Internal problem ID [1133]
Internal file name [OUTPUT/1134_Sunday_June_05_2022_02_03_17_AM_32416935/index.tex]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 5 linear second order equations. Section 5.6 Reduction or order. Page
253
Problem number: 27.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "reduction_of_order", "second_order_bessel_ode", "second_order_change_of_variable_on_y_method_1"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {4 x^{2} y^{\prime \prime }-4 y^{\prime } x +\left (-16 x^{2}+3\right ) y=0} \] Given that one solution of the ode is \begin {align*} y_1 &= {\mathrm e}^{2 x} \sqrt {x} \end {align*}
Given one basis solution \(y_{1}\left (x \right )\), then the second basis solution is given by \[ y_{2}\left (x \right ) = y_{1} \left (\int \frac {{\mathrm e}^{-\left (\int p d x \right )}}{y_{1}^{2}}d x \right ) \] Where \(p(x)\) is the coefficient of \(y^{\prime }\) when the ode is written in the normal form \[ y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y = f \left (x \right ) \] Looking at the ode to solve shows that \[ p \left (x \right ) = -\frac {1}{x} \] Therefore \begin{align*} y_{2}\left (x \right ) &= {\mathrm e}^{2 x} \sqrt {x}\, \left (\int \frac {{\mathrm e}^{-\left (\int -\frac {1}{x}d x \right )} {\mathrm e}^{-4 x}}{x}d x \right ) \\ y_{2}\left (x \right ) &= {\mathrm e}^{2 x} \sqrt {x} \int \frac {x}{{\mathrm e}^{4 x} x} , dx \\ y_{2}\left (x \right ) &= {\mathrm e}^{2 x} \sqrt {x}\, \left (\int {\mathrm e}^{-4 x}d x \right ) \\ y_{2}\left (x \right ) &= -\frac {{\mathrm e}^{2 x} \sqrt {x}\, {\mathrm e}^{-4 x}}{4} \\ \end{align*} Hence the solution is \begin{align*} y &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= {\mathrm e}^{2 x} \sqrt {x}\, c_{1} -\frac {c_{2} {\mathrm e}^{2 x} \sqrt {x}\, {\mathrm e}^{-4 x}}{4} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{2 x} \sqrt {x}\, c_{1} -\frac {c_{2} {\mathrm e}^{2 x} \sqrt {x}\, {\mathrm e}^{-4 x}}{4} \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{2 x} \sqrt {x}\, c_{1} -\frac {c_{2} {\mathrm e}^{2 x} \sqrt {x}\, {\mathrm e}^{-4 x}}{4} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 x^{2} y^{\prime \prime }-4 y^{\prime } x +\left (-16 x^{2}+3\right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {\left (16 x^{2}-3\right ) y}{4 x^{2}}+\frac {y^{\prime }}{x} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime }-\frac {y^{\prime }}{x}-\frac {\left (16 x^{2}-3\right ) y}{4 x^{2}}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=-\frac {1}{x}, P_{3}\left (x \right )=-\frac {16 x^{2}-3}{4 x^{2}}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=-1 \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=\frac {3}{4} \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 4 x^{2} y^{\prime \prime }-4 y^{\prime } x +\left (-16 x^{2}+3\right ) y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x \cdot y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{2}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (-1+2 r \right ) \left (-3+2 r \right ) x^{r}+a_{1} \left (1+2 r \right ) \left (-1+2 r \right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (2 k +2 r -1\right ) \left (2 k +2 r -3\right )-16 a_{k -2}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (-1+2 r \right ) \left (-3+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{\frac {1}{2}, \frac {3}{2}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+2 r \right ) \left (-1+2 r \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 4 \left (k +r -\frac {3}{2}\right ) \left (k +r -\frac {1}{2}\right ) a_{k}-16 a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & 4 \left (k +\frac {1}{2}+r \right ) \left (k +\frac {3}{2}+r \right ) a_{k +2}-16 a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {16 a_{k}}{\left (2 k +1+2 r \right ) \left (2 k +3+2 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & a_{k +2}=\frac {16 a_{k}}{\left (2 k +2\right ) \left (2 k +4\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}}, a_{k +2}=\frac {16 a_{k}}{\left (2 k +2\right ) \left (2 k +4\right )}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {3}{2} \\ {} & {} & a_{k +2}=\frac {16 a_{k}}{\left (2 k +4\right ) \left (2 k +6\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {3}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {3}{2}}, a_{k +2}=\frac {16 a_{k}}{\left (2 k +4\right ) \left (2 k +6\right )}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {3}{2}}\right ), a_{k +2}=\frac {16 a_{k}}{\left (2 k +2\right ) \left (2 k +4\right )}, a_{1}=0, b_{k +2}=\frac {16 b_{k}}{\left (2 k +4\right ) \left (2 k +6\right )}, b_{1}=0\right ] \end {array} \]
Maple trace Kovacic algorithm successful
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature 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`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 21
dsolve([4*x^2*diff(y(x),x$2)-4*x*diff(y(x),x)+(3-16*x^2)*y(x)=0,sqrt(x)*exp(2*x)],singsol=all)
\[ y \left (x \right ) = \sqrt {x}\, \left (c_{1} \sinh \left (2 x \right )+c_{2} \cosh \left (2 x \right )\right ) \]
✓ Solution by Mathematica
Time used: 0.033 (sec). Leaf size: 32
DSolve[4*x^2*y''[x]-4*x*y'[x]+(3-16*x^2)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{4} e^{-2 x} \sqrt {x} \left (c_2 e^{4 x}+4 c_1\right ) \]