Internal problem ID [15434]
Internal file name [OUTPUT/15435_Wednesday_May_08_2024_03_58_46_PM_11576560/index.tex]
Book: A book of problems in ordinary differential equations. M.L. KRASNOV, A.L. KISELYOV,
G.I. MARKARENKO. MIR, MOSCOW. 1983
Section: Chapter 2 (Higher order ODE’s). Section 15.5 Linear equations with variable coefficients.
The Lagrange method. Exercises page 148
Problem number: 670.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "kovacic", "second_order_change_of_variable_on_y_method_2", "second_order_ode_non_constant_coeff_transformation_on_B"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
Unable to solve or complete the solution.
\[ \boxed {\left (1-x \right ) y^{\prime \prime }+x y^{\prime }-y=\left (x -1\right )^{2} {\mathrm e}^{x}} \] With initial conditions \begin {align*} [y \left (-\infty \right ) = 0, y^{\prime }\left (0\right ) = 1] \end {align*}
This is second order non-homogeneous ODE. In standard form the ODE is \[ A y''(x) + B y'(x) + C y(x) = f(x) \] Where \(A=1-x, B=x, C=-1, f(x)=\left (x -1\right )^{2} {\mathrm e}^{x}\). Let the solution be \[ y = y_h + y_p \] Where \(y_h\) is the solution to the homogeneous ODE \( A y''(x) + B y'(x) + C y(x) = 0\), and \(y_p\) is a particular solution to the non-homogeneous ODE \(A y''(x) + B y'(x) + C y(x) = f(x)\). Solving for \(y_h\) from \[ \left (1-x \right ) y^{\prime \prime }+x y^{\prime }-y = 0 \] In normal form the ode \begin {align*} \left (1-x \right ) y^{\prime \prime }+x y^{\prime }-y&=0 \tag {1} \end {align*}
Becomes \begin {align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=0 \tag {2} \end {align*}
Where \begin {align*} p \left (x \right )&=-\frac {x}{x -1}\\ q \left (x \right )&=\frac {1}{x -1} \end {align*}
Applying change of variables on the depndent variable \(y = v \left (x \right ) x^{n}\) to (2) gives the following ode where the dependent variables is \(v \left (x \right )\) and not \(y\). \begin {align*} v^{\prime \prime }\left (x \right )+\left (\frac {2 n}{x}+p \right ) v^{\prime }\left (x \right )+\left (\frac {n \left (n -1\right )}{x^{2}}+\frac {n p}{x}+q \right ) v \left (x \right )&=0 \tag {3} \end {align*}
Let the coefficient of \(v \left (x \right )\) above be zero. Hence \begin {align*} \frac {n \left (n -1\right )}{x^{2}}+\frac {n p}{x}+q&=0 \tag {4} \end {align*}
Substituting the earlier values found for \(p \left (x \right )\) and \(q \left (x \right )\) into (4) gives \begin {align*} \frac {n \left (n -1\right )}{x^{2}}-\frac {n}{x -1}+\frac {1}{x -1}&=0 \tag {5} \end {align*}
Solving (5) for \(n\) gives \begin {align*} n&=1 \tag {6} \end {align*}
Substituting this value in (3) gives \begin {align*} v^{\prime \prime }\left (x \right )+\left (\frac {2}{x}-\frac {x}{x -1}\right ) v^{\prime }\left (x \right )&=0 \\ v^{\prime \prime }\left (x \right )+\left (\frac {2}{x}-\frac {x}{x -1}\right ) v^{\prime }\left (x \right )&=0 \tag {7} \\ \end {align*}
Using the substitution \begin {align*} u \left (x \right ) = v^{\prime }\left (x \right ) \end {align*}
Then (7) becomes \begin {align*} u^{\prime }\left (x \right )+\left (\frac {2}{x}-\frac {x}{x -1}\right ) u \left (x \right ) = 0 \tag {8} \\ \end {align*}
The above is now solved for \(u \left (x \right )\). In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {u \left (x^{2}-2 x +2\right )}{x \left (x -1\right )} \end {align*}
Where \(f(x)=\frac {x^{2}-2 x +2}{x \left (x -1\right )}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= \frac {x^{2}-2 x +2}{x \left (x -1\right )} \,d x\\ \int { \frac {1}{u} \,du} &= \int {\frac {x^{2}-2 x +2}{x \left (x -1\right )} \,d x}\\ \ln \left (u \right )&=x -2 \ln \left (x \right )+\ln \left (x -1\right )+c_{1}\\ u&={\mathrm e}^{x -2 \ln \left (x \right )+\ln \left (x -1\right )+c_{1}}\\ &=c_{1} {\mathrm e}^{x -2 \ln \left (x \right )+\ln \left (x -1\right )} \end {align*}
Which simplifies to \[ u \left (x \right ) = c_{1} \left (\frac {{\mathrm e}^{x}}{x}-\frac {{\mathrm e}^{x}}{x^{2}}\right ) \] Now that \(u \left (x \right )\) is known, then \begin {align*} v^{\prime }\left (x \right )&= u \left (x \right )\\ v \left (x \right )&= \int u \left (x \right )d x +c_{2}\\ &= \frac {{\mathrm e}^{x} c_{1}}{x}+c_{2} \end {align*}
Hence \begin {align*} y&= v \left (x \right ) x^{n}\\ &= \left (\frac {{\mathrm e}^{x} c_{1}}{x}+c_{2} \right ) x\\ &= {\mathrm e}^{x} c_{1} +c_{2} x\\ \end {align*}
Now the particular solution to this ODE is found \[ \left (1-x \right ) y^{\prime \prime }+x y^{\prime }-y = \left (x -1\right )^{2} {\mathrm e}^{x} \] The particular solution \(y_p\) can be found using either the method of undetermined coefficients, or the method of variation of parameters. The method of variation of parameters will be used as it is more general and can be used when the coefficients of the ODE depend on \(x\) as well. Let \begin{equation} \tag{1} y_p(x) = u_1 y_1 + u_2 y_2 \end{equation} Where \(u_1,u_2\) to be determined, and \(y_1,y_2\) are the two basis solutions (the two linearly independent solutions of the homogeneous ODE) found earlier when solving the homogeneous ODE as \begin{align*} y_1 &= x \\ y_2 &= {\mathrm e}^{x} \\ \end{align*} In the Variation of parameters \(u_1,u_2\) are found using \begin{align*} \tag{2} u_1 &= -\int \frac {y_2 f(x)}{a W(x)} \\ \tag{3} u_2 &= \int \frac {y_1 f(x)}{a W(x)} \\ \end{align*} Where \(W(x)\) is the Wronskian and \(a\) is the coefficient in front of \(y''\) in the given ODE. The Wronskian is given by \(W= \begin {vmatrix} y_1 & y_{2} \\ y_{1}^{\prime } & y_{2}^{\prime } \end {vmatrix} \). Hence \[ W = \begin {vmatrix} x & {\mathrm e}^{x} \\ \frac {d}{dx}\left (x\right ) & \frac {d}{dx}\left ({\mathrm e}^{x}\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} x & {\mathrm e}^{x} \\ 1 & {\mathrm e}^{x} \end {vmatrix} \] Therefore \[ W = \left (x\right )\left ({\mathrm e}^{x}\right ) - \left ({\mathrm e}^{x}\right )\left (1\right ) \] Which simplifies to \[ W = {\mathrm e}^{x} x -{\mathrm e}^{x} \] Which simplifies to \[ W = \left (x -1\right ) {\mathrm e}^{x} \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {\left (x -1\right )^{2} {\mathrm e}^{2 x}}{\left (1-x \right ) \left (x -1\right ) {\mathrm e}^{x}}\,dx \] Which simplifies to \[ u_1 = - \int -{\mathrm e}^{x}d x \] Hence \[ u_1 = {\mathrm e}^{x} \] And Eq. (3) becomes \[ u_2 = \int \frac {x \left (x -1\right )^{2} {\mathrm e}^{x}}{\left (1-x \right ) \left (x -1\right ) {\mathrm e}^{x}}\,dx \] Which simplifies to \[ u_2 = \int -x d x \] Hence \[ u_2 = -\frac {x^{2}}{2} \] Therefore the particular solution, from equation (1) is \[ y_p(x) = {\mathrm e}^{x} x -\frac {{\mathrm e}^{x} x^{2}}{2} \] Which simplifies to \[ y_p(x) = -\frac {{\mathrm e}^{x} x \left (x -2\right )}{2} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (\left (\frac {{\mathrm e}^{x} c_{1}}{x}+c_{2} \right ) x\right ) + \left (-\frac {{\mathrm e}^{x} x \left (x -2\right )}{2}\right ) \\ &= -\frac {{\mathrm e}^{x} x \left (x -2\right )}{2}+\left (\frac {{\mathrm e}^{x} c_{1}}{x}+c_{2} \right ) x \\ \end{align*} Which simplifies to \[ y = \frac {\left (-x^{2}+2 c_{1} +2 x \right ) {\mathrm e}^{x}}{2}+c_{2} x \] Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = \frac {\left (-x^{2}+2 c_{1} +2 x \right ) {\mathrm e}^{x}}{2}+c_{2} x \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = 0\) and \(x = -\infty \) in the above gives \begin {align*} 0 = -\operatorname {signum}\left (c_{2} \right ) \infty \tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \frac {\left (-2 x +2\right ) {\mathrm e}^{x}}{2}+\frac {\left (-x^{2}+2 c_{1} +2 x \right ) {\mathrm e}^{x}}{2}+c_{2} \end {align*}
substituting \(y^{\prime } = 1\) and \(x = 0\) in the above gives \begin {align*} 1 = 1+c_{1} +c_{2}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Warning, unable to solve for constants of integrations.
Verification of solutions N/A
Given an ode of the form \begin {align*} A y^{\prime \prime } + B y^{\prime } + C y &= F(x) \end {align*}
This method reduces the order ode the ODE by one by applying the transformation \begin {align*} y&= B v \end {align*}
This results in \begin {align*} y' &=B' v+ v' B \\ y'' &=B'' v+ B' v' +v'' B + v' B' \\ &=v'' B+2 v'+ B'+B'' v \end {align*}
And now the original ode becomes \begin {align*} A\left ( v'' B+2v'B'+ B'' v\right )+B\left ( B'v+ v' B\right ) +CBv & =0\\ ABv'' +\left ( 2AB'+B^{2}\right ) v'+\left (AB''+BB'+CB\right ) v & =0 \tag {1} \end {align*}
If the term \(AB''+BB'+CB\) is zero, then this method works and can be used to solve \[ ABv''+\left ( 2AB' +B^{2}\right ) v'=0 \] By Using \(u=v'\) which reduces the order of the above ode to one. The new ode is \[ ABu'+\left ( 2AB'+B^{2}\right ) u=0 \] The above ode is first order ode which is solved for \(u\). Now a new ode \(v'=u\) is solved for \(v\) as first order ode. Then the final solution is obtain from \(y=Bv\).
This method works only if the term \(AB''+BB'+CB\) is zero. The given ODE shows that \begin {align*} A &= 1-x\\ B &= x\\ C &= -1\\ F &= \left (x -1\right )^{2} {\mathrm e}^{x} \end {align*}
The above shows that for this ode \begin {align*} AB''+BB'+CB &= \left (1-x\right ) \left (0\right ) + \left (x\right ) \left (1\right ) + \left (-1\right ) \left (x\right ) \\ &=0 \end {align*}
Hence the ode in \(v\) given in (1) now simplifies to \begin {align*} -x \left (x -1\right ) v'' +\left ( x^{2}-2 x +2\right ) v' & =0 \end {align*}
Now by applying \(v'=u\) the above becomes \begin {align*} \left (-x^{2}+x \right ) u^{\prime }\left (x \right )+\left (x^{2}-2 x +2\right ) u \left (x \right ) = 0 \end {align*}
Which is now solved for \(u\). In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {u \left (x^{2}-2 x +2\right )}{x \left (x -1\right )} \end {align*}
Where \(f(x)=\frac {x^{2}-2 x +2}{x \left (x -1\right )}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= \frac {x^{2}-2 x +2}{x \left (x -1\right )} \,d x\\ \int { \frac {1}{u} \,du} &= \int {\frac {x^{2}-2 x +2}{x \left (x -1\right )} \,d x}\\ \ln \left (u \right )&=x -2 \ln \left (x \right )+\ln \left (x -1\right )+c_{1}\\ u&={\mathrm e}^{x -2 \ln \left (x \right )+\ln \left (x -1\right )+c_{1}}\\ &=c_{1} {\mathrm e}^{x -2 \ln \left (x \right )+\ln \left (x -1\right )} \end {align*}
Which simplifies to \[ u \left (x \right ) = c_{1} \left (\frac {{\mathrm e}^{x}}{x}-\frac {{\mathrm e}^{x}}{x^{2}}\right ) \] The ode for \(v\) now becomes \begin {align*} v' &= u\\ &=c_{1} \left (\frac {{\mathrm e}^{x}}{x}-\frac {{\mathrm e}^{x}}{x^{2}}\right ) \end {align*}
Which is now solved for \(v\). Integrating both sides gives \begin {align*} v \left (x \right ) &= \int { \frac {c_{1} {\mathrm e}^{x} \left (x -1\right )}{x^{2}}\,\mathop {\mathrm {d}x}}\\ &= \frac {{\mathrm e}^{x} c_{1}}{x}+c_{2} \end {align*}
Therefore the homogeneous solution is \begin {align*} y_h(x) &= B v\\ &= \left (x\right ) \left (\frac {{\mathrm e}^{x} c_{1}}{x}+c_{2}\right ) \\ &= {\mathrm e}^{x} c_{1} +c_{2} x \end {align*}
And now the particular solution \(y_p(x)\) will be found. The particular solution \(y_p\) can be found using either the method of undetermined coefficients, or the method of variation of parameters. The method of variation of parameters will be used as it is more general and can be used when the coefficients of the ODE depend on \(x\) as well. Let \begin{equation} \tag{1} y_p(x) = u_1 y_1 + u_2 y_2 \end{equation} Where \(u_1,u_2\) to be determined, and \(y_1,y_2\) are the two basis solutions (the two linearly independent solutions of the homogeneous ODE) found earlier when solving the homogeneous ODE as \begin{align*} y_1 &= x \\ y_2 &= {\mathrm e}^{x} \\ \end{align*} In the Variation of parameters \(u_1,u_2\) are found using \begin{align*} \tag{2} u_1 &= -\int \frac {y_2 f(x)}{a W(x)} \\ \tag{3} u_2 &= \int \frac {y_1 f(x)}{a W(x)} \\ \end{align*} Where \(W(x)\) is the Wronskian and \(a\) is the coefficient in front of \(y''\) in the given ODE. The Wronskian is given by \(W= \begin {vmatrix} y_1 & y_{2} \\ y_{1}^{\prime } & y_{2}^{\prime } \end {vmatrix} \). Hence \[ W = \begin {vmatrix} x & {\mathrm e}^{x} \\ \frac {d}{dx}\left (x\right ) & \frac {d}{dx}\left ({\mathrm e}^{x}\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} x & {\mathrm e}^{x} \\ 1 & {\mathrm e}^{x} \end {vmatrix} \] Therefore \[ W = \left (x\right )\left ({\mathrm e}^{x}\right ) - \left ({\mathrm e}^{x}\right )\left (1\right ) \] Which simplifies to \[ W = {\mathrm e}^{x} x -{\mathrm e}^{x} \] Which simplifies to \[ W = \left (x -1\right ) {\mathrm e}^{x} \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {\left (x -1\right )^{2} {\mathrm e}^{2 x}}{\left (1-x \right ) \left (x -1\right ) {\mathrm e}^{x}}\,dx \] Which simplifies to \[ u_1 = - \int -{\mathrm e}^{x}d x \] Hence \[ u_1 = {\mathrm e}^{x} \] And Eq. (3) becomes \[ u_2 = \int \frac {x \left (x -1\right )^{2} {\mathrm e}^{x}}{\left (1-x \right ) \left (x -1\right ) {\mathrm e}^{x}}\,dx \] Which simplifies to \[ u_2 = \int -x d x \] Hence \[ u_2 = -\frac {x^{2}}{2} \] Therefore the particular solution, from equation (1) is \[ y_p(x) = {\mathrm e}^{x} x -\frac {{\mathrm e}^{x} x^{2}}{2} \] Which simplifies to \[ y_p(x) = -\frac {{\mathrm e}^{x} x \left (x -2\right )}{2} \] Hence the complete solution is \begin {align*} y(x) &= y_h + y_p \\ &= \left ({\mathrm e}^{x} c_{1} +c_{2} x\right ) + \left (-\frac {{\mathrm e}^{x} x \left (x -2\right )}{2}\right )\\ &= \frac {\left (-x^{2}+2 c_{1} +2 x \right ) {\mathrm e}^{x}}{2}+c_{2} x \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = \frac {\left (-x^{2}+2 c_{1} +2 x \right ) {\mathrm e}^{x}}{2}+c_{2} x \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = 0\) and \(x = -\infty \) in the above gives \begin {align*} 0 = -\operatorname {signum}\left (c_{2} \right ) \infty \tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \frac {\left (-2 x +2\right ) {\mathrm e}^{x}}{2}+\frac {\left (-x^{2}+2 c_{1} +2 x \right ) {\mathrm e}^{x}}{2}+c_{2} \end {align*}
substituting \(y^{\prime } = 1\) and \(x = 0\) in the above gives \begin {align*} 1 = 1+c_{1} +c_{2}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Warning, unable to solve for constants of integrations.
Verification of solutions N/A
Writing the ode as \begin {align*} \left (1-x \right ) y^{\prime \prime }+x y^{\prime }-y &= 0 \tag {1} \\ A y^{\prime \prime } + B y^{\prime } + C y &= 0 \tag {2} \end {align*}
Comparing (1) and (2) shows that \begin {align*} A &= 1-x \\ B &= x\tag {3} \\ C &= -1 \end {align*}
Applying the Liouville transformation on the dependent variable gives \begin {align*} z(x) &= y e^{\int \frac {B}{2 A} \,dx} \end {align*}
Then (2) becomes \begin {align*} z''(x) = r z(x)\tag {4} \end {align*}
Where \(r\) is given by \begin {align*} r &= \frac {s}{t}\tag {5} \\ &= \frac {2 A B' - 2 B A' + B^2 - 4 A C}{4 A^2} \end {align*}
Substituting the values of \(A,B,C\) from (3) in the above and simplifying gives \begin {align*} r &= \frac {x^{2}-4 x +6}{4 \left (x -1\right )^{2}}\tag {6} \end {align*}
Comparing the above to (5) shows that \begin {align*} s &= x^{2}-4 x +6\\ t &= 4 \left (x -1\right )^{2} \end {align*}
Therefore eq. (4) becomes \begin {align*} z''(x) &= \left ( \frac {x^{2}-4 x +6}{4 \left (x -1\right )^{2}}\right ) z(x)\tag {7} \end {align*}
Equation (7) is now solved. After finding \(z(x)\) then \(y\) is found using the inverse transformation \begin {align*} y &= z \left (x \right ) e^{-\int \frac {B}{2 A} \,dx} \end {align*}
The first step is to determine the case of Kovacic algorithm this ode belongs to. There are 3 cases depending on the order of poles of \(r\) and the order of \(r\) at \(\infty \). The following table summarizes these cases.
| Case |
Allowed pole order for \(r\) |
Allowed value for \(\mathcal {O}(\infty )\) |
| 1 |
\(\left \{ 0,1,2,4,6,8,\cdots \right \} \) |
\(\left \{ \cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \right \} \) |
|
2 |
Need to have at least one pole that is either order \(2\) or odd order greater than \(2\). Any other pole order is allowed as long as the above condition is satisfied. Hence the following set of pole orders are all allowed. \(\{1,2\}\),\(\{1,3\}\),\(\{2\}\),\(\{3\}\),\(\{3,4\}\),\(\{1,2,5\}\). |
no condition |
| 3 |
\(\left \{ 1,2\right \} \) |
\(\left \{ 2,3,4,5,6,7,\cdots \right \} \) |
The order of \(r\) at \(\infty \) is the degree of \(t\) minus the degree of \(s\). Therefore \begin {align*} O\left (\infty \right ) &= \text {deg}(t) - \text {deg}(s) \\ &= 2 - 2 \\ &= 0 \end {align*}
The poles of \(r\) in eq. (7) and the order of each pole are determined by solving for the roots of \(t=4 \left (x -1\right )^{2}\). There is a pole at \(x=1\) of order \(2\). Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is \(0\) then the necessary conditions for case one are met. Since there is a pole of order \(2\) then necessary conditions for case two are met. Therefore \begin {align*} L &= [1, 2] \end {align*}
Attempting to find a solution using case \(n=1\).
Looking at poles of order 2. The partial fractions decomposition of \(r\) is \[ r = \frac {1}{4}+\frac {3}{4 \left (x -1\right )^{2}}-\frac {1}{2 \left (x -1\right )} \] For the pole at \(x=1\) let \(b\) be the coefficient of \(\frac {1}{ \left (x -1\right )^{2}}\) in the partial fractions decomposition of \(r\) given above. Therefore \(b={\frac {3}{4}}\). Hence \begin {alignat*} {2} [\sqrt r]_c &= 0 \\ \alpha _c^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= {\frac {3}{2}}\\ \alpha _c^{-} &= \frac {1}{2} - \sqrt {1+4 b} &&= -{\frac {1}{2}} \end {alignat*}
Since the order of \(r\) at \(\infty \) is \(O_r(\infty ) = 0\) then \begin {alignat*} {3} v &= \frac {-O_r(\infty )}{2} &&= \frac {0}{2} &&= 0 \end {alignat*}
\([\sqrt r]_\infty \) is the sum of terms involving \(x^i\) for \(0\leq i \leq v\) in the Laurent series for \(\sqrt r\) at \(\infty \). Therefore \begin {align*} [\sqrt r]_\infty &= \sum _{i=0}^{v} a_i x^i \\ &= \sum _{i=0}^{0} a_i x^i \tag {8} \end {align*}
Let \(a\) be the coefficient of \(x^v=x^0\) in the above sum. The Laurent series of \(\sqrt r\) at \(\infty \) is \[ \sqrt r \approx \frac {1}{2}-\frac {1}{2 x}+\frac {1}{x^{3}}+\frac {11}{4 x^{4}}+\frac {21}{4 x^{5}}+\frac {15}{2 x^{6}}+\frac {6}{x^{7}}-\frac {117}{16 x^{8}} + \dots \tag {9} \] Comparing Eq. (9) with Eq. (8) shows that \[ a = {\frac {1}{2}} \] From Eq. (9) the sum up to \(v=0\) gives \begin {align*} [\sqrt r]_\infty &= \sum _{i=0}^{0} a_i x^i \\ &= {\frac {1}{2}} \tag {10} \end {align*}
Now we need to find \(b\), where \(b\) be the coefficient of \(x^{v-1} = x^{-1}=\frac {1}{x}\) in \(r\) minus the coefficient of same term but in \(\left ( [\sqrt r]_\infty \right )^2 \) where \([\sqrt r]_\infty \) was found above in Eq (10). Hence \[ \left ( [\sqrt r]_\infty \right )^2 = {\frac {1}{4}} \] This shows that the coefficient of \(\frac {1}{x}\) in the above is \(0\). Now we need to find the coefficient of \(\frac {1}{x}\) in \(r\). How this is done depends on if \(v=0\) or not. Since \(v=0\) then starting from \(r=\frac {s}{t}\) and doing long division in the form \[ r = Q + \frac {R}{t} \] Where \(Q\) is the quotient and \(R\) is the remainder. Then the coefficient of \(\frac {1}{x}\) in \(r\) will be the coefficient in \(R\) of the term in \(x\) of degree of \(t\) minus one, divided by the leading coefficient in \(t\). Doing long division gives \begin {align*} r &= \frac {s}{t} \\ &= \frac {x^{2}-4 x +6}{4 x^{2}-8 x +4} \\ &= Q + \frac {R}{4 x^{2}-8 x +4}\\ &= \left ({\frac {1}{4}}\right ) + \left ( \frac {-2 x +5}{4 x^{2}-8 x +4}\right ) \\ &= \frac {1}{4}+\frac {-2 x +5}{4 x^{2}-8 x +4} \end {align*}
Since the degree of \(t\) is \(2\), then we see that the coefficient of the term \(x\) in the remainder \(R\) is \(-2\). Dividing this by leading coefficient in \(t\) which is \(4\) gives \(-{\frac {1}{2}}\). Now \(b\) can be found. \begin {align*} b &= \left (-{\frac {1}{2}}\right )-\left (0\right )\\ &= -{\frac {1}{2}} \end {align*}
Hence \begin {alignat*} {3} [\sqrt r]_\infty &= {\frac {1}{2}}\\ \alpha _{\infty }^{+} &= \frac {1}{2} \left ( \frac {b}{a} - v \right ) &&= \frac {1}{2} \left ( \frac {-{\frac {1}{2}}}{{\frac {1}{2}}} - 0 \right ) &&= -{\frac {1}{2}}\\ \alpha _{\infty }^{-} &= \frac {1}{2} \left ( -\frac {b}{a} - v \right ) &&= \frac {1}{2} \left ( -\frac {-{\frac {1}{2}}}{{\frac {1}{2}}} - 0 \right ) &&= {\frac {1}{2}} \end {alignat*}
The following table summarizes the findings so far for poles and for the order of \(r\) at \(\infty \) where \(r\) is \[ r=\frac {x^{2}-4 x +6}{4 \left (x -1\right )^{2}} \]
| pole \(c\) location | pole order | \([\sqrt r]_c\) | \(\alpha _c^{+}\) | \(\alpha _c^{-}\) |
| \(1\) | \(2\) | \(0\) | \(\frac {3}{2}\) | \(-{\frac {1}{2}}\) |
| Order of \(r\) at \(\infty \) | \([\sqrt r]_\infty \) | \(\alpha _\infty ^{+}\) | \(\alpha _\infty ^{-}\) |
| \(0\) | \(\frac {1}{2}\) | \(-{\frac {1}{2}}\) | \(\frac {1}{2}\) |
Now that the all \([\sqrt r]_c\) and its associated \(\alpha _c^{\pm }\) have been determined for all the poles in the set \(\Gamma \) and \([\sqrt r]_\infty \) and its associated \(\alpha _\infty ^{\pm }\) have also been found, the next step is to determine possible non negative integer \(d\) from these using \begin {align*} d &= \alpha _\infty ^{s(\infty )} - \sum _{c \in \Gamma } \alpha _c^{s(c)} \end {align*}
Where \(s(c)\) is either \(+\) or \(-\) and \(s(\infty )\) is the sign of \(\alpha _\infty ^{\pm }\). This is done by trial over all set of families \(s=(s(c))_{c \in \Gamma \cup {\infty }}\) until such \(d\) is found to work in finding candidate \(\omega \). Trying \(\alpha _\infty ^{+} = -{\frac {1}{2}}\) then \begin {align*} d &= \alpha _\infty ^{+} - \left ( \alpha _{c_1}^{-} \right ) \\ &= -{\frac {1}{2}} - \left ( -{\frac {1}{2}} \right ) \\ &= 0 \end {align*}
Since \(d\) an integer and \(d \geq 0\) then it can be used to find \(\omega \) using \begin {align*} \omega &= \sum _{c \in \Gamma } \left ( s(c) [\sqrt r]_c + \frac {\alpha _c^{s(c)}}{x-c} \right ) + s(\infty ) [\sqrt r]_\infty \end {align*}
Substituting the above values in the above results in \begin {align*} \omega &= \left ( (-)[\sqrt r]_{c_1} + \frac { \alpha _{c_1}^{-} }{x- c_1}\right ) + (+) [\sqrt r]_\infty \\ &= -\frac {1}{2 \left (x -1\right )} + \left ( {\frac {1}{2}} \right ) \\ &= -\frac {1}{2 \left (x -1\right )}+\frac {1}{2}\\ &= \frac {x -2}{2 x -2} \end {align*}
Now that \(\omega \) is determined, the next step is find a corresponding minimal polynomial \(p(x)\) of degree \(d=0\) to solve the ode. The polynomial \(p(x)\) needs to satisfy the equation \begin {align*} p'' + 2 \omega p' + \left ( \omega ' +\omega ^2 -r\right ) p = 0 \tag {1A} \end {align*}
Let \begin {align*} p(x) &= 1\tag {2A} \end {align*}
Substituting the above in eq. (1A) gives \begin {align*} \left (0\right ) + 2 \left (-\frac {1}{2 \left (x -1\right )}+\frac {1}{2}\right ) \left (0\right ) + \left ( \left (\frac {1}{2 \left (x -1\right )^{2}}\right ) + \left (-\frac {1}{2 \left (x -1\right )}+\frac {1}{2}\right )^2 - \left (\frac {x^{2}-4 x +6}{4 \left (x -1\right )^{2}}\right ) \right ) &= 0\\ 0 = 0 \end {align*}
The equation is satisfied since both sides are zero. Therefore the first solution to the ode \(z'' = r z\) is \begin {align*} z_1(x) &= p e^{ \int \omega \,dx} \\ &= {\mathrm e}^{\int \left (-\frac {1}{2 \left (x -1\right )}+\frac {1}{2}\right )d x}\\ &= \frac {{\mathrm e}^{\frac {x}{2}}}{\sqrt {x -1}} \end {align*}
The first solution to the original ode in \(y\) is found from \begin{align*} y_1 &= z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dx} \\ &= z_1 e^{ -\int \frac {1}{2} \frac {x}{1-x} \,dx} \\ &= z_1 e^{\frac {x}{2}+\frac {\ln \left (x -1\right )}{2}} \\ &= z_1 \left (\sqrt {x -1}\, {\mathrm e}^{\frac {x}{2}}\right ) \\ \end{align*} Which simplifies to \[ y_1 = {\mathrm e}^{x} \] The second solution \(y_2\) to the original ode is found using reduction of order \[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,dx}}{y_1^2} \,dx \] Substituting gives \begin{align*} y_2 &= y_1 \int \frac { e^{\int -\frac {x}{1-x} \,dx}}{\left (y_1\right )^2} \,dx \\ &= y_1 \int \frac { e^{x +\ln \left (x -1\right )}}{\left (y_1\right )^2} \,dx \\ &= y_1 \left (-x \,{\mathrm e}^{-x}\right ) \\ \end{align*} Therefore the solution is
\begin{align*} y &= c_{1} y_1 + c_{2} y_2 \\ &= c_{1} \left ({\mathrm e}^{x}\right ) + c_{2} \left ({\mathrm e}^{x}\left (-x \,{\mathrm e}^{-x}\right )\right ) \\ \end{align*} This is second order nonhomogeneous ODE. Let the solution be \[ y = y_h + y_p \] Where \(y_h\) is the solution to the homogeneous ODE \( A y''(x) + B y'(x) + C y(x) = 0\), and \(y_p\) is a particular solution to the nonhomogeneous ODE \(A y''(x) + B y'(x) + C y(x) = f(x)\). \(y_h\) is the solution to \[ \left (1-x \right ) y^{\prime \prime }+x y^{\prime }-y = 0 \] The homogeneous solution is found using the Kovacic algorithm which results in \[ y_h = {\mathrm e}^{x} c_{1} -c_{2} x \] The particular solution \(y_p\) can be found using either the method of undetermined coefficients, or the method of variation of parameters. The method of variation of parameters will be used as it is more general and can be used when the coefficients of the ODE depend on \(x\) as well. Let \begin{equation} \tag{1} y_p(x) = u_1 y_1 + u_2 y_2 \end{equation} Where \(u_1,u_2\) to be determined, and \(y_1,y_2\) are the two basis solutions (the two linearly independent solutions of the homogeneous ODE) found earlier when solving the homogeneous ODE as \begin{align*} y_1 &= {\mathrm e}^{x} \\ y_2 &= -x \\ \end{align*} In the Variation of parameters \(u_1,u_2\) are found using \begin{align*} \tag{2} u_1 &= -\int \frac {y_2 f(x)}{a W(x)} \\ \tag{3} u_2 &= \int \frac {y_1 f(x)}{a W(x)} \\ \end{align*} Where \(W(x)\) is the Wronskian and \(a\) is the coefficient in front of \(y''\) in the given ODE. The Wronskian is given by \(W= \begin {vmatrix} y_1 & y_{2} \\ y_{1}^{\prime } & y_{2}^{\prime } \end {vmatrix} \). Hence \[ W = \begin {vmatrix} {\mathrm e}^{x} & -x \\ \frac {d}{dx}\left ({\mathrm e}^{x}\right ) & \frac {d}{dx}\left (-x\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} {\mathrm e}^{x} & -x \\ {\mathrm e}^{x} & -1 \end {vmatrix} \] Therefore \[ W = \left ({\mathrm e}^{x}\right )\left (-1\right ) - \left (-x\right )\left ({\mathrm e}^{x}\right ) \] Which simplifies to \[ W = {\mathrm e}^{x} x -{\mathrm e}^{x} \] Which simplifies to \[ W = \left (x -1\right ) {\mathrm e}^{x} \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {-x \left (x -1\right )^{2} {\mathrm e}^{x}}{\left (1-x \right ) \left (x -1\right ) {\mathrm e}^{x}}\,dx \] Which simplifies to \[ u_1 = - \int x d x \] Hence \[ u_1 = -\frac {x^{2}}{2} \] And Eq. (3) becomes \[ u_2 = \int \frac {\left (x -1\right )^{2} {\mathrm e}^{2 x}}{\left (1-x \right ) \left (x -1\right ) {\mathrm e}^{x}}\,dx \] Which simplifies to \[ u_2 = \int -{\mathrm e}^{x}d x \] Hence \[ u_2 = -{\mathrm e}^{x} \] Therefore the particular solution, from equation (1) is \[ y_p(x) = {\mathrm e}^{x} x -\frac {{\mathrm e}^{x} x^{2}}{2} \] Which simplifies to \[ y_p(x) = -\frac {{\mathrm e}^{x} x \left (x -2\right )}{2} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left ({\mathrm e}^{x} c_{1} -c_{2} x\right ) + \left (-\frac {{\mathrm e}^{x} x \left (x -2\right )}{2}\right ) \\ \end{align*} Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = {\mathrm e}^{x} c_{1} -c_{2} x -\frac {{\mathrm e}^{x} x \left (x -2\right )}{2} \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = 0\) and \(x = -\infty \) in the above gives \begin {align*} 0 = \operatorname {signum}\left (c_{2} \right ) \infty \tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = {\mathrm e}^{x} c_{1} -c_{2} -\frac {{\mathrm e}^{x} x \left (x -2\right )}{2}-\frac {\left (x -2\right ) {\mathrm e}^{x}}{2}-\frac {{\mathrm e}^{x} x}{2} \end {align*}
substituting \(y^{\prime } = 1\) and \(x = 0\) in the above gives \begin {align*} 1 = c_{1} -c_{2} +1\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Warning, unable to solve for constants of integrations.
Verification of solutions N/A
Maple trace Kovacic algorithm successful
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable trying differential order: 2; linear nonhomogeneous with symmetry [0,1] trying a double symmetry of the form [xi=0, eta=F(x)] -> Try solving first the homogeneous part of the ODE checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm A Liouvillian solution exists Reducible group (found an exponential solution) Reducible group (found another exponential solution) <- Kovacics algorithm successful <- solving first the homogeneous part of the ODE successful`
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 16
dsolve([(1-x)*diff(y(x),x$2)+x*diff(y(x),x)-y(x)=(x-1)^2*exp(x),y(-infinity) = 0, D(y)(0) = 1],y(x), singsol=all)
\[ y \left (x \right ) = -\frac {x \left (x -2\right ) {\mathrm e}^{x}}{2} \]
✓ Solution by Mathematica
Time used: 0.083 (sec). Leaf size: 16
DSolve[{(1-x)*y''[x]+x*y'[x]-y[x]==(x-1)^2*Exp[x],{y[-Infinity]==0,y'[0]==1}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -\frac {1}{2} e^x (x-2) x \]