Internal problem ID [6244]
Internal file name [OUTPUT/5492_Sunday_June_05_2022_03_41_26_PM_64351548/index.tex]
Book: Differential Equations: Theory, Technique, and Practice by George Simmons, Steven
Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 1. What is a differential equation. Section 1.9. Reduction of Order. Page
38
Problem number: 2(a).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_integrable_as_is", "second_order_ode_missing_y"
Maple gives the following as the ode type
[[_2nd_order, _missing_y], [_2nd_order, _exact, _nonlinear], [_2nd_order, _reducible, _mu_poly_yn]]
\[ \boxed {\left (x^{2}+2 y^{\prime }\right ) y^{\prime \prime }+2 y^{\prime } x=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = 0] \end {align*}
Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (\left (x^{2}+2 y^{\prime }\right ) y^{\prime \prime }+2 y^{\prime } x \right )d x &= 0 \\ x^{2} y^{\prime }+{y^{\prime }}^{2} = c_{1} \end {align*}
Which is now solved for \(y\). Solving the given ode for \(y^{\prime }\) results in \(2\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=-\frac {x^{2}}{2}+\frac {\sqrt {x^{4}+4 c_{1}}}{2} \tag {1} \\ y^{\prime }&=-\frac {x^{2}}{2}-\frac {\sqrt {x^{4}+4 c_{1}}}{2} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin {align*} y &= \int { -\frac {x^{2}}{2}+\frac {\sqrt {x^{4}+4 c_{1}}}{2}\,\mathop {\mathrm {d}x}}\\ &= \frac {x \sqrt {x^{4}+4 c_{1}}}{6}+\frac {c_{1} \sqrt {2}\, \sqrt {4-\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \sqrt {4+\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {i}{\sqrt {c_{1}}}}}{2}, i\right )}{3 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {x^{4}+4 c_{1}}}-\frac {x^{3}}{6}+c_{2} \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} y &= \int { -\frac {x^{2}}{2}-\frac {\sqrt {x^{4}+4 c_{1}}}{2}\,\mathop {\mathrm {d}x}}\\ &= -\frac {x^{3}}{6}-\frac {x \sqrt {x^{4}+4 c_{1}}}{6}-\frac {c_{1} \sqrt {2}\, \sqrt {4-\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \sqrt {4+\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {i}{\sqrt {c_{1}}}}}{2}, i\right )}{3 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {x^{4}+4 c_{1}}}+c_{3} \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the First solution \begin {align*} y = \frac {x \sqrt {x^{4}+4 c_{1}}}{6}+\frac {c_{1} \sqrt {2}\, \sqrt {4-\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \sqrt {4+\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {i}{\sqrt {c_{1}}}}}{2}, i\right )}{3 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {x^{4}+4 c_{1}}}-\frac {x^{3}}{6}+c_{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 \(y = 1\) and \(x = 0\) in the above gives \begin {align*} 1 = c_{2}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \frac {\sqrt {x^{4}+4 c_{1}}}{6}+\frac {x^{4}}{3 \sqrt {x^{4}+4 c_{1}}}-\frac {2 i \sqrt {c_{1}}\, \sqrt {2}\, \sqrt {4+\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {i}{\sqrt {c_{1}}}}}{2}, i\right ) x}{3 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {4-\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \sqrt {x^{4}+4 c_{1}}}+\frac {2 i \sqrt {c_{1}}\, \sqrt {2}\, \sqrt {4-\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {i}{\sqrt {c_{1}}}}}{2}, i\right ) x}{3 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {4+\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \sqrt {x^{4}+4 c_{1}}}-\frac {2 c_{1} \sqrt {2}\, \sqrt {4-\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \sqrt {4+\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {i}{\sqrt {c_{1}}}}}{2}, i\right ) x^{3}}{3 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \left (x^{4}+4 c_{1} \right )^{\frac {3}{2}}}+\frac {4 c_{1}}{3 \sqrt {x^{4}+4 c_{1}}}-\frac {x^{2}}{2} \end {align*}
substituting \(y^{\prime } = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = \sqrt {c_{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}&=0\\ c_{2}&=1 \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \munderset {c_{1} \rightarrow 0}{\operatorname {lim}}\left (-\frac {-\sqrt {\frac {i}{\sqrt {c_{1}}}}\, x^{5}+x^{3} \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {x^{4}+4 c_{1}}-4 c_{1} \sqrt {2}\, \sqrt {\frac {-i x^{2}+2 \sqrt {c_{1}}}{\sqrt {c_{1}}}}\, \sqrt {\frac {i x^{2}+2 \sqrt {c_{1}}}{\sqrt {c_{1}}}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {i}{\sqrt {c_{1}}}}}{2}, i\right )-4 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, c_{1} x -6 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {x^{4}+4 c_{1}}}{6 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {x^{4}+4 c_{1}}}\right ) \end {align*}
Looking at the Second solution \begin {align*} y = -\frac {x^{3}}{6}-\frac {x \sqrt {x^{4}+4 c_{1}}}{6}-\frac {c_{1} \sqrt {2}\, \sqrt {4-\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \sqrt {4+\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {i}{\sqrt {c_{1}}}}}{2}, i\right )}{3 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {x^{4}+4 c_{1}}}+c_{3} \tag {2} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = 1\) and \(x = 0\) in the above gives \begin {align*} 1 = c_{3}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = -\frac {x^{2}}{2}-\frac {\sqrt {x^{4}+4 c_{1}}}{6}-\frac {x^{4}}{3 \sqrt {x^{4}+4 c_{1}}}+\frac {2 i \sqrt {c_{1}}\, \sqrt {2}\, \sqrt {4+\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {i}{\sqrt {c_{1}}}}}{2}, i\right ) x}{3 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {4-\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \sqrt {x^{4}+4 c_{1}}}-\frac {2 i \sqrt {c_{1}}\, \sqrt {2}\, \sqrt {4-\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {i}{\sqrt {c_{1}}}}}{2}, i\right ) x}{3 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {4+\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \sqrt {x^{4}+4 c_{1}}}+\frac {2 c_{1} \sqrt {2}\, \sqrt {4-\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \sqrt {4+\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {i}{\sqrt {c_{1}}}}}{2}, i\right ) x^{3}}{3 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \left (x^{4}+4 c_{1} \right )^{\frac {3}{2}}}-\frac {4 c_{1}}{3 \sqrt {x^{4}+4 c_{1}}} \end {align*}
substituting \(y^{\prime } = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = -\sqrt {c_{1}}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{3}\}\). Solving for the constants gives \begin {align*} c_{1}&=0\\ c_{3}&=1 \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \munderset {c_{1} \rightarrow 0}{\operatorname {lim}}\left (-\frac {\sqrt {\frac {i}{\sqrt {c_{1}}}}\, x^{5}+x^{3} \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {x^{4}+4 c_{1}}+4 c_{1} \sqrt {2}\, \sqrt {\frac {-i x^{2}+2 \sqrt {c_{1}}}{\sqrt {c_{1}}}}\, \sqrt {\frac {i x^{2}+2 \sqrt {c_{1}}}{\sqrt {c_{1}}}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {i}{\sqrt {c_{1}}}}}{2}, i\right )+4 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, c_{1} x -6 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {x^{4}+4 c_{1}}}{6 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {x^{4}+4 c_{1}}}\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\left (\munderset {c_{1} \rightarrow 0}{\operatorname {lim}}\left (-\frac {4 \left (\left (1+i\right ) \operatorname {EllipticF}\left (\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) x}{c_{1}^{\frac {1}{4}}}, i\right ) c_{1}^{\frac {5}{4}} \sqrt {\frac {i x^{2}+2 \sqrt {c_{1}}}{\sqrt {c_{1}}}}\, \sqrt {-\frac {i x^{2}-2 \sqrt {c_{1}}}{\sqrt {c_{1}}}}+\left (-\frac {x^{3}}{4}+\frac {3}{2}\right ) \sqrt {x^{4}+4 c_{1}}+\frac {x^{5}}{4}+c_{1} x \right )}{\sqrt {x^{4}+4 c_{1}}}\right )\right )}{6} \\ \tag{2} y &= -\frac {\left (\munderset {c_{1} \rightarrow 0}{\operatorname {lim}}\frac {4 \left (1+i\right ) \operatorname {EllipticF}\left (\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) x}{c_{1}^{\frac {1}{4}}}, i\right ) c_{1}^{\frac {5}{4}} \sqrt {\frac {i x^{2}+2 \sqrt {c_{1}}}{\sqrt {c_{1}}}}\, \sqrt {-\frac {i x^{2}-2 \sqrt {c_{1}}}{\sqrt {c_{1}}}}+4 \left (\frac {x^{3}}{4}-\frac {3}{2}\right ) \sqrt {x^{4}+4 c_{1}}+x^{5}+4 c_{1} x}{\sqrt {x^{4}+4 c_{1}}}\right )}{6} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\left (\munderset {c_{1} \rightarrow 0}{\operatorname {lim}}\left (-\frac {4 \left (\left (1+i\right ) \operatorname {EllipticF}\left (\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) x}{c_{1}^{\frac {1}{4}}}, i\right ) c_{1}^{\frac {5}{4}} \sqrt {\frac {i x^{2}+2 \sqrt {c_{1}}}{\sqrt {c_{1}}}}\, \sqrt {-\frac {i x^{2}-2 \sqrt {c_{1}}}{\sqrt {c_{1}}}}+\left (-\frac {x^{3}}{4}+\frac {3}{2}\right ) \sqrt {x^{4}+4 c_{1}}+\frac {x^{5}}{4}+c_{1} x \right )}{\sqrt {x^{4}+4 c_{1}}}\right )\right )}{6} \] Verified OK.
\[ y = -\frac {\left (\munderset {c_{1} \rightarrow 0}{\operatorname {lim}}\frac {4 \left (1+i\right ) \operatorname {EllipticF}\left (\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) x}{c_{1}^{\frac {1}{4}}}, i\right ) c_{1}^{\frac {5}{4}} \sqrt {\frac {i x^{2}+2 \sqrt {c_{1}}}{\sqrt {c_{1}}}}\, \sqrt {-\frac {i x^{2}-2 \sqrt {c_{1}}}{\sqrt {c_{1}}}}+4 \left (\frac {x^{3}}{4}-\frac {3}{2}\right ) \sqrt {x^{4}+4 c_{1}}+x^{5}+4 c_{1} x}{\sqrt {x^{4}+4 c_{1}}}\right )}{6} \] Verified OK.
This is second order ode with missing dependent variable \(y\). Let \begin {align*} p(x) &= y^{\prime } \end {align*}
Then \begin {align*} p'(x) &= y^{\prime \prime } \end {align*}
Hence the ode becomes \begin {align*} \left (x^{2}+2 p \left (x \right )\right ) p^{\prime }\left (x \right )+2 p \left (x \right ) x = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. Writing the ode as \begin {align*} p^{\prime }\left (x \right )&=-\frac {2 p \left (x \right ) x}{x^{2}+2 p \left (x \right )}\tag {1} \end {align*}
Which becomes \begin {align*} \left (2 p\right ) dp &= \left (-x^{2}\right ) dp + \left (-2 p x\right ) dx\tag {2} \end {align*}
But the RHS is complete differential because \begin {align*} \left (-x^{2}\right ) dp + \left (-2 p x\right ) dx &= d\left (-p \,x^{2}\right ) \end {align*}
Hence (2) becomes \begin {align*} \left (2 p\right ) dp &= d\left (-p \,x^{2}\right ) \end {align*}
Integrating both sides gives gives these solutions \begin {align*} p \left (x \right )&=-\frac {x^{2}}{2}+\frac {\sqrt {x^{4}+4 c_{1}}}{2}+c_{1}\\ p \left (x \right )&=-\frac {x^{2}}{2}-\frac {\sqrt {x^{4}+4 c_{1}}}{2}+c_{1} \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(x=0\) and \(p=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 0 = -\sqrt {c_{1}}+c_{1} \end {align*}
The solutions are \begin {align*} c_{1} = 0\\ c_{1} = 1 \end {align*}
Trying the constant \begin {align*} c_{1} = 0 \end {align*}
Substituting this in the general solution gives \begin {align*} p \left (x \right )&=-\frac {x^{2}}{2}-\frac {\sqrt {x^{4}}}{2} \end {align*}
The constant \(c_{1} = 0\) gives valid solution.
Initial conditions are used to solve for \(c_{1}\). Substituting \(x=0\) and \(p=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 0 = \sqrt {c_{1}}+c_{1} \end {align*}
The solutions are \begin {align*} c_{1} = 0 \end {align*}
Trying the constant \begin {align*} c_{1} = 0 \end {align*}
Substituting this in the general solution gives \begin {align*} p \left (x \right )&=-\frac {x^{2}}{2}+\frac {\sqrt {x^{4}}}{2} \end {align*}
The constant \(c_{1} = 0\) gives valid solution.
For solution (1) found earlier, since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = -\frac {x^{2}}{2}+\frac {\sqrt {x^{4}}}{2} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { -\frac {x^{2}}{2}+\frac {\sqrt {x^{4}}}{2}\,\mathop {\mathrm {d}x}}\\ &= \frac {\sqrt {x^{4}}\, x}{6}-\frac {x^{3}}{6}+c_{2} \end {align*}
Initial conditions are used to solve for \(c_{2}\). Substituting \(x=0\) and \(y=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 1 = c_{2} \end {align*}
The solutions are \begin {align*} c_{2} = 1 \end {align*}
Trying the constant \begin {align*} c_{2} = 1 \end {align*}
Substituting this in the general solution gives \begin {align*} y&=\frac {\sqrt {x^{4}}\, x}{6}-\frac {x^{3}}{6}+1 \end {align*}
The constant \(c_{2} = 1\) gives valid solution.
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = -\frac {x^{2}}{2}-\frac {\sqrt {x^{4}}}{2} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { -\frac {x^{2}}{2}-\frac {\sqrt {x^{4}}}{2}\,\mathop {\mathrm {d}x}}\\ &= -\frac {x^{3}}{6}-\frac {\sqrt {x^{4}}\, x}{6}+c_{3} \end {align*}
Initial conditions are used to solve for \(c_{3}\). Substituting \(x=0\) and \(y=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 1 = c_{3} \end {align*}
The solutions are \begin {align*} c_{3} = 1 \end {align*}
Trying the constant \begin {align*} c_{3} = 1 \end {align*}
Substituting this in the general solution gives \begin {align*} y&=-\frac {x^{3}}{6}-\frac {\sqrt {x^{4}}\, x}{6}+1 \end {align*}
The constant \(c_{3} = 1\) gives valid solution.
Initial conditions are used to solve for the constants of integration.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\sqrt {x^{4}}\, x}{6}-\frac {x^{3}}{6}+1 \\ \tag{2} y &= -\frac {x^{3}}{6}-\frac {\sqrt {x^{4}}\, x}{6}+1 \\ \end{align*}
Verification of solutions
\[ y = \frac {\sqrt {x^{4}}\, x}{6}-\frac {x^{3}}{6}+1 \] Verified OK.
\[ y = -\frac {x^{3}}{6}-\frac {\sqrt {x^{4}}\, x}{6}+1 \] Verified OK.
Writing the ode as \[ \left (x^{2}+2 y^{\prime }\right ) y^{\prime \prime }+2 y^{\prime } x = 0 \] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (\left (x^{2}+2 y^{\prime }\right ) y^{\prime \prime }+2 y^{\prime } x \right )d x &= 0 \\ x^{2} y^{\prime }+{y^{\prime }}^{2} = c_{1} \end {align*}
Which is now solved for \(y\). Solving the given ode for \(y^{\prime }\) results in \(2\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=-\frac {x^{2}}{2}+\frac {\sqrt {x^{4}+4 c_{1}}}{2} \tag {1} \\ y^{\prime }&=-\frac {x^{2}}{2}-\frac {\sqrt {x^{4}+4 c_{1}}}{2} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin {align*} y &= \int { -\frac {x^{2}}{2}+\frac {\sqrt {x^{4}+4 c_{1}}}{2}\,\mathop {\mathrm {d}x}}\\ &= \frac {x \sqrt {x^{4}+4 c_{1}}}{6}+\frac {c_{1} \sqrt {2}\, \sqrt {4-\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \sqrt {4+\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {i}{\sqrt {c_{1}}}}}{2}, i\right )}{3 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {x^{4}+4 c_{1}}}-\frac {x^{3}}{6}+c_{2} \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} y &= \int { -\frac {x^{2}}{2}-\frac {\sqrt {x^{4}+4 c_{1}}}{2}\,\mathop {\mathrm {d}x}}\\ &= -\frac {x^{3}}{6}-\frac {x \sqrt {x^{4}+4 c_{1}}}{6}-\frac {c_{1} \sqrt {2}\, \sqrt {4-\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \sqrt {4+\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {i}{\sqrt {c_{1}}}}}{2}, i\right )}{3 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {x^{4}+4 c_{1}}}+c_{3} \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the First solution \begin {align*} y = \frac {x \sqrt {x^{4}+4 c_{1}}}{6}+\frac {c_{1} \sqrt {2}\, \sqrt {4-\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \sqrt {4+\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {i}{\sqrt {c_{1}}}}}{2}, i\right )}{3 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {x^{4}+4 c_{1}}}-\frac {x^{3}}{6}+c_{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 \(y = 1\) and \(x = 0\) in the above gives \begin {align*} 1 = c_{2}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \frac {\sqrt {x^{4}+4 c_{1}}}{6}+\frac {x^{4}}{3 \sqrt {x^{4}+4 c_{1}}}-\frac {2 i \sqrt {c_{1}}\, \sqrt {2}\, \sqrt {4+\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {i}{\sqrt {c_{1}}}}}{2}, i\right ) x}{3 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {4-\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \sqrt {x^{4}+4 c_{1}}}+\frac {2 i \sqrt {c_{1}}\, \sqrt {2}\, \sqrt {4-\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {i}{\sqrt {c_{1}}}}}{2}, i\right ) x}{3 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {4+\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \sqrt {x^{4}+4 c_{1}}}-\frac {2 c_{1} \sqrt {2}\, \sqrt {4-\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \sqrt {4+\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {i}{\sqrt {c_{1}}}}}{2}, i\right ) x^{3}}{3 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \left (x^{4}+4 c_{1} \right )^{\frac {3}{2}}}+\frac {4 c_{1}}{3 \sqrt {x^{4}+4 c_{1}}}-\frac {x^{2}}{2} \end {align*}
substituting \(y^{\prime } = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = \sqrt {c_{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}&=0\\ c_{2}&=1 \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \munderset {c_{1} \rightarrow 0}{\operatorname {lim}}\left (-\frac {-\sqrt {\frac {i}{\sqrt {c_{1}}}}\, x^{5}+x^{3} \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {x^{4}+4 c_{1}}-4 c_{1} \sqrt {2}\, \sqrt {\frac {-i x^{2}+2 \sqrt {c_{1}}}{\sqrt {c_{1}}}}\, \sqrt {\frac {i x^{2}+2 \sqrt {c_{1}}}{\sqrt {c_{1}}}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {i}{\sqrt {c_{1}}}}}{2}, i\right )-4 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, c_{1} x -6 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {x^{4}+4 c_{1}}}{6 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {x^{4}+4 c_{1}}}\right ) \end {align*}
Looking at the Second solution \begin {align*} y = -\frac {x^{3}}{6}-\frac {x \sqrt {x^{4}+4 c_{1}}}{6}-\frac {c_{1} \sqrt {2}\, \sqrt {4-\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \sqrt {4+\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {i}{\sqrt {c_{1}}}}}{2}, i\right )}{3 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {x^{4}+4 c_{1}}}+c_{3} \tag {2} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = 1\) and \(x = 0\) in the above gives \begin {align*} 1 = c_{3}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = -\frac {x^{2}}{2}-\frac {\sqrt {x^{4}+4 c_{1}}}{6}-\frac {x^{4}}{3 \sqrt {x^{4}+4 c_{1}}}+\frac {2 i \sqrt {c_{1}}\, \sqrt {2}\, \sqrt {4+\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {i}{\sqrt {c_{1}}}}}{2}, i\right ) x}{3 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {4-\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \sqrt {x^{4}+4 c_{1}}}-\frac {2 i \sqrt {c_{1}}\, \sqrt {2}\, \sqrt {4-\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {i}{\sqrt {c_{1}}}}}{2}, i\right ) x}{3 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {4+\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \sqrt {x^{4}+4 c_{1}}}+\frac {2 c_{1} \sqrt {2}\, \sqrt {4-\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \sqrt {4+\frac {2 i x^{2}}{\sqrt {c_{1}}}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {i}{\sqrt {c_{1}}}}}{2}, i\right ) x^{3}}{3 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \left (x^{4}+4 c_{1} \right )^{\frac {3}{2}}}-\frac {4 c_{1}}{3 \sqrt {x^{4}+4 c_{1}}} \end {align*}
substituting \(y^{\prime } = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = -\sqrt {c_{1}}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{3}\}\). Solving for the constants gives \begin {align*} c_{1}&=0\\ c_{3}&=1 \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \munderset {c_{1} \rightarrow 0}{\operatorname {lim}}\left (-\frac {\sqrt {\frac {i}{\sqrt {c_{1}}}}\, x^{5}+x^{3} \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {x^{4}+4 c_{1}}+4 c_{1} \sqrt {2}\, \sqrt {\frac {-i x^{2}+2 \sqrt {c_{1}}}{\sqrt {c_{1}}}}\, \sqrt {\frac {i x^{2}+2 \sqrt {c_{1}}}{\sqrt {c_{1}}}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {i}{\sqrt {c_{1}}}}}{2}, i\right )+4 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, c_{1} x -6 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {x^{4}+4 c_{1}}}{6 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {x^{4}+4 c_{1}}}\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\left (\munderset {c_{1} \rightarrow 0}{\operatorname {lim}}\left (-\frac {4 \left (\left (1+i\right ) \operatorname {EllipticF}\left (\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) x}{c_{1}^{\frac {1}{4}}}, i\right ) c_{1}^{\frac {5}{4}} \sqrt {\frac {i x^{2}+2 \sqrt {c_{1}}}{\sqrt {c_{1}}}}\, \sqrt {-\frac {i x^{2}-2 \sqrt {c_{1}}}{\sqrt {c_{1}}}}+\left (-\frac {x^{3}}{4}+\frac {3}{2}\right ) \sqrt {x^{4}+4 c_{1}}+\frac {x^{5}}{4}+c_{1} x \right )}{\sqrt {x^{4}+4 c_{1}}}\right )\right )}{6} \\ \tag{2} y &= -\frac {\left (\munderset {c_{1} \rightarrow 0}{\operatorname {lim}}\frac {4 \left (1+i\right ) \operatorname {EllipticF}\left (\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) x}{c_{1}^{\frac {1}{4}}}, i\right ) c_{1}^{\frac {5}{4}} \sqrt {\frac {i x^{2}+2 \sqrt {c_{1}}}{\sqrt {c_{1}}}}\, \sqrt {-\frac {i x^{2}-2 \sqrt {c_{1}}}{\sqrt {c_{1}}}}+4 \left (\frac {x^{3}}{4}-\frac {3}{2}\right ) \sqrt {x^{4}+4 c_{1}}+x^{5}+4 c_{1} x}{\sqrt {x^{4}+4 c_{1}}}\right )}{6} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\left (\munderset {c_{1} \rightarrow 0}{\operatorname {lim}}\left (-\frac {4 \left (\left (1+i\right ) \operatorname {EllipticF}\left (\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) x}{c_{1}^{\frac {1}{4}}}, i\right ) c_{1}^{\frac {5}{4}} \sqrt {\frac {i x^{2}+2 \sqrt {c_{1}}}{\sqrt {c_{1}}}}\, \sqrt {-\frac {i x^{2}-2 \sqrt {c_{1}}}{\sqrt {c_{1}}}}+\left (-\frac {x^{3}}{4}+\frac {3}{2}\right ) \sqrt {x^{4}+4 c_{1}}+\frac {x^{5}}{4}+c_{1} x \right )}{\sqrt {x^{4}+4 c_{1}}}\right )\right )}{6} \] Verified OK.
\[ y = -\frac {\left (\munderset {c_{1} \rightarrow 0}{\operatorname {lim}}\frac {4 \left (1+i\right ) \operatorname {EllipticF}\left (\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) x}{c_{1}^{\frac {1}{4}}}, i\right ) c_{1}^{\frac {5}{4}} \sqrt {\frac {i x^{2}+2 \sqrt {c_{1}}}{\sqrt {c_{1}}}}\, \sqrt {-\frac {i x^{2}-2 \sqrt {c_{1}}}{\sqrt {c_{1}}}}+4 \left (\frac {x^{3}}{4}-\frac {3}{2}\right ) \sqrt {x^{4}+4 c_{1}}+x^{5}+4 c_{1} x}{\sqrt {x^{4}+4 c_{1}}}\right )}{6} \] Verified OK.
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville trying 2nd order WeierstrassP trying 2nd order JacobiSN differential order: 2; trying a linearization to 3rd order trying 2nd order ODE linearizable_by_differentiation trying 2nd order, 2 integrating factors of the form mu(x,y) trying differential order: 2; missing variables `, `-> Computing symmetries using: way = 3 -> Calling odsolve with the ODE`, diff(_b(_a), _a) = -2*_b(_a)*_a/(_a^2+2*_b(_a)), _b(_a), HINT = [[_a, 2*_b]]` *** Sublevel 2 *** symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[_a, 2*_b]
✓ Solution by Maple
Time used: 0.079 (sec). Leaf size: 15
dsolve([(x^2+2*diff(y(x),x))*diff(y(x),x$2)+2*x*diff(y(x),x)=0,y(0) = 1, D(y)(0) = 0],y(x), singsol=all)
\begin{align*} y \left (x \right ) &= 1 \\ y \left (x \right ) &= -\frac {x^{3}}{3}+1 \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.252 (sec). Leaf size: 32
DSolve[{(x^2+2*y'[x])*y''[x]+2*x*y'[x]==0,{y[0]==1,y'[0]==0}},y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \text {Indeterminate} \\ y(x)\to 0 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},\text {ComplexInfinity}\right )-\frac {x^3}{6}+1 \\ \end{align*}