Higher Order Euler Equation
Solved Problems
Example 1.
Find the general solution of the equation
\[{x^3}y^{\prime\prime\prime} - 2{x^2}y^{\prime\prime} + 4xy' - 4y = 0\]
for \(x \gt 0.\)
Solution.
Using the substitution \(x = {e^t},\) we pass to the new variable \(t.\) The derivatives are given by
\[y' = {e^{ - t}}Dy,\]
\[y^{\prime\prime} = {e^{ - 2t}}\left[ {D\left( {D - 1} \right)} \right]y = {e^{ - 2t}}\left( {{D^2} - D} \right)y,\]
\[y^{\prime\prime\prime} = {e^{ - 3t}}\left[ {D\left( {D - 1} \right)\left( {D - 2} \right)} \right]y = {e^{ - 3t}}\left( {{D^3} - 3{D^2} + 2D} \right)y.\]
Here, \(D\) denotes the operation of a single differentiation of \(y\) with respect to \(t.\)
Substituting the expressions for the derivatives and given that \(x = {e^t},\) we obtain the following equation:
\[\cancel{e^{3t}}\cancel{e^{ - 3t}}\left( {{D^3} - 3{D^2} + 2D} \right)y
- 2\cancel{e^{2t}}\cancel{e^{ - 2t}}\left( {{D^2} - D} \right)y
+ 4\cancel{e^t}\cancel{e^{ - t}}Dy - 4y = 0,\]
\[\Rightarrow \left( {{D^3} - 3{D^2} + 2D} \right)y
- 2\left( {{D^2} - D} \right)y + 4Dy - 4y = 0,\]
\[\Rightarrow {D^3}y - 3{D^2}y + 2Dy - 2{D^2}y
+ 2Dy + 4Dy - 4y = 0,\]
\[\Rightarrow {D^3}y - 5{D^2}y + 8Dy - 4y = 0,\]
or in standard form:
\[\frac{{{d^3}y}}{{d{t^3}}} - 5\frac{{{d^2}y}}{{d{t^2}}} + 8\frac{{dy}}{{dt}} - 4y = 0.\]
We find the roots of the auxuliary equation, which can be written as
\[{k^3} - 5{k^2} + 8k - 4 = 0.\]
Note that \(k = 2\) is one of the roots of the equation. By factoring the left side of the equation, we have:
\[{k^3} - 2{k^2} - 3{k^2} + 6k + 2k - 4 = 0,\;\; \Rightarrow {k^2}\left( {k - 2} \right) - 3k\left( {k - 2} \right) + 2\left( {k - 2} \right) = 0,\;\; \Rightarrow \left( {k - 2} \right) \left( {{k^2} - 3k + 2} \right) = 0.\]
The roots of the quadratic polynomial, in turn, are equal to: \(k = 1,\) \(k = 2.\) Thus, the auxiliary equation has the roots \({k_1} = 1,\) \({k_{2,3}} = 2.\) The second value has multiplicity \(2.\) Then the general solution of the equation is written as
\[y\left( t \right) = {C_1}{e^t} + \left( {{C_2} + {C_3}t} \right){e^{2t}},\]
where \({C_1},{C_2},{C_3}\) are arbitrary constants.
Now we return from the variable \(t\) to \(x,\) given that \(t = \ln x.\) The final answer is expressed by the formula
\[y\left( x \right) = {C_1}{e^{\ln x}} + \left( {{C_2} + {C_3}\ln x} \right){e^{2\ln x}} = {C_1}x + \left( {{C_2} + {C_3}\ln x} \right){x^2}.\]
Example 2.
Find the general solution of the equation
\[{x^4}{y^{IV}} + 6{x^3}y^{\prime\prime\prime} + 9{x^2}y^{\prime\prime} + 3xy' + y = 0\]
for \(x \gt 0.\)
Solution.
This is a fourth-order homogeneous Euler equation. We construct the general solution by using the trial power function \(y = {x^k}.\) Substitute the derivatives of this function into the differential equation:
\[y' = k{x^{k - 1}},\]
\[y^{\prime\prime} = k\left( {k - 1} \right){x^{k - 2}},\]
\[y^{\prime\prime\prime} = k\left( {k - 1} \right)\left( {k - 2} \right){x^{k - 3}},\]
\[{y^{IV}} = k\left( {k - 1} \right)\left( {k - 2} \right) \left( {k - 3} \right){x^{k - 4}}.\]
We obtain the auxiliary equation of the fourth degree:
\[{x^4}k\left( {k - 1} \right)\left( {k - 2} \right) \left( {k - 3} \right){x^{k - 4}}
+ 6{x^3}k\left( {k - 1} \right)\left( {k - 2} \right){x^{k - 3}}
+ 9{x^2}k\left( {k - 1} \right){x^{k - 2}}
+ 3xk{x^{k - 1}} + {x^k} = 0,\]
We can simplify this equation making standard transformations:
\[{k^4} - \cancel{\color{blue}{k^3}} - \cancel{\color{blue}{5{k^3}}} + \color{red}{5{k^2}} + \color{red}{6{k^2}}
- \cancel{\color{green}{6k}} + \cancel{\color{blue}{6{k^3}}} - \color{red}{6{k^2}} - \color{red}{12{k^2}}
+ \cancel{\color{green}{12k}} + \color{red}{9{k^2}} - \cancel{\color{green}{6k}} + \color{magenta}{1} = \color{black}{0},\;\; \Rightarrow
{k^4} + \color{red}{2{k^2}} + \color{magenta}{1} = \color{black}{0},\;\; \Rightarrow
{\left( {{k^2} + 1} \right)^2} = 0.\]
As it can be seen, the auxiliary equation has two imaginary roots \({k_{1,2}} = \pm i\) of multiplicity \(2.\) Then the general solution is written as
\[y\left( t \right) = \left( {{C_1} + {C_2}t} \right)\cos t + \left( {{C_3} + {C_4}t} \right)\sin t,\]
where \({C_1}, \ldots ,{C_4}\) are arbitrary constants.
Now we apply the reverse substitution \(t = \ln x\) to get the final answer:
\[y\left( x \right) = \left( {{C_1} + {C_2}\ln x} \right)\cos \left( {\ln x} \right)
+ \left( {{C_3} + {C_4}\ln x} \right)\sin \left( {\ln x} \right).\]
Example 3.
Solve the differential equation
\[{x^3}y^{\prime\prime\prime} - 2{x^2}y^{\prime\prime} + 2xy' = x\left( {2\ln x + 1} \right)\]
for \(x \gt 0.\)
Solution.
Here we have a third-order nonhomogeneous Euler equation. We make the substitution \(x = {e^t}.\) Consequently,
\[y' = {e^{ - t}}Dy,\]
\[y^{\prime\prime} = {e^{ - 2t}}D\left( {D - 1} \right)y,\]
\[y^{\prime\prime\prime} = {e^{ - 3t}}D\left( {D - 1} \right) \left( {D - 2} \right)y,\]
where the operator \(D\) denotes differentiation with respect to the variable \(t.\) After substitution into the original equation, we obtain a nonhomogeneous linear equation with constant coefficients:
\[\cancel{e^{3t}}\cancel{e^{ - 3t}}D\left( {D - 1} \right)\left( {D - 2} \right)y
- 2\cancel{e^{2t}}\cancel{e^{ - 2t}}D\left( {D - 1} \right)y
+ 2\cancel{e^t}\cancel{e^{ - t}}Dy
= {e^t}\left( {2t + 1} \right),\]
\[\Rightarrow D\left( {D - 1} \right)\left( {D - 2} \right)y
- 2D\left( {D - 1} \right)y + 2Dy
= {e^t}\left( {2t + 1} \right),\]
\[\Rightarrow \left( {{D^3} - 3{D^2} + 2D} \right)y
- \left( {2{D^2} - 2D} \right)y + 2Dy
= {e^t}\left( {2t + 1} \right),\]
\[\Rightarrow {D^3}y - \color{blue}{3{D^2}y} + \color{red}{2Dy} - \color{blue}{2{D^2}y} + \color{red}{2Dy} + \color{red}{2Dy} = \color{black}{{e^t}}\left( {2t + 1} \right),\]
\[ \Rightarrow \frac{{{d^3}y}}{{d{t^3}}} - \color{blue}{5\frac{{{d^2}y}}{{d{t^2}}}} + \color{red}{6\frac{{dy}}{{dt}}} = \color{black}{0}.\]
We find the general solution of the homogeneous equation
\[\frac{{{d^3}y}}{{d{t^3}}} - 5\frac{{{d^2}y}}{{d{t^2}}} + 6\frac{{dy}}{{dt}} = 0.\]
Write the auxuliary equation:
\[{k^3} - 5{k^2} + 6k = 0.\]
Its roots are
\[k\left( {{k^2} - 5k + 6} \right) = 0,\;\; \Rightarrow k\left( {k - 2} \right)\left( {k - 3} \right) = 0,\;\; \Rightarrow {k_1} = 0,\;\; {k_2} = 2,\;\; {k_3} = 3.\]
Then the general solution of the homogeneous equation can be written as
\[{y_0}\left( t \right) = {C_1} + {C_2}{e^{2t}} + {C_3}{e^{3t}},\]
where \({C_1},{C_2},{C_3}\) are arbitrary constants.
Now we determine a particular solution of the nonhomogeneous equation. Note that the exponential function \({e^t}\) on the right side does not coincide with a solution of the homogeneous equation. Therefore, a particular solution will be sought in the form of
\[{y_1}\left( t \right) = \left( {At + B} \right){e^t}.\]
Find the derivatives of the function \({y_1}:\)
\[{y'_1} = A{e^t} + \left( {At + B} \right){e^t} = \left( {At + A + B} \right){e^t},\]
\[{y^{\prime\prime}_1} = A{e^t} + \left( {At + A + B} \right){e^t} = \left( {At + 2A + B} \right){e^t},\]
\[{y^{\prime\prime\prime}_1} = A{e^t} + \left( {At + 2A + B} \right){e^t} = \left( {At + 3A + B} \right){e^t}.\]
Substituting the derivatives in the nonhomogeneous equation, we calculate the coefficients \(A, B:\)
\[\left( {At + 3A + B} \right){e^t}
- 5\left( {At + 2A + B} \right){e^t}
+ 6\left( {At + A + B} \right){e^t}
\equiv \left( {2t + 1} \right){e^t},\]
\[\Rightarrow \color{blue}{At} + \color{green}{3A} + \color{red}{B} - \color{blue}{5At} - \color{green}{10A} - \color{red}{5B} + \color{blue}{6At} + \color{green}{6A} + \color{red}{6B} \equiv \color{black}{2t + 1},\]
\[\Rightarrow \color{blue}{2At} - \color{green}{A} + \color{red}{2B} \equiv \color{black}{2t + 1},\]
\[\Rightarrow \left\{ {\begin{array}{*{20}{c}}
{2A = 2}\\
{ - A + 2B = 1}
\end{array}} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{c}}
{A = 1}\\
{B = 1}
\end{array}} \right..\]
Thus, a particular solution is given by
\[{y_1}\left( t \right) = \left( {t + 1} \right){e^t}.\]
Consequently, the general solution of the nonhomogeneous equation can be written as
\[y\left( t \right) = {y_0}\left( t \right) + {y_1}\left( t \right) = {C_1} + {C_2}{e^{2t}} + {C_3}{e^{3t}} + \left( {t + 1} \right){e^t}.\]
Returning back to the variable \(x,\) we obtain the final form of the general solution:
\[y\left( x \right) = {C_1} + {C_2}{e^{2\ln x}} + {C_3}{e^{3\ln x}}
+ \left( {\ln x + 1} \right){e^{\ln x}}
= {C_1} + {C_2}{x^2} + {C_3}{x^3} + \left( {\ln x + 1} \right)x.\]
