Internal problem ID [1189]
Internal file name [OUTPUT/1190_Sunday_June_05_2022_02_04_40_AM_39778088/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: 35.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "kovacic", "exact linear second order ode", "second_order_integrable_as_is", "second_order_ode_non_constant_coeff_transformation_on_B"
Maple gives the following as the ode type
[[_2nd_order, _exact, _linear, _nonhomogeneous]]
\[ \boxed {\left (x +1\right ) \left (2 x +3\right ) y^{\prime \prime }+2 \left (x +2\right ) y^{\prime }-2 y=\left (2 x +3\right )^{2}} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 0] \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 +4}{2 x^{2}+5 x +3}\\ q(x) &=-\frac {2}{2 x^{2}+5 x +3}\\ F &=\frac {4 \left (x +\frac {3}{2}\right )^{2}}{2 x^{2}+5 x +3} \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }+\frac {\left (2 x +4\right ) y^{\prime }}{2 x^{2}+5 x +3}-\frac {2 y}{2 x^{2}+5 x +3} = \frac {4 \left (x +\frac {3}{2}\right )^{2}}{2 x^{2}+5 x +3} \end {align*}
The domain of \(p(x)=\frac {2 x +4}{2 x^{2}+5 x +3}\) is \[
\left \{-\infty \le x <-1, -1
Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime \prime } \left (2 x^{2}+5 x +3\right )+\left (2 x +4\right ) y^{\prime }-2 y\right )d x &= \int 4 \left (x +\frac {3}{2}\right )^{2}d x\\ \left (-2 x -1\right ) y+\left (2 x^{2}+5 x +3\right ) y^{\prime } = \frac {4 \left (x +\frac {3}{2}\right )^{3}}{3} + c_{1} \end {align*}
Which is now solved for \(y\).
Entering Linear first order ODE solver. In canonical form a linear first order is
\begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}
Where here \begin {align*} p(x) &=-\frac {12 x +6}{6 \left (2 x^{2}+5 x +3\right )}\\ q(x) &=\frac {8 x^{3}+36 x^{2}+6 c_{1} +54 x +27}{12 x^{2}+30 x +18} \end {align*}
Hence the ode is \begin {align*} y^{\prime }-\frac {\left (12 x +6\right ) y}{6 \left (2 x^{2}+5 x +3\right )} = \frac {8 x^{3}+36 x^{2}+6 c_{1} +54 x +27}{12 x^{2}+30 x +18} \end {align*}
The integrating factor \(\mu \) is \begin{align*}
\mu &= {\mathrm e}^{\int -\frac {12 x +6}{6 \left (2 x^{2}+5 x +3\right )}d x} \\
&= {\mathrm e}^{\ln \left (x +1\right )-2 \ln \left (2 x +3\right )} \\
\end{align*} Which simplifies to \[
\mu = \frac {x +1}{\left (2 x +3\right )^{2}}
\] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {8 x^{3}+36 x^{2}+6 c_{1} +54 x +27}{12 x^{2}+30 x +18}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\frac {\left (x +1\right ) y}{\left (2 x +3\right )^{2}}\right ) &= \left (\frac {x +1}{\left (2 x +3\right )^{2}}\right ) \left (\frac {8 x^{3}+36 x^{2}+6 c_{1} +54 x +27}{12 x^{2}+30 x +18}\right )\\ \mathrm {d} \left (\frac {\left (x +1\right ) y}{\left (2 x +3\right )^{2}}\right ) &= \left (\frac {8 x^{3}+36 x^{2}+6 c_{1} +54 x +27}{6 \left (2 x +3\right )^{3}}\right )\, \mathrm {d} x \end {align*}
Integrating gives \begin {align*} \frac {\left (x +1\right ) y}{\left (2 x +3\right )^{2}} &= \int {\frac {8 x^{3}+36 x^{2}+6 c_{1} +54 x +27}{6 \left (2 x +3\right )^{3}}\,\mathrm {d} x}\\ \frac {\left (x +1\right ) y}{\left (2 x +3\right )^{2}} &= \frac {x}{6}-\frac {c_{1}}{4 \left (2 x +3\right )^{2}} + c_{2} \end {align*}
Dividing both sides by the integrating factor \(\mu =\frac {x +1}{\left (2 x +3\right )^{2}}\) results in \begin {align*} y &= \frac {\left (2 x +3\right )^{2} \left (\frac {x}{6}-\frac {c_{1}}{4 \left (2 x +3\right )^{2}}\right )}{x +1}+\frac {c_{2} \left (2 x +3\right )^{2}}{x +1} \end {align*}
which simplifies to \begin {align*} y &= \frac {8 x^{3}+\left (48 c_{2} +24\right ) x^{2}+\left (144 c_{2} +18\right ) x -3 c_{1} +108 c_{2}}{12 x +12} \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = \frac {8 x^{3}+\left (48 c_{2} +24\right ) x^{2}+\left (144 c_{2} +18\right ) x -3 c_{1} +108 c_{2}}{12 x +12} \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 \(x = 0\) in the above gives
\begin {align*} 0 = -\frac {c_{1}}{4}+9 c_{2}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \frac {24 x^{2}+2 \left (48 c_{2} +24\right ) x +144 c_{2} +18}{12 x +12}-\frac {12 \left (8 x^{3}+\left (48 c_{2} +24\right ) x^{2}+\left (144 c_{2} +18\right ) x -3 c_{1} +108 c_{2} \right )}{\left (12 x +12\right )^{2}} \end {align*}
substituting \(y^{\prime } = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = \frac {c_{1}}{4}+3 c_{2} +\frac {3}{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 {9}{2}}\\ c_{2}&=-{\frac {1}{8}} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \frac {x^{2} \left (4 x +9\right )}{6 x +6} \end {align*}
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {x^{2} \left (4 x +9\right )}{6 x +6} \\
\end{align*} Verification of solutions
\[
y = \frac {x^{2} \left (4 x +9\right )}{6 x +6}
\] 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 &= 2 x^{2}+5 x +3\\ B &= 2 x +4\\ C &= -2\\ F &= 4 \left (x +\frac {3}{2}\right )^{2} \end {align*}
The above shows that for this ode \begin {align*} AB''+BB'+CB &= \left (2 x^{2}+5 x +3\right ) \left (0\right ) + \left (2 x +4\right ) \left (2\right ) + \left (-2\right ) \left (2 x +4\right ) \\ &=0 \end {align*}
Hence the ode in \(v\) given in (1) now simplifies to \begin {align*} 4 x^{3}+18 x^{2}+26 x +12 v'' +\left ( 12 x^{2}+36 x +28\right ) v' & =0 \end {align*}
Now by applying \(v'=u\) the above becomes \begin {align*} \left (4 x^{3}+18 x^{2}+26 x +12\right ) u^{\prime }\left (x \right )+12 u \left (x \right ) \left (x^{2}+3 x +\frac {7}{3}\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 {2 u \left (3 x^{2}+9 x +7\right )}{2 x^{3}+9 x^{2}+13 x +6} \end {align*}
Where \(f(x)=-\frac {2 \left (3 x^{2}+9 x +7\right )}{2 x^{3}+9 x^{2}+13 x +6}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= -\frac {2 \left (3 x^{2}+9 x +7\right )}{2 x^{3}+9 x^{2}+13 x +6} \,d x\\ \int { \frac {1}{u} \,du} &= \int {-\frac {2 \left (3 x^{2}+9 x +7\right )}{2 x^{3}+9 x^{2}+13 x +6} \,d x}\\ \ln \left (u \right )&=-2 \ln \left (x +1\right )+\ln \left (2 x +3\right )-2 \ln \left (x +2\right )+c_{1}\\ u&={\mathrm e}^{-2 \ln \left (x +1\right )+\ln \left (2 x +3\right )-2 \ln \left (x +2\right )+c_{1}}\\ &=c_{1} {\mathrm e}^{-2 \ln \left (x +1\right )+\ln \left (2 x +3\right )-2 \ln \left (x +2\right )} \end {align*}
Which simplifies to \[
u \left (x \right ) = c_{1} \left (\frac {2 x}{\left (x +1\right )^{2} \left (x +2\right )^{2}}+\frac {3}{\left (x +1\right )^{2} \left (x +2\right )^{2}}\right )
\] The ode for \(v\) now becomes \begin {align*} v' &= u\\ &=c_{1} \left (\frac {2 x}{\left (x +1\right )^{2} \left (x +2\right )^{2}}+\frac {3}{\left (x +1\right )^{2} \left (x +2\right )^{2}}\right ) \end {align*}
Which is now solved for \(v\). Integrating both sides gives \begin {align*} v \left (x \right ) &= \int { \frac {c_{1} \left (2 x +3\right )}{\left (x +1\right )^{2} \left (x +2\right )^{2}}\,\mathop {\mathrm {d}x}}\\ &= -\frac {c_{1}}{\left (x +1\right ) \left (x +2\right )}+c_{2} \end {align*}
Therefore the homogeneous solution is \begin {align*} y_h(x) &= B v\\ &= \left (2 x +4\right ) \left (-\frac {c_{1}}{\left (x +1\right ) \left (x +2\right )}+c_{2}\right ) \\ &= \frac {\left (2 x^{2}+6 x +4\right ) c_{2} -2 c_{1}}{x +1} \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 &= \frac {1}{x +1} \\
y_2 &= \frac {2 x^{2}}{x +1}+\frac {6 x}{x +1}+\frac {4}{x +1} \\
\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 \[ W = \begin {vmatrix} \frac {1}{x +1} & \frac {2 x^{2}}{x +1}+\frac {6 x}{x +1}+\frac {4}{x +1} \\ \frac {d}{dx}\left (\frac {1}{x +1}\right ) & \frac {d}{dx}\left (\frac {2 x^{2}}{x +1}+\frac {6 x}{x +1}+\frac {4}{x +1}\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} \frac {1}{x +1} & \frac {2 x^{2}}{x +1}+\frac {6 x}{x +1}+\frac {4}{x +1} \\ -\frac {1}{\left (x +1\right )^{2}} & \frac {4 x}{x +1}-\frac {2 x^{2}}{\left (x +1\right )^{2}}+\frac {6}{x +1}-\frac {6 x}{\left (x +1\right )^{2}}-\frac {4}{\left (x +1\right )^{2}} \end {vmatrix} \] Therefore \[
W = \left (\frac {1}{x +1}\right )\left (\frac {4 x}{x +1}-\frac {2 x^{2}}{\left (x +1\right )^{2}}+\frac {6}{x +1}-\frac {6 x}{\left (x +1\right )^{2}}-\frac {4}{\left (x +1\right )^{2}}\right ) - \left (\frac {2 x^{2}}{x +1}+\frac {6 x}{x +1}+\frac {4}{x +1}\right )\left (-\frac {1}{\left (x +1\right )^{2}}\right )
\] Which simplifies to \[
W = \frac {4 x +6}{\left (x +1\right )^{2}}
\] Which
simplifies to \[
W = \frac {4 x +6}{\left (x +1\right )^{2}}
\] Therefore Eq. (2) becomes \[
u_1 = -\int \frac {4 \left (\frac {2 x^{2}}{x +1}+\frac {6 x}{x +1}+\frac {4}{x +1}\right ) \left (x +\frac {3}{2}\right )^{2}}{\frac {\left (2 x^{2}+5 x +3\right ) \left (4 x +6\right )}{\left (x +1\right )^{2}}}\,dx
\] Which simplifies to \[
u_1 = - \int \left (x +1\right ) \left (x +2\right )d x
\] Hence \[
u_1 = -\frac {1}{3} x^{3}-\frac {3}{2} x^{2}-2 x
\] And Eq. (3)
becomes \[
u_2 = \int \frac {\frac {4 \left (x +\frac {3}{2}\right )^{2}}{x +1}}{\frac {\left (2 x^{2}+5 x +3\right ) \left (4 x +6\right )}{\left (x +1\right )^{2}}}\,dx
\] Which simplifies to \[
u_2 = \int \frac {1}{2}d x
\] Hence \[
u_2 = \frac {x}{2}
\] Which simplifies to \begin{align*}
u_1 &= -\frac {x \left (x^{2}+\frac {9}{2} x +6\right )}{3} \\
u_2 &= \frac {x}{2} \\
\end{align*} Therefore the particular
solution, from equation (1) is \[
y_p(x) = -\frac {x \left (x^{2}+\frac {9}{2} x +6\right )}{3 \left (x +1\right )}+\frac {x \left (\frac {2 x^{2}}{x +1}+\frac {6 x}{x +1}+\frac {4}{x +1}\right )}{2}
\] Which simplifies to \[
y_p(x) = \frac {x^{2} \left (4 x +9\right )}{6 x +6}
\] Hence the complete solution is
\begin {align*} y(x) &= y_h + y_p \\ &= \left (\frac {\left (2 x^{2}+6 x +4\right ) c_{2} -2 c_{1}}{x +1}\right ) + \left (\frac {x^{2} \left (4 x +9\right )}{6 x +6}\right )\\ &= \frac {4 x^{3}+\left (12 c_{2} +9\right ) x^{2}+36 c_{2} x -12 c_{1} +24 c_{2}}{6 x +6} \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = \frac {4 x^{3}+\left (12 c_{2} +9\right ) x^{2}+36 c_{2} x -12 c_{1} +24 c_{2}}{6 x +6} \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 \(x = 0\) in the above gives
\begin {align*} 0 = -2 c_{1} +4 c_{2}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \frac {12 x^{2}+2 \left (12 c_{2} +9\right ) x +36 c_{2}}{6 x +6}-\frac {6 \left (4 x^{3}+\left (12 c_{2} +9\right ) x^{2}+36 c_{2} x -12 c_{1} +24 c_{2} \right )}{\left (6 x +6\right )^{2}} \end {align*}
substituting \(y^{\prime } = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = 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}&=0\\ c_{2}&=0 \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \frac {x^{2} \left (4 x +9\right )}{6 x +6} \end {align*}
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {x^{2} \left (4 x +9\right )}{6 x +6} \\
\end{align*} Verification of solutions
\[
y = \frac {x^{2} \left (4 x +9\right )}{6 x +6}
\] Verified OK. Writing the ode as \[
y^{\prime \prime } \left (2 x^{2}+5 x +3\right )+\left (2 x +4\right ) y^{\prime }-2 y = 4 \left (x +\frac {3}{2}\right )^{2}
\] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime \prime } \left (2 x^{2}+5 x +3\right )+\left (2 x +4\right ) y^{\prime }-2 y\right )d x &= \int 4 \left (x +\frac {3}{2}\right )^{2}d x\\ 4 y+3 y^{\prime }-2 y x +\frac {5 y^{\prime \prime } x^{2}}{2}+2 y^{\prime } x^{2}-\frac {5 y^{\prime \prime \prime } x^{3}}{6} = \frac {\left (2 x +3\right )^{3}}{6} +c_{1} \end {align*}
Which is now solved for \(y\). Unable to solve this ODE.
Writing the ode as \begin {align*} y^{\prime \prime } \left (2 x^{2}+5 x +3\right )+\left (2 x +4\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 &= 2 x^{2}+5 x +3 \\ B &= 2 x +4\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 {3}{\left (2 x +3\right )^{2}}\tag {6} \end {align*}
Comparing the above to (5) shows that \begin {align*} s &= 3\\ t &= \left (2 x +3\right )^{2} \end {align*}
Therefore eq. (4) becomes \begin {align*} z''(x) &= \left ( \frac {3}{\left (2 x +3\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 - 0 \\ &= 2 \end {align*}
The poles of \(r\) in eq. (7) and the order of each pole are determined by solving for the roots of \(t=\left (2 x +3\right )^{2}\).
There is a pole at \(x=-{\frac {3}{2}}\) of order \(2\). Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is
\(2\) 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. Since pole order is not larger than \(2\) and
the order at \(\infty \) is \(2\) then the necessary conditions for case three are met. Therefore
\begin {align*} L &= [1, 2, 4, 6, 12] \end {align*}
Attempting to find a solution using case \(n=1\).
Looking at poles of order 2. The partial fractions decomposition of \(r\) is \[
r = \frac {3}{4 \left (x +\frac {3}{2}\right )^{2}}
\] For the pole at \(x=-{\frac {3}{2}}\) let \(b\)
be the coefficient of \(\frac {1}{ \left (x +\frac {3}{2}\right )^{2}}\) 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 2 then \([\sqrt r]_\infty =0\). Let \(b\) be the coefficient of \(\frac {1}{x^{2}}\) in the Laurent series expansion of \(r\)
at \(\infty \). which can be found by dividing the leading coefficient of \(s\) by the leading coefficient of \(t\)
from \begin {alignat*} {2} r &= \frac {s}{t} &&= \frac {3}{\left (2 x +3\right )^{2}} \end {alignat*}
Since the \(\text {gcd}(s,t)=1\). This gives \(b={\frac {3}{4}}\). Hence \begin {alignat*} {2} [\sqrt r]_\infty &= 0 \\ \alpha _{\infty }^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= {\frac {3}{2}}\\ \alpha _{\infty }^{-} &= \frac {1}{2} - \sqrt {1+4 b} &&= -{\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 \[ r=\frac {3}{\left (2 x +3\right )^{2}} \]
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*}
The above gives \begin {align*} \omega &= \left ( (-)[\sqrt r]_{c_1} + \frac { \alpha _{c_1}^{-} }{x- c_1}\right ) + (-) [\sqrt r]_\infty \\ &= -\frac {1}{2 \left (x +\frac {3}{2}\right )} + (-) \left ( 0 \right ) \\ &= -\frac {1}{2 \left (x +\frac {3}{2}\right )}\\ &= -\frac {1}{2 x +3} \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 +\frac {3}{2}\right )}\right ) \left (0\right ) + \left ( \left (\frac {1}{2 \left (x +\frac {3}{2}\right )^{2}}\right ) + \left (-\frac {1}{2 \left (x +\frac {3}{2}\right )}\right )^2 - \left (\frac {3}{\left (2 x +3\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 -\frac {1}{2 \left (x +\frac {3}{2}\right )}d x}\\ &= \frac {2}{\sqrt {4 x +6}} \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 {2 x +4}{2 x^{2}+5 x +3} \,dx} \\
&= z_1 e^{-\ln \left (x +1\right )+\frac {\ln \left (2 x +3\right )}{2}} \\
&= z_1 \left (\frac {\sqrt {2 x +3}}{x +1}\right ) \\
\end{align*} Which simplifies to \[
y_1 = \frac {\sqrt {2}}{x +1}
\] 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 +4}{2 x^{2}+5 x +3} \,dx}}{\left (y_1\right )^2} \,dx \\
&= y_1 \int \frac { e^{-2 \ln \left (x +1\right )+\ln \left (2 x +3\right )}}{\left (y_1\right )^2} \,dx \\
&= y_1 \left (\frac {x \left (x +3\right )}{2}\right ) \\
\end{align*} Therefore
the solution is
\begin{align*}
y &= c_{1} y_1 + c_{2} y_2 \\
&= c_{1} \left (\frac {\sqrt {2}}{x +1}\right ) + c_{2} \left (\frac {\sqrt {2}}{x +1}\left (\frac {x \left (x +3\right )}{2}\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 \[
y^{\prime \prime } \left (2 x^{2}+5 x +3\right )+\left (2 x +4\right ) y^{\prime }-2 y = 0
\] The homogeneous solution is found using the Kovacic algorithm which results in \[
y_h = \frac {c_{1} \sqrt {2}}{x +1}+\frac {c_{2} \sqrt {2}\, x \left (x +3\right )}{2 x +2}
\]
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 &= \frac {\sqrt {2}}{x +1} \\
y_2 &= \frac {\sqrt {2}\, x \left (x +3\right )}{2 x +2} \\
\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 \[ W = \begin {vmatrix} \frac {\sqrt {2}}{x +1} & \frac {\sqrt {2}\, x \left (x +3\right )}{2 x +2} \\ \frac {d}{dx}\left (\frac {\sqrt {2}}{x +1}\right ) & \frac {d}{dx}\left (\frac {\sqrt {2}\, x \left (x +3\right )}{2 x +2}\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} \frac {\sqrt {2}}{x +1} & \frac {\sqrt {2}\, x \left (x +3\right )}{2 x +2} \\ -\frac {\sqrt {2}}{\left (x +1\right )^{2}} & \frac {\sqrt {2}\, \left (x +3\right )}{2 x +2}+\frac {\sqrt {2}\, x}{2 x +2}-\frac {2 \sqrt {2}\, x \left (x +3\right )}{\left (2 x +2\right )^{2}} \end {vmatrix} \] Therefore \[
W = \left (\frac {\sqrt {2}}{x +1}\right )\left (\frac {\sqrt {2}\, \left (x +3\right )}{2 x +2}+\frac {\sqrt {2}\, x}{2 x +2}-\frac {2 \sqrt {2}\, x \left (x +3\right )}{\left (2 x +2\right )^{2}}\right ) - \left (\frac {\sqrt {2}\, x \left (x +3\right )}{2 x +2}\right )\left (-\frac {\sqrt {2}}{\left (x +1\right )^{2}}\right )
\] Which simplifies to \[
W = \frac {2 x +3}{\left (x +1\right )^{2}}
\] Which
simplifies to \[
W = \frac {2 x +3}{\left (x +1\right )^{2}}
\] Therefore Eq. (2) becomes \[
u_1 = -\int \frac {\frac {4 \sqrt {2}\, x \left (x +3\right ) \left (x +\frac {3}{2}\right )^{2}}{2 x +2}}{\frac {\left (2 x^{2}+5 x +3\right ) \left (2 x +3\right )}{\left (x +1\right )^{2}}}\,dx
\] Which simplifies to \[
u_1 = - \int \frac {\sqrt {2}\, x \left (x +3\right )}{2}d x
\] Hence \[
u_1 = -\frac {\sqrt {2}\, \left (\frac {1}{3} x^{3}+\frac {3}{2} x^{2}\right )}{2}
\] And Eq. (3)
becomes \[
u_2 = \int \frac {\frac {4 \sqrt {2}\, \left (x +\frac {3}{2}\right )^{2}}{x +1}}{\frac {\left (2 x^{2}+5 x +3\right ) \left (2 x +3\right )}{\left (x +1\right )^{2}}}\,dx
\] Which simplifies to \[
u_2 = \int \sqrt {2}d x
\] Hence \[
u_2 = \sqrt {2}\, x
\] Which simplifies to \begin{align*}
u_1 &= -\frac {\sqrt {2}\, x^{2} \left (2 x +9\right )}{12} \\
u_2 &= \sqrt {2}\, x \\
\end{align*} Therefore the particular
solution, from equation (1) is \[
y_p(x) = -\frac {x^{2} \left (2 x +9\right )}{6 \left (x +1\right )}+\frac {2 x^{2} \left (x +3\right )}{2 x +2}
\] Which simplifies to \[
y_p(x) = \frac {x^{2} \left (4 x +9\right )}{6 x +6}
\] Therefore the general solution
is \begin{align*}
y &= y_h + y_p \\
&= \left (\frac {c_{1} \sqrt {2}}{x +1}+\frac {c_{2} \sqrt {2}\, x \left (x +3\right )}{2 x +2}\right ) + \left (\frac {x^{2} \left (4 x +9\right )}{6 x +6}\right ) \\
\end{align*} Which simplifies to \[
y = \frac {\sqrt {2}\, \left (c_{2} x^{2}+3 c_{2} x +2 c_{1} \right )}{2 x +2}+\frac {x^{2} \left (4 x +9\right )}{6 x +6}
\] Initial conditions are used to solve for the constants of
integration.
Looking at the above solution \begin {align*} y = \frac {\sqrt {2}\, \left (c_{2} x^{2}+3 c_{2} x +2 c_{1} \right )}{2 x +2}+\frac {x^{2} \left (4 x +9\right )}{6 x +6} \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 \(x = 0\) in the above gives
\begin {align*} 0 = c_{1} \sqrt {2}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \frac {\sqrt {2}\, \left (2 c_{2} x +3 c_{2} \right )}{2 x +2}-\frac {2 \sqrt {2}\, \left (c_{2} x^{2}+3 c_{2} x +2 c_{1} \right )}{\left (2 x +2\right )^{2}}+\frac {2 x \left (4 x +9\right )}{6 x +6}+\frac {4 x^{2}}{6 x +6}-\frac {6 x^{2} \left (4 x +9\right )}{\left (6 x +6\right )^{2}} \end {align*}
substituting \(y^{\prime } = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = \left (-c_{1} +\frac {3 c_{2}}{2}\right ) \sqrt {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}&=0\\ c_{2}&=0 \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \frac {4 x^{3}+9 x^{2}}{6 x +6} \end {align*}
Which simplifies to \[
y = \frac {x^{2} \left (4 x +9\right )}{6 x +6}
\] The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {x^{2} \left (4 x +9\right )}{6 x +6} \\
\end{align*} Verification of solutions
\[
y = \frac {x^{2} \left (4 x +9\right )}{6 x +6}
\] Verified OK. An ode of the form \begin {align*} p \left (x \right ) y^{\prime \prime }+q \left (x \right ) y^{\prime }+r \left (x \right ) y&=s \left (x \right ) \end {align*}
is exact if \begin {align*} p''(x) - q'(x) + r(x) &= 0 \tag {1} \end {align*}
For the given ode we have \begin {align*} p(x) &= 2 x^{2}+5 x +3\\ q(x) &= 2 x +4\\ r(x) &= -2\\ s(x) &= 4 \left (x +\frac {3}{2}\right )^{2} \end {align*}
Hence \begin {align*} p''(x) &= 4\\ q'(x) &= 2 \end {align*}
Therefore (1) becomes \begin {align*} 4- \left (2\right ) + \left (-2\right )&=0 \end {align*}
Hence the ode is exact. Since we now know the ode is exact, it can be written as
\begin {align*} \left (p \left (x \right ) y^{\prime }+\left (q \left (x \right )-p^{\prime }\left (x \right )\right ) y\right )' &= s(x) \end {align*}
Integrating gives \begin {align*} p \left (x \right ) y^{\prime }+\left (q \left (x \right )-p^{\prime }\left (x \right )\right ) y&=\int {s \left (x \right )\, dx} \end {align*}
Substituting the above values for \(p,q,r,s\) gives \begin {align*} \left (-2 x -1\right ) y+\left (2 x^{2}+5 x +3\right ) y^{\prime }&=\int {4 \left (x +\frac {3}{2}\right )^{2}\, dx} \end {align*}
We now have a first order ode to solve which is \begin {align*} \left (-2 x -1\right ) y+\left (2 x^{2}+5 x +3\right ) y^{\prime } = \frac {4 \left (x +\frac {3}{2}\right )^{3}}{3}+c_{1} \end {align*}
Entering Linear first order ODE solver. In canonical form a linear first order is
\begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}
Where here \begin {align*} p(x) &=-\frac {12 x +6}{6 \left (2 x^{2}+5 x +3\right )}\\ q(x) &=\frac {8 x^{3}+36 x^{2}+6 c_{1} +54 x +27}{12 x^{2}+30 x +18} \end {align*}
Hence the ode is \begin {align*} y^{\prime }-\frac {\left (12 x +6\right ) y}{6 \left (2 x^{2}+5 x +3\right )} = \frac {8 x^{3}+36 x^{2}+6 c_{1} +54 x +27}{12 x^{2}+30 x +18} \end {align*}
The integrating factor \(\mu \) is \begin{align*}
\mu &= {\mathrm e}^{\int -\frac {12 x +6}{6 \left (2 x^{2}+5 x +3\right )}d x} \\
&= {\mathrm e}^{\ln \left (x +1\right )-2 \ln \left (2 x +3\right )} \\
\end{align*} Which simplifies to \[
\mu = \frac {x +1}{\left (2 x +3\right )^{2}}
\] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {8 x^{3}+36 x^{2}+6 c_{1} +54 x +27}{12 x^{2}+30 x +18}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\frac {\left (x +1\right ) y}{\left (2 x +3\right )^{2}}\right ) &= \left (\frac {x +1}{\left (2 x +3\right )^{2}}\right ) \left (\frac {8 x^{3}+36 x^{2}+6 c_{1} +54 x +27}{12 x^{2}+30 x +18}\right )\\ \mathrm {d} \left (\frac {\left (x +1\right ) y}{\left (2 x +3\right )^{2}}\right ) &= \left (\frac {8 x^{3}+36 x^{2}+6 c_{1} +54 x +27}{6 \left (2 x +3\right )^{3}}\right )\, \mathrm {d} x \end {align*}
Integrating gives \begin {align*} \frac {\left (x +1\right ) y}{\left (2 x +3\right )^{2}} &= \int {\frac {8 x^{3}+36 x^{2}+6 c_{1} +54 x +27}{6 \left (2 x +3\right )^{3}}\,\mathrm {d} x}\\ \frac {\left (x +1\right ) y}{\left (2 x +3\right )^{2}} &= \frac {x}{6}-\frac {c_{1}}{4 \left (2 x +3\right )^{2}} + c_{2} \end {align*}
Dividing both sides by the integrating factor \(\mu =\frac {x +1}{\left (2 x +3\right )^{2}}\) results in \begin {align*} y &= \frac {\left (2 x +3\right )^{2} \left (\frac {x}{6}-\frac {c_{1}}{4 \left (2 x +3\right )^{2}}\right )}{x +1}+\frac {c_{2} \left (2 x +3\right )^{2}}{x +1} \end {align*}
which simplifies to \begin {align*} y &= \frac {8 x^{3}+\left (48 c_{2} +24\right ) x^{2}+\left (144 c_{2} +18\right ) x -3 c_{1} +108 c_{2}}{12 x +12} \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = \frac {8 x^{3}+\left (48 c_{2} +24\right ) x^{2}+\left (144 c_{2} +18\right ) x -3 c_{1} +108 c_{2}}{12 x +12} \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 \(x = 0\) in the above gives
\begin {align*} 0 = -\frac {c_{1}}{4}+9 c_{2}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \frac {24 x^{2}+2 \left (48 c_{2} +24\right ) x +144 c_{2} +18}{12 x +12}-\frac {12 \left (8 x^{3}+\left (48 c_{2} +24\right ) x^{2}+\left (144 c_{2} +18\right ) x -3 c_{1} +108 c_{2} \right )}{\left (12 x +12\right )^{2}} \end {align*}
substituting \(y^{\prime } = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = \frac {c_{1}}{4}+3 c_{2} +\frac {3}{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 {9}{2}}\\ c_{2}&=-{\frac {1}{8}} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \frac {x^{2} \left (4 x +9\right )}{6 x +6} \end {align*}
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {x^{2} \left (4 x +9\right )}{6 x +6} \\
\end{align*} Verification of solutions
\[
y = \frac {x^{2} \left (4 x +9\right )}{6 x +6}
\] Verified OK. Maple trace
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 20
\[
y \left (x \right ) = \frac {x^{2} \left (4 x +9\right )}{6 x +6}
\]
✓ Solution by Mathematica
Time used: 0.694 (sec). Leaf size: 22
\[
y(x)\to \frac {x^2 (4 x+9)}{6 (x+1)}
\]
10.35.2 Solving as second order integrable as is ode
10.35.3 Solving as second order ode non constant coeff transformation on B
ode
10.35.4 Solving as type second_order_integrable_as_is (not using ABC version)
10.35.5 Solving using Kovacic algorithm
pole \(c\) location
pole order
\([\sqrt r]_c\)
\(\alpha _c^{+}\)
\(\alpha _c^{-}\)
\(-{\frac {3}{2}}\) \(2\) \(0\) \(\frac {3}{2}\) \(-{\frac {1}{2}}\)
Order of \(r\) at \(\infty \)
\([\sqrt r]_\infty \)
\(\alpha _\infty ^{+}\)
\(\alpha _\infty ^{-}\) \(2\)
\(0\) \(\frac {3}{2}\) \(-{\frac {1}{2}}\)
10.35.6 Solving as exact linear second order ode ode
`Methods for second order ODEs:
--- Trying classification methods ---
trying a quadrature
trying high order exact linear fully integrable
<- high order exact linear fully integrable successful`
dsolve([(x+1)*(2*x+3)*diff(y(x),x$2)+2*(x+2)*diff(y(x),x)-2*y(x)=(2*x+3)^2,y(0) = 0, D(y)(0) = 0],y(x), singsol=all)
DSolve[{(x+1)*(2*x+3)*y''[x]+2*(x+2)*y'[x]-2*y[x]==(2*x+3)^2,{y[0]==0,y'[0]==0}},y[x],x,IncludeSingularSolutions -> True]