Internal problem ID [4891]
Internal file name [OUTPUT/4384_Sunday_June_05_2022_01_12_24_PM_52737621/index.tex]
Book: Mathematical Methods in the Physical Sciences. third edition. Mary L. Boas. John Wiley.
2006
Section: Chapter 8, Ordinary differential equations. Section 13. Miscellaneous problems. page
466
Problem number: 28.
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]]
\[ \boxed {y y^{\prime \prime }+{y^{\prime }}^{2}=-4} \] With initial conditions \begin {align*} [y \left (1\right ) = 3, y^{\prime }\left (1\right ) = 0] \end {align*}
Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y y^{\prime \prime }+{y^{\prime }}^{2}\right )d x &= \int \left (-4\right )d x\\ y y^{\prime } = -4 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 {-4 x +c_{1}}{y} \end {align*}
Where \(f(x)=-4 x +c_{1}\) and \(g(y)=\frac {1}{y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{y}} \,dy &= -4 x +c_{1} \,d x \\ \int { \frac {1}{\frac {1}{y}} \,dy} &= \int {-4 x +c_{1} \,d x} \\ \frac {y^{2}}{2}&=c_{1} x -2 x^{2}+c_{2} \\ \end{align*} The solution is \[ \frac {y^{2}}{2}-c_{1} x +2 x^{2}-c_{2} = 0 \] Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} \frac {y^{2}}{2}-c_{1} x +2 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 \(y = 3\) and \(x = 1\) in the above gives \begin {align*} \frac {13}{2}-c_{1} -c_{2} = 0\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \frac {2 c_{1} -8 x}{2 \sqrt {2 c_{1} x -4 x^{2}+2 c_{2}}} \end {align*}
substituting \(y^{\prime } = 0\) and \(x = 1\) in the above gives \begin {align*} 0 = \frac {c_{1} -4}{\sqrt {2 c_{1} -4+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}&=4\\ c_{2}&={\frac {5}{2}} \end {align*}
Substituting these values back in above solution results in \begin {align*} \frac {y^{2}}{2}-4 x +2 x^{2}-\frac {5}{2} = 0 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \frac {y^{2}}{2}-4 x +2 x^{2}-\frac {5}{2} &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {y^{2}}{2}-4 x +2 x^{2}-\frac {5}{2} = 0 \] 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*} y p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+p \left (y \right )^{2} = -4 \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}+4}{y p} \end {align*}
Where \(f(y)=-\frac {1}{y}\) and \(g(p)=\frac {p^{2}+4}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {p^{2}+4}{p}} \,dp &= -\frac {1}{y} \,d y \\ \int { \frac {1}{\frac {p^{2}+4}{p}} \,dp} &= \int {-\frac {1}{y} \,d y} \\ \frac {\ln \left (p^{2}+4\right )}{2}&=-\ln \left (y \right )+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} \sqrt {p^{2}+4} &= {\mathrm e}^{-\ln \left (y \right )+c_{1}} \end {align*}
Which simplifies to \begin {align*} \sqrt {p^{2}+4} &= \frac {c_{2}}{y} \end {align*}
Which can be simplified to become \[ \sqrt {p \left (y \right )^{2}+4} = \frac {c_{2} {\mathrm e}^{c_{1}}}{y} \] The solution is \[ \sqrt {p \left (y \right )^{2}+4} = \frac {c_{2} {\mathrm e}^{c_{1}}}{y} \] Initial conditions are used to solve for \(c_{1}\). Substituting \(y=3\) and \(p=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 2 = \frac {c_{2} {\mathrm e}^{c_{1}}}{3} \end {align*}
The solutions are \begin {align*} c_{1} = \ln \left (\frac {6}{c_{2}}\right ) \end {align*}
Trying the constant \begin {align*} c_{1} = \ln \left (\frac {6}{c_{2}}\right ) \end {align*}
Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \sqrt {p^{2}+4} = \frac {6}{y} \end {align*}
The constant \(c_{1} = \ln \left (\frac {6}{c_{2}}\right )\) 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*} \sqrt {{y^{\prime }}^{2}+4} = \frac {6}{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 {2 \sqrt {9-y^{2}}}{y} \tag {1} \\ y^{\prime }&=-\frac {2 \sqrt {9-y^{2}}}{y} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Since ode has form \(y^{\prime }= f(y)\) and initial conditions \(y = 3\) is verified to satisfy the ode, then the solution is \begin {align*} y&=y_0 \\ &=3 \end {align*}
Solving equation (2)
Since ode has form \(y^{\prime }= f(y)\) and initial conditions \(y = 3\) is verified to satisfy the ode, then the solution is \begin {align*} y&=y_0 \\ &=3 \end {align*}
Initial conditions are used to solve for the constants of integration.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= 3 \\ \end{align*}
Verification of solutions
\[ y = 3 \] Warning, solution could not be verified
Writing the ode as \[ y y^{\prime \prime }+{y^{\prime }}^{2} = -4 \] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y y^{\prime \prime }+{y^{\prime }}^{2}\right )d x &= \int \left (-4\right )d x\\ y y^{\prime } = -4 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 {-4 x +c_{1}}{y} \end {align*}
Where \(f(x)=-4 x +c_{1}\) and \(g(y)=\frac {1}{y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{y}} \,dy &= -4 x +c_{1} \,d x \\ \int { \frac {1}{\frac {1}{y}} \,dy} &= \int {-4 x +c_{1} \,d x} \\ \frac {y^{2}}{2}&=c_{1} x -2 x^{2}+c_{2} \\ \end{align*} The solution is \[ \frac {y^{2}}{2}-c_{1} x +2 x^{2}-c_{2} = 0 \] Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} \frac {y^{2}}{2}-c_{1} x +2 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 \(y = 3\) and \(x = 1\) in the above gives \begin {align*} \frac {13}{2}-c_{1} -c_{2} = 0\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \frac {2 c_{1} -8 x}{2 \sqrt {2 c_{1} x -4 x^{2}+2 c_{2}}} \end {align*}
substituting \(y^{\prime } = 0\) and \(x = 1\) in the above gives \begin {align*} 0 = \frac {c_{1} -4}{\sqrt {2 c_{1} -4+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}&=4\\ c_{2}&={\frac {5}{2}} \end {align*}
Substituting these values back in above solution results in \begin {align*} \frac {y^{2}}{2}-4 x +2 x^{2}-\frac {5}{2} = 0 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \frac {y^{2}}{2}-4 x +2 x^{2}-\frac {5}{2} &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {y^{2}}{2}-4 x +2 x^{2}-\frac {5}{2} = 0 \] Verified OK.
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 &= 4 \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 {4\,d x} &= c_{1} \end {align*}
Which results in \begin {align*} 2 y y^{\prime }+4 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 {-2 x +\frac {c_{1}}{2}}{y} \end {align*}
Where \(f(x)=-2 x +\frac {c_{1}}{2}\) and \(g(y)=\frac {1}{y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{y}} \,dy &= -2 x +\frac {c_{1}}{2} \,d x \\ \int { \frac {1}{\frac {1}{y}} \,dy} &= \int {-2 x +\frac {c_{1}}{2} \,d x} \\ \frac {y^{2}}{2}&=-x^{2}+\frac {1}{2} c_{1} x +c_{2} \\ \end{align*} The solution is \[ \frac {y^{2}}{2}+x^{2}-\frac {c_{1} x}{2}-c_{2} = 0 \] Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} \frac {y^{2}}{2}+x^{2}-\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 \(y = 3\) and \(x = 1\) in the above gives \begin {align*} \frac {11}{2}-\frac {c_{1}}{2}-c_{2} = 0\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \frac {-4 x +c_{1}}{2 \sqrt {c_{1} x -2 x^{2}+2 c_{2}}} \end {align*}
substituting \(y^{\prime } = 0\) and \(x = 1\) in the above gives \begin {align*} 0 = \frac {c_{1} -4}{2 \sqrt {c_{1} -2+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}&=4\\ c_{2}&={\frac {7}{2}} \end {align*}
Substituting these values back in above solution results in \begin {align*} \frac {y^{2}}{2}+x^{2}-2 x -\frac {7}{2} = 0 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \frac {y^{2}}{2}+x^{2}-2 x -\frac {7}{2} &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {y^{2}}{2}+x^{2}-2 x -\frac {7}{2} = 0 \] Warning, solution could not be verified
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y y^{\prime \prime }+{y^{\prime }}^{2}=-4, y \left (1\right )=3, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=0\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} \\ {} & {} & y u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )+u \left (y \right )^{2}=-4 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=\frac {-u \left (y \right )^{2}-4}{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}-4}=\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}-4}d y =\int \frac {1}{y}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {\ln \left (u \left (y \right )^{2}+4\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 {1-4 y^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}}}{{\mathrm e}^{c_{1}} y}, u \left (y \right )=-\frac {\sqrt {1-4 y^{2} \left ({\mathrm e}^{c_{1}}\right )^{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 {1-4 y^{2} \left ({\mathrm e}^{c_{1}}\right )^{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 {1-4 y^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}}}{{\mathrm e}^{c_{1}} y} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\sqrt {1-4 y^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}}}{{\mathrm e}^{c_{1}} y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime } y}{\sqrt {1-4 y^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}}}=\frac {1}{{\mathrm e}^{c_{1}}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime } y}{\sqrt {1-4 y^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}}}d x =\int \frac {1}{{\mathrm e}^{c_{1}}}d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {\sqrt {1-4 y^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}}}{4 \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 {1-16 \left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}-32 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} x -16 \left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}}{2 \,{\mathrm e}^{c_{1}}}, y=\frac {\sqrt {1-16 \left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}-32 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} x -16 \left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}}{2 \,{\mathrm e}^{c_{1}}}\right \} \\ \bullet & {} & \textrm {Solve 2nd ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-\frac {\sqrt {1-4 y^{2} \left ({\mathrm e}^{c_{1}}\right )^{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 {1-4 y^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}}}{{\mathrm e}^{c_{1}} y} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {\sqrt {1-4 y^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}}}{{\mathrm e}^{c_{1}} y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime } y}{\sqrt {1-4 y^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}}}=-\frac {1}{{\mathrm e}^{c_{1}}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime } y}{\sqrt {1-4 y^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}}}d x =\int -\frac {1}{{\mathrm e}^{c_{1}}}d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {\sqrt {1-4 y^{2} \left ({\mathrm e}^{c_{1}}\right )^{2}}}{4 \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 {1-16 \left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}+32 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} x -16 \left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}}{2 \,{\mathrm e}^{c_{1}}}, y=\frac {\sqrt {1-16 \left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}+32 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} x -16 \left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}}{2 \,{\mathrm e}^{c_{1}}}\right \} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=-\frac {\sqrt {1-16 \left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}-32 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} x -16 \left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}}{2 {\mathrm e}^{c_{1}}} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=3 \\ {} & {} & 3=-\frac {\sqrt {1-16 \left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}-32 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} -16 \left ({\mathrm e}^{c_{1}}\right )^{2}}}{2 \,{\mathrm e}^{c_{1}}} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {-32 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} -32 \left ({\mathrm e}^{c_{1}}\right )^{2} x}{4 \sqrt {1-16 \left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}-32 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} x -16 \left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}\, {\mathrm e}^{c_{1}}} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=0 \\ {} & {} & 0=-\frac {-32 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} -32 \left ({\mathrm e}^{c_{1}}\right )^{2}}{4 \sqrt {1-16 \left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}-32 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} -16 \left ({\mathrm e}^{c_{1}}\right )^{2}}\, {\mathrm e}^{c_{1}}} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-\ln \left (6\right )+\mathrm {I} \pi , c_{2} =6\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\sqrt {-4 x^{2}+8 x +5} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=\frac {\sqrt {1-16 \left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}-32 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} x -16 \left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}}{2 {\mathrm e}^{c_{1}}} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=3 \\ {} & {} & 3=\frac {\sqrt {1-16 \left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}-32 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} -16 \left ({\mathrm e}^{c_{1}}\right )^{2}}}{2 \,{\mathrm e}^{c_{1}}} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {-32 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} -32 \left ({\mathrm e}^{c_{1}}\right )^{2} x}{4 \sqrt {1-16 \left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}-32 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} x -16 \left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}\, {\mathrm e}^{c_{1}}} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=0 \\ {} & {} & 0=\frac {-32 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} -32 \left ({\mathrm e}^{c_{1}}\right )^{2}}{4 \sqrt {1-16 \left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}-32 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} -16 \left ({\mathrm e}^{c_{1}}\right )^{2}}\, {\mathrm e}^{c_{1}}} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-\ln \left (6\right ), c_{2} =-6\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\sqrt {-4 x^{2}+8 x +5} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=-\frac {\sqrt {1-16 \left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}+32 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} x -16 \left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}}{2 {\mathrm e}^{c_{1}}} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=3 \\ {} & {} & 3=-\frac {\sqrt {1-16 \left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}+32 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} -16 \left ({\mathrm e}^{c_{1}}\right )^{2}}}{2 \,{\mathrm e}^{c_{1}}} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {32 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} -32 \left ({\mathrm e}^{c_{1}}\right )^{2} x}{4 \sqrt {1-16 \left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}+32 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} x -16 \left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}\, {\mathrm e}^{c_{1}}} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=0 \\ {} & {} & 0=-\frac {32 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} -32 \left ({\mathrm e}^{c_{1}}\right )^{2}}{4 \sqrt {1-16 \left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}+32 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} -16 \left ({\mathrm e}^{c_{1}}\right )^{2}}\, {\mathrm e}^{c_{1}}} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-\ln \left (6\right )+\mathrm {I} \pi , c_{2} =-6\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\sqrt {-4 x^{2}+8 x +5} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=\frac {\sqrt {1-16 \left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}+32 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} x -16 \left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}}{2 {\mathrm e}^{c_{1}}} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=3 \\ {} & {} & 3=\frac {\sqrt {1-16 \left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}+32 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} -16 \left ({\mathrm e}^{c_{1}}\right )^{2}}}{2 \,{\mathrm e}^{c_{1}}} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {32 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} -32 \left ({\mathrm e}^{c_{1}}\right )^{2} x}{4 \sqrt {1-16 \left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}+32 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} x -16 \left ({\mathrm e}^{c_{1}}\right )^{2} x^{2}}\, {\mathrm e}^{c_{1}}} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=0 \\ {} & {} & 0=\frac {32 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} -32 \left ({\mathrm e}^{c_{1}}\right )^{2}}{4 \sqrt {1-16 \left ({\mathrm e}^{c_{1}}\right )^{4} c_{2}^{2}+32 \left ({\mathrm e}^{c_{1}}\right )^{3} c_{2} -16 \left ({\mathrm e}^{c_{1}}\right )^{2}}\, {\mathrm e}^{c_{1}}} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-\ln \left (6\right ), c_{2} =6\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\sqrt {-4 x^{2}+8 x +5} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\sqrt {-4 x^{2}+8 x +5} \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: 0.891 (sec). Leaf size: 16
dsolve([y(x)*diff(y(x),x$2)+diff(y(x),x)^2+4=0,y(1) = 3, D(y)(1) = 0],y(x), singsol=all)
\[ y \left (x \right ) = \sqrt {-4 x^{2}+8 x +5} \]
✓ Solution by Mathematica
Time used: 31.559 (sec). Leaf size: 19
DSolve[{y[x]*y''[x]+y'[x]^2+4==0,{y[1]==3,y'[1]==0}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \sqrt {-4 x^2+8 x+5} \]