Internal problem ID [6098]
Internal file name [OUTPUT/5346_Sunday_June_05_2022_03_34_53_PM_64136479/index.tex]
Book: An introduction to Ordinary Differential Equations. Earl A. Coddington. Dover. NY
1961
Section: Chapter 6. Existence and uniqueness of solutions to systems and nth order equations.
Page 238
Problem number: 3.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_integrable_as_is", "second_order_ode_missing_x", "second_order_ode_missing_y", "exact nonlinear second order ode"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_poly_yn]]
\[ \boxed {y^{\prime \prime }+\frac {1}{2 {y^{\prime }}^{2}}=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = -1] \end {align*}
Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int 2 y^{\prime \prime } {y^{\prime }}^{2}d x &= \int \left (-1\right )d x\\ \frac {2 {y^{\prime }}^{3}}{3} = -x + c_{1} \end {align*}
Which is now solved for \(y\). Solving the given ode for \(y^{\prime }\) results in \(3\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=\frac {\left (-12 x +12 c_{1} \right )^{\frac {1}{3}}}{2} \tag {1} \\ y^{\prime }&=-\frac {\left (-12 x +12 c_{1} \right )^{\frac {1}{3}}}{4}+\frac {i \sqrt {3}\, \left (-12 x +12 c_{1} \right )^{\frac {1}{3}}}{4} \tag {2} \\ y^{\prime }&=-\frac {\left (-12 x +12 c_{1} \right )^{\frac {1}{3}}}{4}-\frac {i \sqrt {3}\, \left (-12 x +12 c_{1} \right )^{\frac {1}{3}}}{4} \tag {3} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin {align*} y &= \int { \frac {\left (-12 x +12 c_{1} \right )^{\frac {1}{3}}}{2}\,\mathop {\mathrm {d}x}}\\ &= -\frac {\left (-12 x +12 c_{1} \right )^{\frac {4}{3}}}{32}+c_{2} \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} y &= \int { -\frac {\left (-12 x +12 c_{1} \right )^{\frac {1}{3}}}{4}+\frac {i \sqrt {3}\, \left (-12 x +12 c_{1} \right )^{\frac {1}{3}}}{4}\,\mathop {\mathrm {d}x}}\\ &= -\frac {\left (-12 x +12 c_{1} \right )^{\frac {4}{3}} \left (i \sqrt {3}-1\right )}{64}+c_{3} \end {align*}
Solving equation (3)
Integrating both sides gives \begin {align*} y &= \int { -\frac {\left (-12 x +12 c_{1} \right )^{\frac {1}{3}}}{4}-\frac {i \sqrt {3}\, \left (-12 x +12 c_{1} \right )^{\frac {1}{3}}}{4}\,\mathop {\mathrm {d}x}}\\ &= \frac {\left (-12 x +12 c_{1} \right )^{\frac {4}{3}} \left (1+i \sqrt {3}\right )}{64}+c_{4} \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the First solution \begin {align*} y = -\frac {\left (-12 x +12 c_{1} \right )^{\frac {4}{3}}}{32}+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 = -\frac {3 c_{1}^{\frac {4}{3}} 12^{\frac {1}{3}}}{8}+c_{2}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \frac {\left (-12 x +12 c_{1} \right )^{\frac {1}{3}}}{2} \end {align*}
substituting \(y^{\prime } = -1\) and \(x = 0\) in the above gives \begin {align*} -1 = \frac {c_{1}^{\frac {1}{3}} 12^{\frac {1}{3}}}{2}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). There is no solution for the constants of integrations. This solution is removed.
Looking at the Second solution \begin {align*} y = -\frac {\left (-12 x +12 c_{1} \right )^{\frac {4}{3}} \left (i \sqrt {3}-1\right )}{64}+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 = -\frac {3 \,2^{\frac {2}{3}} \left (i 3^{\frac {5}{6}}-3^{\frac {1}{3}}\right ) c_{1}^{\frac {4}{3}}}{16}+c_{3}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \frac {\left (-12 x +12 c_{1} \right )^{\frac {1}{3}} \left (i \sqrt {3}-1\right )}{4} \end {align*}
substituting \(y^{\prime } = -1\) and \(x = 0\) in the above gives \begin {align*} -1 = \frac {2^{\frac {2}{3}} c_{1}^{\frac {1}{3}} \left (i 3^{\frac {5}{6}}-3^{\frac {1}{3}}\right )}{4}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{3}\}\). Solving for the constants gives \begin {align*} c_{1}&=-{\frac {2}{3}}\\ c_{3}&={\frac {3}{2}} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = -\frac {i \left (-12 x -8\right )^{\frac {4}{3}} \sqrt {3}}{64}+\frac {\left (-12 x -8\right )^{\frac {4}{3}}}{64}+\frac {3}{2} \end {align*}
Looking at the Third solution \begin {align*} y = \frac {\left (-12 x +12 c_{1} \right )^{\frac {4}{3}} \left (1+i \sqrt {3}\right )}{64}+c_{4} \tag {3} \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 = \frac {3 \,2^{\frac {2}{3}} \left (i 3^{\frac {5}{6}}+3^{\frac {1}{3}}\right ) c_{1}^{\frac {4}{3}}}{16}+c_{4}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = -\frac {\left (-12 x +12 c_{1} \right )^{\frac {1}{3}} \left (1+i \sqrt {3}\right )}{4} \end {align*}
substituting \(y^{\prime } = -1\) and \(x = 0\) in the above gives \begin {align*} -1 = -\frac {2^{\frac {2}{3}} c_{1}^{\frac {1}{3}} \left (i 3^{\frac {5}{6}}+3^{\frac {1}{3}}\right )}{4}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{4}\}\). There is no solution for the constants of integrations. This solution is removed.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {3}{2}+\frac {3 \left (x +\frac {2}{3}\right ) \left (i \sqrt {3}-1\right ) \left (-12 x -8\right )^{\frac {1}{3}}}{16} \\ \end{align*}
Verification of solutions
\[ y = \frac {3}{2}+\frac {3 \left (x +\frac {2}{3}\right ) \left (i \sqrt {3}-1\right ) \left (-12 x -8\right )^{\frac {1}{3}}}{16} \] 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*} 2 p^{\prime }\left (x \right ) p \left (x \right )^{2}+1 = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. Integrating both sides gives \begin {align*} \int -2 p^{2}d p &= \int {dx}\\ -\frac {2 p^{3}}{3}&= c_{1} +x \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(x=0\) and \(p=-1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} {\frac {2}{3}} = c_{1} \end {align*}
The solutions are \begin {align*} c_{1} = {\frac {2}{3}} \end {align*}
Trying the constant \begin {align*} c_{1} = {\frac {2}{3}} \end {align*}
Substituting \(c_{1}\) found above in the general solution gives \begin {align*} -\frac {2 p^{3}}{3} = x +\frac {2}{3} \end {align*}
The constant \(c_{1} = {\frac {2}{3}}\) gives valid solution.
For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} -\frac {2 {y^{\prime }}^{3}}{3} = x +\frac {2}{3} \end {align*}
Solving the given ode for \(y^{\prime }\) results in \(3\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=\frac {\left (-12 x -8\right )^{\frac {1}{3}}}{2} \tag {1} \\ y^{\prime }&=-\frac {\left (-12 x -8\right )^{\frac {1}{3}}}{4}-\frac {i \sqrt {3}\, \left (-12 x -8\right )^{\frac {1}{3}}}{4} \tag {2} \\ y^{\prime }&=-\frac {\left (-12 x -8\right )^{\frac {1}{3}}}{4}+\frac {i \sqrt {3}\, \left (-12 x -8\right )^{\frac {1}{3}}}{4} \tag {3} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin {align*} y &= \int { \frac {\left (-12 x -8\right )^{\frac {1}{3}}}{2}\,\mathop {\mathrm {d}x}}\\ &= -\frac {\left (-12 x -8\right )^{\frac {4}{3}}}{32}+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 = \frac {1}{4}+\frac {i \sqrt {3}}{4}+c_{2} \end {align*}
The solutions are \begin {align*} c_{2} = \frac {3}{4}-\frac {i \sqrt {3}}{4} \end {align*}
Trying the constant \begin {align*} c_{2} = \frac {3}{4}-\frac {i \sqrt {3}}{4} \end {align*}
Substituting this in the general solution gives \begin {align*} y&=-\frac {\left (-12 x -8\right )^{\frac {4}{3}}}{32}+\frac {3}{4}-\frac {i \sqrt {3}}{4} \end {align*}
The constant \(c_{2} = \frac {3}{4}-\frac {i \sqrt {3}}{4}\) gives valid solution.
Solving equation (2)
Integrating both sides gives \begin {align*} y &= \int { -\frac {\left (-12 x -8\right )^{\frac {1}{3}}}{4}-\frac {i \sqrt {3}\, \left (-12 x -8\right )^{\frac {1}{3}}}{4}\,\mathop {\mathrm {d}x}}\\ &= \frac {\left (-12 x -8\right )^{\frac {4}{3}} \left (1+i \sqrt {3}\right )}{64}+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 = \frac {1}{4}-\frac {i \sqrt {3}}{4}+c_{3} \end {align*}
The solutions are \begin {align*} c_{3} = \frac {3}{4}+\frac {i \sqrt {3}}{4} \end {align*}
Trying the constant \begin {align*} c_{3} = \frac {3}{4}+\frac {i \sqrt {3}}{4} \end {align*}
Substituting this in the general solution gives \begin {align*} y&=\frac {\left (-12 x -8\right )^{\frac {4}{3}}}{64}+\frac {i \left (-12 x -8\right )^{\frac {4}{3}} \sqrt {3}}{64}+\frac {3}{4}+\frac {i \sqrt {3}}{4} \end {align*}
The constant \(c_{3} = \frac {3}{4}+\frac {i \sqrt {3}}{4}\) gives valid solution.
Solving equation (3)
Integrating both sides gives \begin {align*} y &= \int { -\frac {\left (-12 x -8\right )^{\frac {1}{3}}}{4}+\frac {i \sqrt {3}\, \left (-12 x -8\right )^{\frac {1}{3}}}{4}\,\mathop {\mathrm {d}x}}\\ &= -\frac {\left (-12 x -8\right )^{\frac {4}{3}} \left (i \sqrt {3}-1\right )}{64}+c_{4} \end {align*}
Initial conditions are used to solve for \(c_{4}\). Substituting \(x=0\) and \(y=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 1 = -\frac {1}{2}+c_{4} \end {align*}
The solutions are \begin {align*} c_{4} = {\frac {3}{2}} \end {align*}
Trying the constant \begin {align*} c_{4} = {\frac {3}{2}} \end {align*}
Substituting this in the general solution gives \begin {align*} y&=-\frac {i \left (-12 x -8\right )^{\frac {4}{3}} \sqrt {3}}{64}+\frac {\left (-12 x -8\right )^{\frac {4}{3}}}{64}+\frac {3}{2} \end {align*}
The constant \(c_{4} = {\frac {3}{2}}\) 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 {\left (-12 x -8\right )^{\frac {4}{3}}}{32}+\frac {3}{4}-\frac {i \sqrt {3}}{4} \\ \tag{2} y &= \frac {\left (-12 x -8\right )^{\frac {4}{3}}}{64}+\frac {i \left (-12 x -8\right )^{\frac {4}{3}} \sqrt {3}}{64}+\frac {3}{4}+\frac {i \sqrt {3}}{4} \\ \tag{3} y &= -\frac {i \left (-12 x -8\right )^{\frac {4}{3}} \sqrt {3}}{64}+\frac {\left (-12 x -8\right )^{\frac {4}{3}}}{64}+\frac {3}{2} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\left (-12 x -8\right )^{\frac {4}{3}}}{32}+\frac {3}{4}-\frac {i \sqrt {3}}{4} \] Warning, solution could not be verified
\[ y = \frac {\left (-12 x -8\right )^{\frac {4}{3}}}{64}+\frac {i \left (-12 x -8\right )^{\frac {4}{3}} \sqrt {3}}{64}+\frac {3}{4}+\frac {i \sqrt {3}}{4} \] Warning, solution could not be verified
\[ y = -\frac {i \left (-12 x -8\right )^{\frac {4}{3}} \sqrt {3}}{64}+\frac {\left (-12 x -8\right )^{\frac {4}{3}}}{64}+\frac {3}{2} \] Verified OK.
This is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable \(y\) an independent variable. Using \begin {align*} y' &= p(y) \end {align*}
Then \begin {align*} y'' &= \frac {dp}{dx}\\ &= \frac {dy}{dx} \frac {dp}{dy}\\ &= p \frac {dp}{dy} \end {align*}
Hence the ode becomes \begin {align*} 2 p \left (y \right )^{3} \left (\frac {d}{d y}p \left (y \right )\right ) = -1 \end {align*}
Which is now solved as first order ode for \(p(y)\). Integrating both sides gives \begin {align*} \int -2 p^{3}d p &= \int {dy}\\ -\frac {p^{4}}{2}&= y +c_{1} \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(y=1\) and \(p=-1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -{\frac {1}{2}} = 1+c_{1} \end {align*}
The solutions are \begin {align*} c_{1} = -{\frac {3}{2}} \end {align*}
Trying the constant \begin {align*} c_{1} = -{\frac {3}{2}} \end {align*}
Substituting \(c_{1}\) found above in the general solution gives \begin {align*} -\frac {p^{4}}{2} = y -\frac {3}{2} \end {align*}
The constant \(c_{1} = -{\frac {3}{2}}\) gives valid solution.
For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} -\frac {{y^{\prime }}^{4}}{2} = y-\frac {3}{2} \end {align*}
Solving the given ode for \(y^{\prime }\) results in \(4\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=\left (-2 y+3\right )^{\frac {1}{4}} \tag {1} \\ y^{\prime }&=i \left (-2 y+3\right )^{\frac {1}{4}} \tag {2} \\ y^{\prime }&=-\left (-2 y+3\right )^{\frac {1}{4}} \tag {3} \\ y^{\prime }&=-i \left (-2 y+3\right )^{\frac {1}{4}} \tag {4} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin {align*} \int \frac {1}{\left (-2 y +3\right )^{\frac {1}{4}}}d y &= \int {dx}\\ -\frac {2 \left (-2 y +3\right )^{\frac {3}{4}}}{3}&= x +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*} -{\frac {2}{3}} = c_{2} \end {align*}
The solutions are \begin {align*} c_{2} = -{\frac {2}{3}} \end {align*}
Trying the constant \begin {align*} c_{2} = -{\frac {2}{3}} \end {align*}
Substituting \(c_{2}\) found above in the general solution gives \begin {align*} -\frac {2 \left (-2 y +3\right )^{\frac {3}{4}}}{3} = x -\frac {2}{3} \end {align*}
The constant \(c_{2} = -{\frac {2}{3}}\) gives valid solution.
Solving equation (2)
Integrating both sides gives \begin {align*} \int -\frac {i}{\left (-2 y +3\right )^{\frac {1}{4}}}d y &= \int {dx}\\ \frac {2 i \left (-2 y +3\right )^{\frac {3}{4}}}{3}&= x +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*} \frac {2 i}{3} = c_{3} \end {align*}
The solutions are \begin {align*} c_{3} = \frac {2 i}{3} \end {align*}
Trying the constant \begin {align*} c_{3} = \frac {2 i}{3} \end {align*}
Substituting \(c_{3}\) found above in the general solution gives \begin {align*} \frac {2 i \left (-2 y +3\right )^{\frac {3}{4}}}{3} = x +\frac {2 i}{3} \end {align*}
The constant \(c_{3} = \frac {2 i}{3}\) gives valid solution.
Solving equation (3)
Integrating both sides gives \begin {align*} \int -\frac {1}{\left (-2 y +3\right )^{\frac {1}{4}}}d y &= \int {dx}\\ \frac {2 \left (-2 y +3\right )^{\frac {3}{4}}}{3}&= x +c_{4} \end {align*}
Initial conditions are used to solve for \(c_{4}\). Substituting \(x=0\) and \(y=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} {\frac {2}{3}} = c_{4} \end {align*}
The solutions are \begin {align*} c_{4} = {\frac {2}{3}} \end {align*}
Trying the constant \begin {align*} c_{4} = {\frac {2}{3}} \end {align*}
Substituting \(c_{4}\) found above in the general solution gives \begin {align*} \frac {2 \left (-2 y +3\right )^{\frac {3}{4}}}{3} = x +\frac {2}{3} \end {align*}
The constant \(c_{4} = {\frac {2}{3}}\) gives valid solution.
Solving equation (4)
Integrating both sides gives \begin {align*} \int \frac {i}{\left (-2 y +3\right )^{\frac {1}{4}}}d y &= \int {dx}\\ -\frac {2 i \left (-2 y +3\right )^{\frac {3}{4}}}{3}&= x +c_{5} \end {align*}
Initial conditions are used to solve for \(c_{5}\). Substituting \(x=0\) and \(y=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -\frac {2 i}{3} = c_{5} \end {align*}
The solutions are \begin {align*} c_{5} = -\frac {2 i}{3} \end {align*}
Trying the constant \begin {align*} c_{5} = -\frac {2 i}{3} \end {align*}
Substituting \(c_{5}\) found above in the general solution gives \begin {align*} -\frac {2 i \left (-2 y +3\right )^{\frac {3}{4}}}{3} = x -\frac {2 i}{3} \end {align*}
The constant \(c_{5} = -\frac {2 i}{3}\) 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 {3}{2}-\frac {\left (-\frac {3 x}{2}+1\right )^{\frac {4}{3}}}{2} \\ \tag{2} y &= \frac {3}{2}-\frac {\left (-\frac {i \left (2 i+3 x \right )}{2}\right )^{\frac {4}{3}}}{2} \\ \tag{3} y &= \frac {3}{2}-\frac {\left (\frac {3 x}{2}+1\right )^{\frac {4}{3}}}{2} \\ \tag{4} y &= \frac {3}{2}-\frac {\left (-\frac {i \left (2 i-3 x \right )}{2}\right )^{\frac {4}{3}}}{2} \\ \end{align*}
Verification of solutions
\[ y = \frac {3}{2}-\frac {\left (-\frac {3 x}{2}+1\right )^{\frac {4}{3}}}{2} \] Warning, solution could not be verified
\[ y = \frac {3}{2}-\frac {\left (-\frac {i \left (2 i+3 x \right )}{2}\right )^{\frac {4}{3}}}{2} \] Warning, solution could not be verified
\[ y = \frac {3}{2}-\frac {\left (\frac {3 x}{2}+1\right )^{\frac {4}{3}}}{2} \] Verified OK.
\[ y = \frac {3}{2}-\frac {\left (-\frac {i \left (2 i-3 x \right )}{2}\right )^{\frac {4}{3}}}{2} \] Warning, solution could not be verified
An exact non-linear second order ode has the form \begin {align*} a_{2} \left (x , y, y^{\prime }\right ) y^{\prime \prime }+a_{1} \left (x , y, y^{\prime }\right ) y^{\prime }+a_{0} \left (x , y, y^{\prime }\right )&=0 \end {align*}
Where the following conditions are satisfied \begin {align*} \frac {\partial a_2}{\partial y} &= \frac {\partial a_1}{\partial y'}\\ \frac {\partial a_2}{\partial x} &= \frac {\partial a_0}{\partial y'}\\ \frac {\partial a_1}{\partial x} &= \frac {\partial a_0}{\partial y} \end {align*}
Looking at the the ode given we see that \begin {align*} a_2 &= 2 {y^{\prime }}^{2}\\ a_1 &= 0\\ a_0 &= 1 \end {align*}
Applying the conditions to the above shows this is a nonlinear exact second order ode. Therefore it can be reduced to first order ode given by \begin {align*} \int {a_2\,d y'} + \int {a_1\,d y} + \int {a_0\,d x} &= c_{1}\\ \int {2 {y^{\prime }}^{2}\,d y'} + \int {0\,d y} + \int {1\,d x} &= c_{1} \end {align*}
Which results in \begin {align*} \frac {2 {y^{\prime }}^{3}}{3}+x = c_{1} \end {align*}
Which is now solved Solving the given ode for \(y^{\prime }\) results in \(3\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=\frac {\left (-12 x +12 c_{1} \right )^{\frac {1}{3}}}{2} \tag {1} \\ y^{\prime }&=-\frac {\left (-12 x +12 c_{1} \right )^{\frac {1}{3}}}{4}+\frac {i \sqrt {3}\, \left (-12 x +12 c_{1} \right )^{\frac {1}{3}}}{4} \tag {2} \\ y^{\prime }&=-\frac {\left (-12 x +12 c_{1} \right )^{\frac {1}{3}}}{4}-\frac {i \sqrt {3}\, \left (-12 x +12 c_{1} \right )^{\frac {1}{3}}}{4} \tag {3} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin {align*} y &= \int { \frac {\left (-12 x +12 c_{1} \right )^{\frac {1}{3}}}{2}\,\mathop {\mathrm {d}x}}\\ &= -\frac {\left (-12 x +12 c_{1} \right )^{\frac {4}{3}}}{32}+c_{2} \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} y &= \int { -\frac {\left (-12 x +12 c_{1} \right )^{\frac {1}{3}}}{4}+\frac {i \sqrt {3}\, \left (-12 x +12 c_{1} \right )^{\frac {1}{3}}}{4}\,\mathop {\mathrm {d}x}}\\ &= -\frac {\left (-12 x +12 c_{1} \right )^{\frac {4}{3}} \left (i \sqrt {3}-1\right )}{64}+c_{3} \end {align*}
Solving equation (3)
Integrating both sides gives \begin {align*} y &= \int { -\frac {\left (-12 x +12 c_{1} \right )^{\frac {1}{3}}}{4}-\frac {i \sqrt {3}\, \left (-12 x +12 c_{1} \right )^{\frac {1}{3}}}{4}\,\mathop {\mathrm {d}x}}\\ &= \frac {\left (-12 x +12 c_{1} \right )^{\frac {4}{3}} \left (1+i \sqrt {3}\right )}{64}+c_{4} \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the First solution \begin {align*} y = -\frac {\left (-12 x +12 c_{1} \right )^{\frac {4}{3}}}{32}+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 = -\frac {3 c_{1}^{\frac {4}{3}} 12^{\frac {1}{3}}}{8}+c_{2}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \frac {\left (-12 x +12 c_{1} \right )^{\frac {1}{3}}}{2} \end {align*}
substituting \(y^{\prime } = -1\) and \(x = 0\) in the above gives \begin {align*} -1 = \frac {c_{1}^{\frac {1}{3}} 12^{\frac {1}{3}}}{2}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). There is no solution for the constants of integrations. This solution is removed.
Looking at the Second solution \begin {align*} y = -\frac {\left (-12 x +12 c_{1} \right )^{\frac {4}{3}} \left (i \sqrt {3}-1\right )}{64}+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 = -\frac {3 \,2^{\frac {2}{3}} \left (i 3^{\frac {5}{6}}-3^{\frac {1}{3}}\right ) c_{1}^{\frac {4}{3}}}{16}+c_{3}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \frac {\left (-12 x +12 c_{1} \right )^{\frac {1}{3}} \left (i \sqrt {3}-1\right )}{4} \end {align*}
substituting \(y^{\prime } = -1\) and \(x = 0\) in the above gives \begin {align*} -1 = \frac {2^{\frac {2}{3}} c_{1}^{\frac {1}{3}} \left (i 3^{\frac {5}{6}}-3^{\frac {1}{3}}\right )}{4}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{3}\}\). Solving for the constants gives \begin {align*} c_{1}&=-{\frac {2}{3}}\\ c_{3}&={\frac {3}{2}} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = -\frac {i \left (-12 x -8\right )^{\frac {4}{3}} \sqrt {3}}{64}+\frac {\left (-12 x -8\right )^{\frac {4}{3}}}{64}+\frac {3}{2} \end {align*}
Looking at the Third solution \begin {align*} y = \frac {\left (-12 x +12 c_{1} \right )^{\frac {4}{3}} \left (1+i \sqrt {3}\right )}{64}+c_{4} \tag {3} \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 = \frac {3 \,2^{\frac {2}{3}} \left (i 3^{\frac {5}{6}}+3^{\frac {1}{3}}\right ) c_{1}^{\frac {4}{3}}}{16}+c_{4}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = -\frac {\left (-12 x +12 c_{1} \right )^{\frac {1}{3}} \left (1+i \sqrt {3}\right )}{4} \end {align*}
substituting \(y^{\prime } = -1\) and \(x = 0\) in the above gives \begin {align*} -1 = -\frac {2^{\frac {2}{3}} c_{1}^{\frac {1}{3}} \left (i 3^{\frac {5}{6}}+3^{\frac {1}{3}}\right )}{4}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{4}\}\). There is no solution for the constants of integrations. This solution is removed.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {3}{2}+\frac {3 \left (x +\frac {2}{3}\right ) \left (i \sqrt {3}-1\right ) \left (-12 x -8\right )^{\frac {1}{3}}}{16} \\ \end{align*}
Verification of solutions
\[ y = \frac {3}{2}+\frac {3 \left (x +\frac {2}{3}\right ) \left (i \sqrt {3}-1\right ) \left (-12 x -8\right )^{\frac {1}{3}}}{16} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [2 \left (\frac {d}{d x}y^{\prime }\right ) {y^{\prime }}^{2}=-1, y \left (0\right )=1, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=-1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime }\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & 2 u^{\prime }\left (x \right ) u \left (x \right )^{2}=-1 \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right ) u \left (x \right )^{2}=-\frac {1}{2} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int u^{\prime }\left (x \right ) u \left (x \right )^{2}d x =\int -\frac {1}{2}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {u \left (x \right )^{3}}{3}=-\frac {x}{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\frac {\left (24 c_{1} -12 x \right )^{\frac {1}{3}}}{2} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\frac {\left (24 c_{1} -12 x \right )^{\frac {1}{3}}}{2} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=\frac {\left (24 c_{1} -12 x \right )^{\frac {1}{3}}}{2} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int \frac {\left (24 c_{1} -12 x \right )^{\frac {1}{3}}}{2}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=\frac {3 \left (-2 c_{1} +x \right ) \left (24 c_{1} -12 x \right )^{\frac {1}{3}}}{8}+c_{2} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=\frac {3 \left (-2 c_{1} +x \right ) \left (24 c_{1} -12 x \right )^{\frac {1}{3}}}{8}+c_{2} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=-\frac {3 c_{1}^{\frac {4}{3}} 24^{\frac {1}{3}}}{4}+c_{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {3 \left (24 c_{1} -12 x \right )^{\frac {1}{3}}}{8}-\frac {3 \left (-2 c_{1} +x \right )}{2 \left (24 c_{1} -12 x \right )^{\frac {2}{3}}} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=-1 \\ {} & {} & -1=\frac {24^{\frac {1}{3}} c_{1}^{\frac {1}{3}}}{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & \circ & \textrm {The solution does not satisfy the initial conditions}\hspace {3pt} \end {array} \]
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) = -(1/2)/_b(_a)^2, _b(_a), HINT = [[1, 0], [_a, (1/3)*_b]]` *** Sublevel 2 *** symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[1, 0], [_a, 1/3*_b]
✓ Solution by Maple
Time used: 0.469 (sec). Leaf size: 26
dsolve([diff(y(x),x$2)=-1/(2*diff(y(x),x)^2),y(0) = 1, D(y)(0) = -1],y(x), singsol=all)
\[ y \left (x \right ) = \frac {3 \left (x +\frac {2}{3}\right ) \left (-12 x -8\right )^{\frac {1}{3}} \left (i \sqrt {3}-1\right )}{16}+\frac {3}{2} \]
✓ Solution by Mathematica
Time used: 0.013 (sec). Leaf size: 27
DSolve[{y''[x]==-1/(2*(y'[x])^2),{y[0]==1,y'[0]==-1}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{8} \left (12-(-2)^{2/3} (-3 x-2)^{4/3}\right ) \]