22.5 problem 710

Internal problem ID [15453]
Internal file name [OUTPUT/15454_Wednesday_May_08_2024_03_59_09_PM_89884284/index.tex]

Book: A book of problems in ordinary differential equations. M.L. KRASNOV, A.L. KISELYOV, G.I. MARKARENKO. MIR, MOSCOW. 1983
Section: Chapter 2 (Higher order ODE’s). Section 17. Boundary value problems. Exercises page 163
Problem number: 710.
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", "exact nonlinear second order ode"

Maple gives the following as the ode type

[[_2nd_order, _missing_x], [_2nd_order, _exact, _nonlinear], [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_xy]]

\[ \boxed {{y^{\prime }}^{2}+y y^{\prime \prime }=-1} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y \left (1\right ) = 2] \end {align*}

22.5.1 Solving as second order integrable as is ode

Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left ({y^{\prime }}^{2}+y y^{\prime \prime }\right )d x &= \int \left (-1\right )d x\\ y y^{\prime } = -x + c_{1} \end {align*}

Which is now solved for \(y\). In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {-x +c_{1}}{y} \end {align*}

Where \(f(x)=-x +c_{1}\) and \(g(y)=\frac {1}{y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{y}} \,dy &= -x +c_{1} \,d x \\ \int { \frac {1}{\frac {1}{y}} \,dy} &= \int {-x +c_{1} \,d x} \\ \frac {y^{2}}{2}&=-\frac {1}{2} x^{2}+c_{1} x +c_{2} \\ \end{align*} The solution is \[ \frac {y^{2}}{2}+\frac {x^{2}}{2}-c_{1} x -c_{2} = 0 \] Initial conditions are used to solve for the constants of integration.

Looking at the above solution \begin {align*} \frac {y^{2}}{2}+\frac {x^{2}}{2}-c_{1} x -c_{2} = 0 \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 \(y = 2\) and \(x = 1\) in the above gives \begin {align*} \frac {5}{2}-c_{1} -c_{2} = 0\tag {1A} \end {align*}

substituting \(y = 1\) and \(x = 0\) in the above gives \begin {align*} \frac {1}{2}-c_{2} = 0\tag {2A} \end {align*}

Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&=2\\ c_{2}&={\frac {1}{2}} \end {align*}

Substituting these values back in above solution results in \begin {align*} \frac {y^{2}}{2}+\frac {x^{2}}{2}-2 x -\frac {1}{2} = 0 \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} \frac {y^{2}}{2}+\frac {x^{2}}{2}-2 x -\frac {1}{2} &= 0 \\ \end{align*}

Verification of solutions

\[ \frac {y^{2}}{2}+\frac {x^{2}}{2}-2 x -\frac {1}{2} = 0 \] Verified OK.

22.5.2 Solving as second order ode missing x ode

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*} p \left (y \right )^{2}+y p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right ) = -1 \end {align*}

Which is now solved as first order ode for \(p(y)\). In canonical form the ODE is \begin {align*} p' &= F(y,p)\\ &= f( y) g(p)\\ &= -\frac {p^{2}+1}{y p} \end {align*}

Where \(f(y)=-\frac {1}{y}\) and \(g(p)=\frac {p^{2}+1}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {p^{2}+1}{p}} \,dp &= -\frac {1}{y} \,d y \\ \int { \frac {1}{\frac {p^{2}+1}{p}} \,dp} &= \int {-\frac {1}{y} \,d y} \\ \frac {\ln \left (p^{2}+1\right )}{2}&=-\ln \left (y \right )+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} \sqrt {p^{2}+1} &= {\mathrm e}^{-\ln \left (y \right )+c_{1}} \end {align*}

Which simplifies to \begin {align*} \sqrt {p^{2}+1} &= \frac {c_{2}}{y} \end {align*}

Which simplifies to \[ \sqrt {p \left (y \right )^{2}+1} = \frac {c_{2} {\mathrm e}^{c_{1}}}{y} \] The solution is \[ \sqrt {p \left (y \right )^{2}+1} = \frac {c_{2} {\mathrm e}^{c_{1}}}{y} \] For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} \sqrt {1+{y^{\prime }}^{2}} = \frac {c_{2} {\mathrm e}^{c_{1}}}{y} \end {align*}

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 {\sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}-y^{2}}}{y} \tag {1} \\ y^{\prime }&=-\frac {\sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}-y^{2}}}{y} \tag {2} \end {align*}

Now each one of the above ODE is solved.

Solving equation (1)

Integrating both sides gives \begin {align*} \int \frac {y}{\sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}-y^{2}}}d y &= \int {dx}\\ -\sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}-y^{2}}&= c_{3} +x \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(x=1\) and \(y=2\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -\sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}-4} = c_{3} +1 \end {align*}

The solutions are \begin {align*} c_{1} = \frac {\ln \left (\frac {c_{3}^{2}+2 c_{3} +5}{c_{2}^{2}}\right )}{2} \end {align*}

Trying the constant \begin {align*} c_{1} = \frac {\ln \left (\frac {c_{3}^{2}+2 c_{3} +5}{c_{2}^{2}}\right )}{2} \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} -\sqrt {c_{3}^{2}-y^{2}+2 c_{3} +5} = c_{3} +x \end {align*}

The constant \(c_{1} = \frac {\ln \left (\frac {c_{3}^{2}+2 c_{3} +5}{c_{2}^{2}}\right )}{2}\) does not give valid solution.

Warning: Unable to solve for constant of integration. Unable to determine ODE type.

Solving equation (2)

Integrating both sides gives \begin {align*} \int -\frac {y}{\sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}-y^{2}}}d y &= \int {dx}\\ \sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}-y^{2}}&= x +c_{4} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(x=1\) and \(y=2\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} \sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}-4} = 1+c_{4} \end {align*}

The solutions are \begin {align*} c_{1} = \frac {\ln \left (\frac {c_{4}^{2}+2 c_{4} +5}{c_{2}^{2}}\right )}{2} \end {align*}

Trying the constant \begin {align*} c_{1} = \frac {\ln \left (\frac {c_{4}^{2}+2 c_{4} +5}{c_{2}^{2}}\right )}{2} \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \sqrt {c_{4}^{2}-y^{2}+2 c_{4} +5} = x +c_{4} \end {align*}

The constant \(c_{1} = \frac {\ln \left (\frac {c_{4}^{2}+2 c_{4} +5}{c_{2}^{2}}\right )}{2}\) does not give valid solution.

Which is valid for any constant of integration. Therefore keeping the constant in place. Initial conditions are used to solve for the constants of integration.

Looking at the above solution \begin {align*} \sqrt {c_{4}^{2}-y^{2}+2 c_{4} +5} = x +c_{4} \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 \(y = 2\) and \(x = 1\) in the above gives \begin {align*} \operatorname {csgn}\left (1+c_{4} \right ) \left (1+c_{4} \right ) = 1+c_{4}\tag {1A} \end {align*}

substituting \(y = 1\) and \(x = 0\) in the above gives \begin {align*} \sqrt {c_{4}^{2}+2 c_{4} +4} = c_{4}\tag {2A} \end {align*}

Equations {1A,2A} are now solved for \(\{c_{4}\}\). There is no solution for the constants of integrations. This solution is removed.

Verification of solutions N/A

22.5.3 Solving as type second_order_integrable_as_is (not using ABC version)

Writing the ode as \[ {y^{\prime }}^{2}+y y^{\prime \prime } = -1 \] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left ({y^{\prime }}^{2}+y y^{\prime \prime }\right )d x &= \int \left (-1\right )d x\\ y y^{\prime } = -x +c_{1} \end {align*}

Which is now solved for \(y\). In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {-x +c_{1}}{y} \end {align*}

Where \(f(x)=-x +c_{1}\) and \(g(y)=\frac {1}{y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{y}} \,dy &= -x +c_{1} \,d x \\ \int { \frac {1}{\frac {1}{y}} \,dy} &= \int {-x +c_{1} \,d x} \\ \frac {y^{2}}{2}&=-\frac {1}{2} x^{2}+c_{1} x +c_{2} \\ \end{align*} The solution is \[ \frac {y^{2}}{2}+\frac {x^{2}}{2}-c_{1} x -c_{2} = 0 \] Initial conditions are used to solve for the constants of integration.

Looking at the above solution \begin {align*} \frac {y^{2}}{2}+\frac {x^{2}}{2}-c_{1} x -c_{2} = 0 \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 \(y = 2\) and \(x = 1\) in the above gives \begin {align*} \frac {5}{2}-c_{1} -c_{2} = 0\tag {1A} \end {align*}

substituting \(y = 1\) and \(x = 0\) in the above gives \begin {align*} \frac {1}{2}-c_{2} = 0\tag {2A} \end {align*}

Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&=2\\ c_{2}&={\frac {1}{2}} \end {align*}

Substituting these values back in above solution results in \begin {align*} \frac {y^{2}}{2}+\frac {x^{2}}{2}-2 x -\frac {1}{2} = 0 \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} \frac {y^{2}}{2}+\frac {x^{2}}{2}-2 x -\frac {1}{2} &= 0 \\ \end{align*}

Verification of solutions

\[ \frac {y^{2}}{2}+\frac {x^{2}}{2}-2 x -\frac {1}{2} = 0 \] Verified OK.

22.5.4 Solving as exact nonlinear second order ode ode

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 &= y\\ a_1 &= y^{\prime }\\ 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 {y\,d y'} + \int {y^{\prime }\,d y} + \int {1\,d x} &= c_{1} \end {align*}

Which results in \begin {align*} 2 y y^{\prime }+x = c_{1} \end {align*}

Which is now solved In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {-\frac {x}{2}+\frac {c_{1}}{2}}{y} \end {align*}

Where \(f(x)=-\frac {x}{2}+\frac {c_{1}}{2}\) and \(g(y)=\frac {1}{y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{y}} \,dy &= -\frac {x}{2}+\frac {c_{1}}{2} \,d x \\ \int { \frac {1}{\frac {1}{y}} \,dy} &= \int {-\frac {x}{2}+\frac {c_{1}}{2} \,d x} \\ \frac {y^{2}}{2}&=-\frac {1}{4} x^{2}+\frac {1}{2} c_{1} x +c_{2} \\ \end{align*} The solution is \[ \frac {y^{2}}{2}+\frac {x^{2}}{4}-\frac {c_{1} x}{2}-c_{2} = 0 \] Initial conditions are used to solve for the constants of integration.

Looking at the above solution \begin {align*} \frac {y^{2}}{2}+\frac {x^{2}}{4}-\frac {c_{1} x}{2}-c_{2} = 0 \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 \(y = 2\) and \(x = 1\) in the above gives \begin {align*} \frac {9}{4}-\frac {c_{1}}{2}-c_{2} = 0\tag {1A} \end {align*}

substituting \(y = 1\) and \(x = 0\) in the above gives \begin {align*} \frac {1}{2}-c_{2} = 0\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 {7}{2}}\\ c_{2}&={\frac {1}{2}} \end {align*}

Substituting these values back in above solution results in \begin {align*} \frac {y^{2}}{2}+\frac {x^{2}}{4}-\frac {7 x}{4}-\frac {1}{2} = 0 \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} \frac {y^{2}}{2}+\frac {x^{2}}{4}-\frac {7 x}{4}-\frac {1}{2} &= 0 \\ \end{align*}

Verification of solutions

\[ \frac {y^{2}}{2}+\frac {x^{2}}{4}-\frac {7 x}{4}-\frac {1}{2} = 0 \] Warning, solution could not be verified

22.5.5 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [{y^{\prime }}^{2}+y y^{\prime \prime }=-1, y \left (0\right )=1, y \left (1\right )=2\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (x \right )=y^{\prime } \\ \bullet & {} & \textrm {Compute}\hspace {3pt} y^{\prime \prime } \\ {} & {} & u^{\prime }\left (x \right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Use chain rule on the lhs}\hspace {3pt} \\ {} & {} & y^{\prime } \left (\frac {d}{d y}u \left (y \right )\right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Substitute in the definition of}\hspace {3pt} u \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Make substitutions}\hspace {3pt} y^{\prime }=u \left (y \right ),y^{\prime \prime }=u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & u \left (y \right )^{2}+y u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=-1 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=\frac {-u \left (y \right )^{2}-1}{y u \left (y \right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\left (\frac {d}{d y}u \left (y \right )\right ) u \left (y \right )}{-u \left (y \right )^{2}-1}=\frac {1}{y} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int \frac {\left (\frac {d}{d y}u \left (y \right )\right ) u \left (y \right )}{-u \left (y \right )^{2}-1}d y =\int \frac {1}{y}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {\ln \left (u \left (y \right )^{2}+1\right )}{2}=\ln \left (y \right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & \left \{u \left (y \right )=\frac {\sqrt {1-\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}}}{{\mathrm e}^{c_{1}} y}, u \left (y \right )=-\frac {\sqrt {1-\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}}}{{\mathrm e}^{c_{1}} y}\right \} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\frac {\sqrt {1-\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}}}{{\mathrm e}^{c_{1}} y} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=\frac {\sqrt {1-\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}}}{{\mathrm e}^{c_{1}} y} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\sqrt {1-\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}}}{{\mathrm e}^{c_{1}} y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime } y}{\sqrt {1-\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}}}=\frac {1}{{\mathrm e}^{c_{1}}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime } y}{\sqrt {1-\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}}}d x =\int \frac {1}{{\mathrm e}^{c_{1}}}d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {\sqrt {1-\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}}}{\left ({\mathrm e}^{c_{1}}\right )^{2}}=\frac {x}{{\mathrm e}^{c_{1}}}+c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=\frac {\sqrt {1-\left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}-2 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} x -\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}}{{\mathrm e}^{c_{1}}}, y=-\frac {\sqrt {1-\left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}-2 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} x -\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}}{{\mathrm e}^{c_{1}}}\right \} \\ \bullet & {} & \textrm {Solve 2nd ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-\frac {\sqrt {1-\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}}}{{\mathrm e}^{c_{1}} y} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=-\frac {\sqrt {1-\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}}}{{\mathrm e}^{c_{1}} y} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {\sqrt {1-\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}}}{{\mathrm e}^{c_{1}} y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime } y}{\sqrt {1-\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}}}=-\frac {1}{{\mathrm e}^{c_{1}}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime } y}{\sqrt {1-\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}}}d x =\int -\frac {1}{{\mathrm e}^{c_{1}}}d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {\sqrt {1-\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}}}{\left ({\mathrm e}^{c_{1}}\right )^{2}}=-\frac {x}{{\mathrm e}^{c_{1}}}+c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=\frac {\sqrt {1-\left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}+2 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} x -\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}}{{\mathrm e}^{c_{1}}}, y=-\frac {\sqrt {1-\left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}+2 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} x -\left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}}{{\mathrm e}^{c_{1}}}\right \} \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 a quadrature 
<- quadrature successful 
<- 2nd order, 2 integrating factors of the form mu(x,y) successful`
 

✓ Solution by Maple

Time used: 0.781 (sec). Leaf size: 16

dsolve([y(x)*diff(y(x),x$2)+diff(y(x),x)^2+1=0,y(0) = 1, y(1) = 2],y(x), singsol=all)
 

\[ y \left (x \right ) = \sqrt {-x^{2}+4 x +1} \]

✓ Solution by Mathematica

Time used: 12.271 (sec). Leaf size: 19

DSolve[{y[x]*y''[x]+y'[x]^2+1==0,{y[0]==1,y[1]==2}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \sqrt {-x^2+4 x+1} \]