Internal problem ID [1185]
Internal file name [OUTPUT/1186_Sunday_June_05_2022_02_04_29_AM_25254914/index.tex
]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 5 linear second order equations. Section 5.7 Variation of Parameters. Page
262
Problem number: 31.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "kovacic", "second_order_change_of_variable_on_x_method_2", "linear_second_order_ode_solved_by_an_integrating_factor", "second_order_ode_non_constant_coeff_transformation_on_B"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {\left (x -1\right )^{2} y^{\prime \prime }-2 \left (x -1\right ) y^{\prime }+2 y=\left (x -1\right )^{2}} \] With initial conditions \begin {align*} [y \left (0\right ) = 3, y^{\prime }\left (0\right ) = -6] \end {align*}
This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime \prime } + p(x)y^{\prime } + q(x) y &= F \end {align*}
Where here \begin {align*} p(x) &=\frac {-2 x +2}{\left (x -1\right )^{2}}\\ q(x) &=\frac {2}{\left (x -1\right )^{2}}\\ F &=1 \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }+\frac {\left (-2 x +2\right ) y^{\prime }}{\left (x -1\right )^{2}}+\frac {2 y}{\left (x -1\right )^{2}} = 1 \end {align*}
The domain of \(p(x)=\frac {-2 x +2}{\left (x -1\right )^{2}}\) is \[
\{x <1\boldsymbol {\lor }1
The ode satisfies this form \[ y^{\prime \prime }+p \left (x \right ) y^{\prime }+\frac {\left (p \left (x \right )^{2}+p^{\prime }\left (x \right )\right ) y}{2} = f \left (x \right ) \] Where \( p(x) = -\frac {2}{x -1}\). Therefore, there is an integrating factor given by
\begin {align*} M(x) &= e^{\frac {1}{2} \int p \, dx} \\ &= e^{ \int -\frac {2}{x -1} \, dx} \\ &= \frac {1}{x -1} \end {align*}
Multiplying both sides of the ODE by the integrating factor \(M(x)\) makes the left side of
the ODE a complete differential \begin{align*}
\left ( M(x) y \right )'' &= \frac {1}{x -1} \\
\left ( \frac {y}{x -1} \right )'' &= \frac {1}{x -1} \\
\end{align*} Integrating once gives \[ \left ( \frac {y}{x -1} \right )' = \ln \left (x -1\right )+c_{1} \] Integrating again gives \[ \left ( \frac {y}{x -1} \right ) = \ln \left (x -1\right ) \left (x -1\right )+1+\left (c_{1} -1\right ) x +c_{2} \]
Hence the solution is \begin{align*}
y &= \frac {\ln \left (x -1\right ) \left (x -1\right )+1+\left (c_{1} -1\right ) x +c_{2}}{\frac {1}{x -1}} \\
\end{align*} Or \[
y = x^{2} \ln \left (x -1\right )-x^{2}-2 x \ln \left (x -1\right )+\left (x^{2}-x \right ) c_{1} +\left (x -1\right ) c_{2} +2 x +\ln \left (x -1\right )-1
\] Initial conditions are used to solve for the constants of
integration.
Looking at the above solution \begin {align*} y = x^{2} \ln \left (x -1\right )-x^{2}-2 x \ln \left (x -1\right )+\left (x^{2}-x \right ) c_{1} +\left (x -1\right ) c_{2} +2 x +\ln \left (x -1\right )-1 \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 \(y = 3\) and \(x = 0\) in the above gives
\begin {align*} 3 = i \pi -c_{2} -1\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = 2 x \ln \left (x -1\right )+\frac {x^{2}}{x -1}-2 x -2 \ln \left (x -1\right )-\frac {2 x}{x -1}+\left (2 x -1\right ) c_{1} +c_{2} +2+\frac {1}{x -1} \end {align*}
substituting \(y^{\prime } = -6\) and \(x = 0\) in the above gives \begin {align*} -6 = -2 i \pi -c_{1} +c_{2} +1\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&=-i \pi +3\\ c_{2}&=i \pi -4 \end {align*}
Substituting these values back in above solution results in \begin {align*} y = x^{2} \ln \left (x -1\right )+2 x^{2}-2 x \ln \left (x -1\right )-5 x -i \pi \,x^{2}+2 i x \pi -i \pi +3+\ln \left (x -1\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= -\left (x -1\right ) \left (\left (1-x \right ) \ln \left (x -1\right )+i x \pi -i \pi -2 x +3\right ) \\
\end{align*} Verification of solutions
\[
y = -\left (x -1\right ) \left (\left (1-x \right ) \ln \left (x -1\right )+i x \pi -i \pi -2 x +3\right )
\] Verified OK.
This is second order non-homogeneous 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 non-homogeneous ODE \(A y''(x) + B y'(x) + C y(x) = f(x)\). \(y_h\) is the
solution to \[
\left (x -1\right )^{2} y^{\prime \prime }+\left (-2 x +2\right ) y^{\prime }+2 y = 0
\] In normal form the ode \begin {align*} \left (x -1\right )^{2} y^{\prime \prime }+\left (-2 x +2\right ) y^{\prime }+2 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 {2}{1-x}\\ q \left (x \right )&=\frac {2}{\left (x -1\right )^{2}} \end {align*}
Applying change of variables \(\tau = g \left (x \right )\) to (2) gives \begin {align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+p_{1} \left (\frac {d}{d \tau }y \left (\tau \right )\right )+q_{1} y \left (\tau \right )&=0 \tag {3} \end {align*}
Where \(\tau \) is the new independent variable, and \begin {align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {5} \end {align*}
Let \(p_{1} = 0\). Eq (4) simplifies to \begin {align*} \tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )&=0 \end {align*}
This ode is solved resulting in \begin {align*} \tau &= \int {\mathrm e}^{-\left (\int p \left (x \right )d x \right )}d x\\ &= \int {\mathrm e}^{-\left (\int \frac {2}{1-x}d x \right )}d x\\ &= \int e^{2 \ln \left (1-x \right )} \,dx\\ &= \int \left (x -1\right )^{2}d x\\ &= \frac {\left (x -1\right )^{3}}{3}\tag {6} \end {align*}
Using (6) to evaluate \(q_{1}\) from (5) gives \begin {align*} q_{1} \left (\tau \right ) &= \frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\\ &= \frac {\frac {2}{\left (x -1\right )^{2}}}{\left (x -1\right )^{4}}\\ &= \frac {2}{\left (x -1\right )^{6}}\tag {7} \end {align*}
Substituting the above in (3) and noting that now \(p_{1} = 0\) results in \begin {align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+q_{1} y \left (\tau \right )&=0 \\ \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+\frac {2 y \left (\tau \right )}{\left (x -1\right )^{6}}&=0 \\ \end {align*}
But in terms of \(\tau \) \begin {align*} \frac {2}{\left (x -1\right )^{6}}&=\frac {2}{9 \tau ^{2}} \end {align*}
Hence the above ode becomes \begin {align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+\frac {2 y \left (\tau \right )}{9 \tau ^{2}}&=0 \end {align*}
The above ode is now solved for \(y \left (\tau \right )\). The ode can be written as \[ 9 \left (\frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )\right ) \tau ^{2}+2 y \left (\tau \right ) = 0 \] Which shows it is a
Euler ODE. This is Euler second order ODE. Let the solution be \(y \left (\tau \right ) = \tau ^r\), then \(y'=r \tau ^{r-1}\) and \(y''=r(r-1) \tau ^{r-2}\).
Substituting these back into the given ODE gives \[ 9 \tau ^{2}(r(r-1))\tau ^{r-2}+0 r \tau ^{r-1}+2 \tau ^{r} = 0 \] Simplifying gives \[ 9 r \left (r -1\right )\tau ^{r}+0\,\tau ^{r}+2 \tau ^{r} = 0 \] Since \(\tau ^{r}\neq 0\) then dividing
throughout by \(\tau ^{r}\) gives \[ 9 r \left (r -1\right )+0+2 = 0 \] Or \[ 9 r^{2}-9 r +2 = 0 \tag {1} \] Equation (1) is the characteristic equation. Its roots
determine the form of the general solution. Using the quadratic equation the roots are
\begin {align*} r_1 &= {\frac {1}{3}}\\ r_2 &= {\frac {2}{3}} \end {align*}
Since the roots are real and distinct, then the general solution is \[ y \left (\tau \right )= c_{1} y_1 + c_{2} y_2 \] Where \(y_1 = \tau ^{r_1}\) and \(y_2 = \tau ^{r_2} \).
Hence \[ y \left (\tau \right ) = c_{1} \tau ^{\frac {1}{3}}+c_{2} \tau ^{\frac {2}{3}} \] The above solution is now transformed back to \(y\) using (6) which results in
\begin {align*} y &= \frac {c_{1} 3^{\frac {2}{3}} \left (\left (x -1\right )^{3}\right )^{\frac {1}{3}}}{3}+\frac {c_{2} 3^{\frac {1}{3}} \left (\left (x -1\right )^{3}\right )^{\frac {2}{3}}}{3} \end {align*}
Therefore the homogeneous solution \(y_h\) is \[
y_h = \frac {c_{1} 3^{\frac {2}{3}} \left (\left (x -1\right )^{3}\right )^{\frac {1}{3}}}{3}+\frac {c_{2} 3^{\frac {1}{3}} \left (\left (x -1\right )^{3}\right )^{\frac {2}{3}}}{3}
\] 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 &= \left (x^{3}-3 x^{2}+3 x -1\right )^{\frac {1}{3}} \\
y_2 &= \left (x^{3}-3 x^{2}+3 x -1\right )^{\frac {2}{3}} \\
\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 \[ W = \begin {vmatrix} \left (x^{3}-3 x^{2}+3 x -1\right )^{\frac {1}{3}} & \left (x^{3}-3 x^{2}+3 x -1\right )^{\frac {2}{3}} \\ \frac {d}{dx}\left (\left (x^{3}-3 x^{2}+3 x -1\right )^{\frac {1}{3}}\right ) & \frac {d}{dx}\left (\left (x^{3}-3 x^{2}+3 x -1\right )^{\frac {2}{3}}\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} \left (x^{3}-3 x^{2}+3 x -1\right )^{\frac {1}{3}} & \left (x^{3}-3 x^{2}+3 x -1\right )^{\frac {2}{3}} \\ \frac {3 x^{2}-6 x +3}{3 \left (x^{3}-3 x^{2}+3 x -1\right )^{\frac {2}{3}}} & \frac {2 x^{2}-4 x +2}{\left (x^{3}-3 x^{2}+3 x -1\right )^{\frac {1}{3}}} \end {vmatrix} \]
Therefore \[
W = \left (\left (x^{3}-3 x^{2}+3 x -1\right )^{\frac {1}{3}}\right )\left (\frac {2 x^{2}-4 x +2}{\left (x^{3}-3 x^{2}+3 x -1\right )^{\frac {1}{3}}}\right ) - \left (\left (x^{3}-3 x^{2}+3 x -1\right )^{\frac {2}{3}}\right )\left (\frac {3 x^{2}-6 x +3}{3 \left (x^{3}-3 x^{2}+3 x -1\right )^{\frac {2}{3}}}\right )
\] Which simplifies to \[
W = x^{2}-2 x +1
\] Which simplifies to \[
W = \left (x -1\right )^{2}
\] Therefore Eq. (2) becomes \[
u_1 = -\int \frac {\left (x^{3}-3 x^{2}+3 x -1\right )^{\frac {2}{3}} \left (x -1\right )^{2}}{\left (x -1\right )^{4}}\,dx
\]
Which simplifies to \[
u_1 = - \int \frac {\left (\left (x -1\right )^{3}\right )^{\frac {2}{3}}}{\left (x -1\right )^{2}}d x
\] Hence \[
u_1 = -\frac {\left (\left (x -1\right )^{3}\right )^{\frac {2}{3}} x}{\left (x -1\right )^{2}}
\] And Eq. (3) becomes \[
u_2 = \int \frac {\left (x^{3}-3 x^{2}+3 x -1\right )^{\frac {1}{3}} \left (x -1\right )^{2}}{\left (x -1\right )^{4}}\,dx
\] Which simplifies to \[
u_2 = \int \frac {\left (\left (x -1\right )^{3}\right )^{\frac {1}{3}}}{\left (x -1\right )^{2}}d x
\] Hence \[
u_2 = \frac {\left (\left (x -1\right )^{3}\right )^{\frac {1}{3}} \ln \left (x -1\right )}{x -1}
\]
Therefore the particular solution, from equation (1) is \[
y_p(x) = -\frac {\left (\left (x -1\right )^{3}\right )^{\frac {2}{3}} x \left (x^{3}-3 x^{2}+3 x -1\right )^{\frac {1}{3}}}{\left (x -1\right )^{2}}+\frac {\left (\left (x -1\right )^{3}\right )^{\frac {1}{3}} \ln \left (x -1\right ) \left (x^{3}-3 x^{2}+3 x -1\right )^{\frac {2}{3}}}{x -1}
\] Which simplifies to \[
y_p(x) = \left (x -1\right ) \left (\ln \left (x -1\right ) \left (x -1\right )-x \right )
\] Therefore
the general solution is \begin{align*}
y &= y_h + y_p \\
&= \left (\frac {c_{1} 3^{\frac {2}{3}} \left (\left (x -1\right )^{3}\right )^{\frac {1}{3}}}{3}+\frac {c_{2} 3^{\frac {1}{3}} \left (\left (x -1\right )^{3}\right )^{\frac {2}{3}}}{3}\right ) + \left (\left (x -1\right ) \left (\ln \left (x -1\right ) \left (x -1\right )-x \right )\right ) \\
\end{align*} Initial conditions are used to solve for the constants of
integration.
Looking at the above solution \begin {align*} y = \frac {c_{1} 3^{\frac {2}{3}} \left (\left (x -1\right )^{3}\right )^{\frac {1}{3}}}{3}+\frac {c_{2} 3^{\frac {1}{3}} \left (\left (x -1\right )^{3}\right )^{\frac {2}{3}}}{3}+\left (x -1\right ) \left (\ln \left (x -1\right ) \left (x -1\right )-x \right ) \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 \(y = 3\) and \(x = 0\) in the above gives
\begin {align*} 3 = \frac {3^{\frac {2}{3}} c_{1}}{6}+\frac {i 3^{\frac {1}{6}} c_{1}}{2}-\frac {3^{\frac {1}{3}} c_{2}}{6}+\frac {i 3^{\frac {5}{6}} c_{2}}{6}+i \pi \tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \frac {c_{1} 3^{\frac {2}{3}} \left (x -1\right )^{2}}{3 \left (\left (x -1\right )^{3}\right )^{\frac {2}{3}}}+\frac {2 c_{2} 3^{\frac {1}{3}} \left (x -1\right )^{2}}{3 \left (\left (x -1\right )^{3}\right )^{\frac {1}{3}}}+2 \ln \left (x -1\right ) \left (x -1\right )-x \end {align*}
substituting \(y^{\prime } = -6\) and \(x = 0\) in the above gives \begin {align*} -6 = -\frac {3^{\frac {2}{3}} c_{1}}{6}-\frac {i 3^{\frac {1}{6}} c_{1}}{2}+\frac {3^{\frac {1}{3}} c_{2}}{3}-\frac {i 3^{\frac {5}{6}} c_{2}}{3}-2 i \pi \tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&=0\\ c_{2}&=-\frac {6 \left (3 i+\pi \right )}{3^{\frac {5}{6}}+i 3^{\frac {1}{3}}} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \frac {\ln \left (x -1\right ) 3^{\frac {5}{6}} x^{2}-2 \ln \left (x -1\right ) 3^{\frac {5}{6}} x -3^{\frac {5}{6}} x^{2}+\ln \left (x -1\right ) 3^{\frac {5}{6}}+3^{\frac {5}{6}} x +i \ln \left (x -1\right ) 3^{\frac {1}{3}} x^{2}-2 i \ln \left (x -1\right ) 3^{\frac {1}{3}} x -6 i \left (x^{3}-3 x^{2}+3 x -1\right )^{\frac {2}{3}} 3^{\frac {1}{3}}-i 3^{\frac {1}{3}} x^{2}-2 \pi \left (x^{3}-3 x^{2}+3 x -1\right )^{\frac {2}{3}} 3^{\frac {1}{3}}+i \ln \left (x -1\right ) 3^{\frac {1}{3}}+i 3^{\frac {1}{3}} x}{3^{\frac {5}{6}}+i 3^{\frac {1}{3}}} \end {align*}
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {-6 \,3^{\frac {1}{3}} \left (i+\frac {\pi }{3}\right ) \left (\left (x -1\right )^{3}\right )^{\frac {2}{3}}+\left (x -1\right ) \left (3^{\frac {5}{6}}+i 3^{\frac {1}{3}}\right ) \left (\ln \left (x -1\right ) \left (x -1\right )-x \right )}{3^{\frac {5}{6}}+i 3^{\frac {1}{3}}} \\
\end{align*} Verification of solutions
\[
y = \frac {-6 \,3^{\frac {1}{3}} \left (i+\frac {\pi }{3}\right ) \left (\left (x -1\right )^{3}\right )^{\frac {2}{3}}+\left (x -1\right ) \left (3^{\frac {5}{6}}+i 3^{\frac {1}{3}}\right ) \left (\ln \left (x -1\right ) \left (x -1\right )-x \right )}{3^{\frac {5}{6}}+i 3^{\frac {1}{3}}}
\] Verified OK.
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 &= \left (x -1\right )^{2}\\ B &= -2 x +2\\ C &= 2\\ F &= \left (x -1\right )^{2} \end {align*}
The above shows that for this ode \begin {align*} AB''+BB'+CB &= \left (\left (x -1\right )^{2}\right ) \left (0\right ) + \left (-2 x +2\right ) \left (-2\right ) + \left (2\right ) \left (-2 x +2\right ) \\ &=0 \end {align*}
Hence the ode in \(v\) given in (1) now simplifies to \begin {align*} -2 \left (x -1\right )^{3} v'' +\left ( 0\right ) v' & =0 \end {align*}
Now by applying \(v'=u\) the above becomes \begin {align*} -2 \left (x -1\right )^{3} u^{\prime }\left (x \right ) = 0 \end {align*}
Which is now solved for \(u\). Integrating both sides gives \begin {align*} u \left (x \right ) &= \int { 0\,\mathop {\mathrm {d}x}}\\ &= c_{1} \end {align*}
The ode for \(v\) now becomes \begin {align*} v' &= u\\ &=c_{1} \end {align*}
Which is now solved for \(v\). Integrating both sides gives \begin {align*} v \left (x \right ) &= \int { c_{1}\,\mathop {\mathrm {d}x}}\\ &= c_{1} x +c_{2} \end {align*}
Therefore the homogeneous solution is \begin {align*} y_h(x) &= B v\\ &= \left (-2 x +2\right ) \left (c_{1} x +c_{2}\right ) \\ &= -2 \left (x -1\right ) \left (c_{1} x +c_{2} \right ) \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 &= -2 x +2 \\
y_2 &= -2 x^{2}+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 \[ W = \begin {vmatrix} -2 x +2 & -2 x^{2}+2 x \\ \frac {d}{dx}\left (-2 x +2\right ) & \frac {d}{dx}\left (-2 x^{2}+2 x\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} -2 x +2 & -2 x^{2}+2 x \\ -2 & -4 x +2 \end {vmatrix} \] Therefore \[
W = \left (-2 x +2\right )\left (-4 x +2\right ) - \left (-2 x^{2}+2 x\right )\left (-2\right )
\] Which
simplifies to \[
W = 4 x^{2}-8 x +4
\] Which simplifies to \[
W = 4 \left (x -1\right )^{2}
\] Therefore Eq. (2) becomes \[
u_1 = -\int \frac {\left (-2 x^{2}+2 x \right ) \left (x -1\right )^{2}}{4 \left (x -1\right )^{4}}\,dx
\] Which simplifies to \[
u_1 = - \int -\frac {x}{2 x -2}d x
\]
Hence \[
u_1 = \frac {x}{2}+\frac {\ln \left (x -1\right )}{2}
\] And Eq. (3) becomes \[
u_2 = \int \frac {\left (x -1\right )^{2} \left (-2 x +2\right )}{4 \left (x -1\right )^{4}}\,dx
\] Which simplifies to \[
u_2 = \int -\frac {1}{2 x -2}d x
\] Hence \[
u_2 = -\frac {\ln \left (x -1\right )}{2}
\] Therefore the particular
solution, from equation (1) is \[
y_p(x) = \left (\frac {x}{2}+\frac {\ln \left (x -1\right )}{2}\right ) \left (-2 x +2\right )-\frac {\ln \left (x -1\right ) \left (-2 x^{2}+2 x \right )}{2}
\] Which simplifies to \[
y_p(x) = \left (x -1\right ) \left (\ln \left (x -1\right ) \left (x -1\right )-x \right )
\] Hence the complete solution is
\begin {align*} y(x) &= y_h + y_p \\ &= \left (-2 \left (x -1\right ) \left (c_{1} x +c_{2} \right )\right ) + \left (\left (x -1\right ) \left (\ln \left (x -1\right ) \left (x -1\right )-x \right )\right )\\ &= \left (x -1\right )^{2} \ln \left (x -1\right )-\left (x -1\right ) \left (2 c_{1} x +2 c_{2} +x \right ) \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = \left (x -1\right )^{2} \ln \left (x -1\right )-\left (x -1\right ) \left (2 c_{1} x +2 c_{2} +x \right ) \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 \(y = 3\) and \(x = 0\) in the above gives
\begin {align*} 3 = i \pi +2 c_{2}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = 2 \ln \left (x -1\right ) \left (x -1\right )-1-2 c_{1} x -2 c_{2} -\left (x -1\right ) \left (2 c_{1} +1\right ) \end {align*}
substituting \(y^{\prime } = -6\) and \(x = 0\) in the above gives \begin {align*} -6 = -2 i \pi +2 c_{1} -2 c_{2}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&=-\frac {3}{2}+\frac {i \pi }{2}\\ c_{2}&=\frac {3}{2}-\frac {i \pi }{2} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = x^{2} \ln \left (x -1\right )+2 x^{2}-2 x \ln \left (x -1\right )-5 x -i \pi \,x^{2}+2 i x \pi -i \pi +3+\ln \left (x -1\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= -\left (x -1\right ) \left (\left (1-x \right ) \ln \left (x -1\right )+i x \pi -i \pi -2 x +3\right ) \\
\end{align*} Verification of solutions
\[
y = -\left (x -1\right ) \left (\left (1-x \right ) \ln \left (x -1\right )+i x \pi -i \pi -2 x +3\right )
\] Verified OK.
Writing the ode as \begin {align*} \left (x -1\right )^{2} y^{\prime \prime }+\left (-2 x +2\right ) y^{\prime }+2 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 &= \left (x -1\right )^{2} \\ B &= -2 x +2\tag {3} \\ C &= 2 \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 {0}{1}\tag {6} \end {align*}
Comparing the above to (5) shows that \begin {align*} s &= 0\\ t &= 1 \end {align*}
Therefore eq. (4) becomes \begin {align*} z''(x) &= 0 \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) \\ &= 0 - -\infty \\ &= \infty \end {align*}
There are no poles in \(r\). Therefore the set of poles \(\Gamma \) is empty. Since there is no odd order pole
larger than \(2\) and the order at \(\infty \) is \(infinity\) then the necessary conditions for case one are met.
Therefore \begin {align*} L &= [1] \end {align*}
Since \(r = 0\) is not a function of \(x\), then there is no need run Kovacic algorithm to obtain a solution
for transformed ode \(z''=r z\) as one solution is \[ z_1(x) = 1 \] Using the above, the solution for the original ode can
now be found. 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 {-2 x +2}{\left (x -1\right )^{2}} \,dx} \\
&= z_1 e^{\ln \left (x -1\right )} \\
&= z_1 \left (x -1\right ) \\
\end{align*} Which simplifies to \[
y_1 = x -1
\]
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 {-2 x +2}{\left (x -1\right )^{2}} \,dx}}{\left (y_1\right )^2} \,dx \\
&= y_1 \int \frac { e^{2 \ln \left (x -1\right )}}{\left (y_1\right )^2} \,dx \\
&= y_1 \left (x\right ) \\
\end{align*}
Therefore the solution is
\begin{align*}
y &= c_{1} y_1 + c_{2} y_2 \\
&= c_{1} \left (x -1\right ) + c_{2} \left (x -1\left (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 \[
\left (x -1\right )^{2} y^{\prime \prime }+\left (-2 x +2\right ) y^{\prime }+2 y = 0
\] The homogeneous solution is found using the Kovacic algorithm which results in \[
y_h = c_{1} \left (x -1\right )+c_{2} x \left (x -1\right )
\]
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 -1 \\
y_2 &= x \left (x -1\right ) \\
\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 \[ W = \begin {vmatrix} x -1 & x \left (x -1\right ) \\ \frac {d}{dx}\left (x -1\right ) & \frac {d}{dx}\left (x \left (x -1\right )\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} x -1 & x \left (x -1\right ) \\ 1 & 2 x -1 \end {vmatrix} \] Therefore \[
W = \left (x -1\right )\left (2 x -1\right ) - \left (x \left (x -1\right )\right )\left (1\right )
\] Which
simplifies to \[
W = x^{2}-2 x +1
\] Which simplifies to \[
W = \left (x -1\right )^{2}
\] Therefore Eq. (2) becomes \[
u_1 = -\int \frac {x \left (x -1\right )^{3}}{\left (x -1\right )^{4}}\,dx
\] Which simplifies to \[
u_1 = - \int \frac {x}{x -1}d x
\]
Hence \[
u_1 = -x -\ln \left (x -1\right )
\] And Eq. (3) becomes \[
u_2 = \int \frac {\left (x -1\right )^{3}}{\left (x -1\right )^{4}}\,dx
\] Which simplifies to \[
u_2 = \int \frac {1}{x -1}d x
\] Hence \[
u_2 = \ln \left (x -1\right )
\] Therefore the particular
solution, from equation (1) is \[
y_p(x) = \left (-x -\ln \left (x -1\right )\right ) \left (x -1\right )+\ln \left (x -1\right ) x \left (x -1\right )
\] Which simplifies to \[
y_p(x) = \left (x -1\right ) \left (\ln \left (x -1\right ) \left (x -1\right )-x \right )
\] Therefore the general solution
is \begin{align*}
y &= y_h + y_p \\
&= \left (c_{1} \left (x -1\right )+c_{2} x \left (x -1\right )\right ) + \left (\left (x -1\right ) \left (\ln \left (x -1\right ) \left (x -1\right )-x \right )\right ) \\
\end{align*} Which simplifies to \[
y = \left (x -1\right ) \left (c_{2} x +c_{1} \right )+\left (x -1\right ) \left (\ln \left (x -1\right ) \left (x -1\right )-x \right )
\] Initial conditions are used to solve for the constants of
integration.
Looking at the above solution \begin {align*} y = \left (x -1\right ) \left (c_{2} x +c_{1} \right )+\left (x -1\right ) \left (\ln \left (x -1\right ) \left (x -1\right )-x \right ) \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 \(y = 3\) and \(x = 0\) in the above gives
\begin {align*} 3 = i \pi -c_{1}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = c_{2} x +c_{1} +\left (x -1\right ) c_{2} +2 \ln \left (x -1\right ) \left (x -1\right )-x \end {align*}
substituting \(y^{\prime } = -6\) and \(x = 0\) in the above gives \begin {align*} -6 = -2 i \pi +c_{1} -c_{2}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&=i \pi -3\\ c_{2}&=-i \pi +3 \end {align*}
Substituting these values back in above solution results in \begin {align*} y = x^{2} \ln \left (x -1\right )+2 x^{2}-2 x \ln \left (x -1\right )-5 x -i \pi \,x^{2}+2 i x \pi -i \pi +3+\ln \left (x -1\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= -\left (x -1\right ) \left (\left (1-x \right ) \ln \left (x -1\right )+i x \pi -i \pi -2 x +3\right ) \\
\end{align*} Verification of solutions
\[
y = -\left (x -1\right ) \left (\left (1-x \right ) \ln \left (x -1\right )+i x \pi -i \pi -2 x +3\right )
\] Verified OK. Maple trace
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 33
\[
y \left (x \right ) = \left (-i \pi x +i \pi +\ln \left (x -1\right ) x -\ln \left (x -1\right )+2 x -3\right ) \left (x -1\right )
\]
✓ Solution by Mathematica
Time used: 0.035 (sec). Leaf size: 30
\[
y(x)\to (x-1) (-i \pi (x-1)+2 x+(x-1) \log (x-1)-3)
\]
10.31.2 Solving as linear second order ode solved by an integrating factor ode
10.31.3 Solving as second order change of variable on x method 2 ode
10.31.4 Solving as second order ode non constant coeff transformation on B
ode
10.31.5 Solving using Kovacic algorithm
`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
<- LODE of Euler type successful
<- solving first the homogeneous part of the ODE successful`
dsolve([(x-1)^2*diff(y(x),x$2)-2*(x-1)*diff(y(x),x)+2*y(x)=(x-1)^2,y(0) = 3, D(y)(0) = -6],y(x), singsol=all)
DSolve[{(x-1)^2*y''[x]-2*(x-1)*y'[x]+2*y[x]==(x-1)^2,{y[0]==3,y'[0]==-6}},y[x],x,IncludeSingularSolutions -> True]