeneous equation:
\[{x^2}y^{\prime\prime} - 2xy' + 2y = 0.\]
By using the substitution \(x = {e^t}\) we can convert the last equation into the constant coefficient equation:
\[\frac{{{d^2}y}}{{d{t^2}}} - 3\frac{{dy}}{{dt}} + 2y = 0.\]
Calculate the roots of the characteristic equation and write the general solution \({y_0}\left( t \right)\) of the homogeneous differential equation:
\[{k^2} - 3k + 2 = 0,\;\; \Rightarrow D = 9 - 4 \cdot 2 = 1,\;\; \Rightarrow {k_{1,2}} = \frac{{3 \pm 1}}{2} = 2,1.\]
Hence,
\[{y_0}\left( t \right) = {C_1}{e^{2t}} + {C_2}{e^t}.\]
Now consider the non-homogeneous equation, which can be written in terms of \(t\) as
\[\frac{{{d^2}y}}{{d{t^2}}} - 3\frac{{dy}}{{dt}} + 2y = 6{e^{2t}} + 4\ln \left( {{e^t}} \right),\;\; \Rightarrow \frac{{{d^2}y}}{{d{t^2}}} - 3\frac{{dy}}{{dt}} + 2y = 6{e^{2t}} + 4t.\]
The coefficient before the exponential function in the right side is equal to \(2\) and, hence it coincides with one of the roots of the characteristic equation. Therefore, we will look for a particular solution of the form
\[{y_1}\left( t \right) = at{e^{2t}} + bt + c,\]
where \(a, b\) and \(c\) are some unknown numbers.
The derivatives of this function are given by
\[\frac{{d{y_1}}}{{dt}} = a{e^{2t}} + 2at{e^{2t}} + b,\]
\[\frac{{{d^2}{y_1}}}{{d{t^2}}} = 2a{e^{2t}} + 2a{e^{2t}} + 4at{e^{2t}} = 4a{e^{2t}} + 4at{e^{2t}}.\]
Plug this into the non-homogeneous equation:
\[4a{e^{2t}} + 4at{e^{2t}}
- 3\left( {a{e^{2t}} + 2at{e^{2t}} + b} \right)
+ 2\left( {at{e^{2t}} + bt + c} \right) = 6{e^{2t}} + 4t,\]
\[\Rightarrow 4a{e^{2t}} + \cancel{4at{e^{2t}}} - 3a{e^{2t}}
- \cancel{6at{e^{2t}}} - 3b
+ \cancel{2at{e^{2t}}} + 2bt + 2c
= 6{e^{2t}} + 4t,\]
The last relation is identical, that is, it is true for all values \(t.\) Equating the coefficients of the similar terms on both sides, we obtain
\[\left\{ \begin{array}{l}
a = 6\\
2b = 4\\
- 3b + 2c = 0
\end{array} \right.,\;\; \Rightarrow
\left\{ \begin{array}{l}
a = 6\\
b = 2\\
c = 3
\end{array} \right..\]
Thus, a particular solution of the non-homogeneous equation is given by the expression
\[{y_1}\left( t \right) = 6t{e^{2t}} + 2t + 3.\]
Now we can write the general solution for the non-homogeneous differential Euler equation:
\[y\left( t \right) = {y_0}\left( t \right) + {y_1}\left( t \right) = {C_1}{e^{2t}} + {C_2}{e^t} + 6t{e^{2t}} + 2t + 3.\]
Since \(t = \ln x,\) the final answer is written as
\[y\left( x \right) = {C_1}{x^2} + {C_2}x + 6{x^2}\ln x + 2\ln x + 3,\]
where \({C_1},\) \({C_2}\) are arbitrary real constants.
