Internal problem ID [4666]
Internal file name [OUTPUT/4159_Sunday_June_05_2022_12_31_07_PM_75810893/index.tex]
Book: Ordinary Differential Equations, By Tenenbaum and Pollard. Dover, NY 1963
Section: Chapter 8. Special second order equations. Lesson 35. Independent variable x
absent
Problem number: Exercise 35.16, page 504.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_x"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], _Liouville, [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {\left (1+y\right ) y^{\prime \prime }-3 {y^{\prime }}^{2}=0} \] With initial conditions \begin {align*} \left [y \left (1\right ) = 0, y^{\prime }\left (1\right ) = -{\frac {1}{2}}\right ] \end {align*}
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*} \left (y +1\right ) p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )-3 p \left (y \right )^{2} = 0 \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 {3 p}{y +1} \end {align*}
Where \(f(y)=\frac {3}{y +1}\) and \(g(p)=p\). Integrating both sides gives \begin {align*} \frac {1}{p} \,dp &= \frac {3}{y +1} \,d y\\ \int { \frac {1}{p} \,dp} &= \int {\frac {3}{y +1} \,d y}\\ \ln \left (p \right )&=3 \ln \left (y +1\right )+c_{1}\\ p&={\mathrm e}^{3 \ln \left (y +1\right )+c_{1}}\\ &=c_{1} \left (y +1\right )^{3} \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(y=0\) and \(p=-{\frac {1}{2}}\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -{\frac {1}{2}} = c_{1} \end {align*}
The solutions are \begin {align*} c_{1} = -{\frac {1}{2}} \end {align*}
Trying the constant \begin {align*} c_{1} = -{\frac {1}{2}} \end {align*}
Substituting this in the general solution gives \begin {align*} p \left (y \right )&=-\frac {\left (y +1\right )^{3}}{2} \end {align*}
The constant \(c_{1} = -{\frac {1}{2}}\) 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*} y^{\prime } = -\frac {\left (1+y\right )^{3}}{2} \end {align*}
Integrating both sides gives \begin {align*} \int -\frac {2}{\left (y +1\right )^{3}}d y &= \int {dx}\\ \frac {1}{\left (y +1\right )^{2}}&= x +c_{2} \end {align*}
Initial conditions are used to solve for \(c_{2}\). Substituting \(x=1\) and \(y=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 1 = 1+c_{2} \end {align*}
The solutions are \begin {align*} c_{2} = 0 \end {align*}
Trying the constant \begin {align*} c_{2} = 0 \end {align*}
Substituting \(c_{2}\) found above in the general solution gives \begin {align*} \frac {1}{\left (y +1\right )^{2}} = x \end {align*}
The constant \(c_{2} = 0\) 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} \frac {1}{\left (1+y\right )^{2}} &= x \\ \end{align*}
Verification of solutions
\[ \frac {1}{\left (1+y\right )^{2}} = x \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\left (1+y\right ) y^{\prime \prime }-3 {y^{\prime }}^{2}=0, y \left (1\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=-\frac {1}{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} \\ {} & {} & \left (y +1\right ) u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )-3 u \left (y \right )^{2}=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=\frac {3 u \left (y \right )}{y +1} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d y}u \left (y \right )}{u \left (y \right )}=\frac {3}{y +1} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int \frac {\frac {d}{d y}u \left (y \right )}{u \left (y \right )}d y =\int \frac {3}{y +1}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (u \left (y \right )\right )=3 \ln \left (y +1\right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )={\mathrm e}^{c_{1}} \left (y +1\right )^{3} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )={\mathrm e}^{c_{1}} \left (y +1\right )^{3} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }={\mathrm e}^{c_{1}} \left (1+y\right )^{3} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }={\mathrm e}^{c_{1}} \left (1+y\right )^{3} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\left (1+y\right )^{3}}={\mathrm e}^{c_{1}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{\left (1+y\right )^{3}}d x =\int {\mathrm e}^{c_{1}}d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{2 \left (1+y\right )^{2}}={\mathrm e}^{c_{1}} x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=-\frac {\sqrt {-2 \,{\mathrm e}^{c_{1}} x -2 c_{2}}-1}{\sqrt {-2 \,{\mathrm e}^{c_{1}} x -2 c_{2}}}, y=-\frac {\sqrt {-2 \,{\mathrm e}^{c_{1}} x -2 c_{2}}+1}{\sqrt {-2 \,{\mathrm e}^{c_{1}} x -2 c_{2}}}\right \} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=-\frac {\sqrt {-2 {\mathrm e}^{c_{1}} x -2 c_{2}}-1}{\sqrt {-2 {\mathrm e}^{c_{1}} x -2 c_{2}}} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=0 \\ {} & {} & 0=-\frac {\sqrt {-2 \,{\mathrm e}^{c_{1}}-2 c_{2}}-1}{\sqrt {-2 \,{\mathrm e}^{c_{1}}-2 c_{2}}} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {{\mathrm e}^{c_{1}}}{-2 \,{\mathrm e}^{c_{1}} x -2 c_{2}}-\frac {\left (\sqrt {-2 \,{\mathrm e}^{c_{1}} x -2 c_{2}}-1\right ) {\mathrm e}^{c_{1}}}{\left (-2 \,{\mathrm e}^{c_{1}} x -2 c_{2} \right )^{\frac {3}{2}}} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=-\frac {1}{2} \\ {} & {} & -\frac {1}{2}=\frac {{\mathrm e}^{c_{1}}}{-2 \,{\mathrm e}^{c_{1}}-2 c_{2}}-\frac {\left (\sqrt {-2 \,{\mathrm e}^{c_{1}}-2 c_{2}}-1\right ) {\mathrm e}^{c_{1}}}{\left (-2 \,{\mathrm e}^{c_{1}}-2 c_{2} \right )^{\frac {3}{2}}} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-\ln \left (2\right )+\mathrm {I} \pi , c_{2} =0\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\frac {\sqrt {x}-1}{\sqrt {x}} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=-\frac {\sqrt {-2 {\mathrm e}^{c_{1}} x -2 c_{2}}+1}{\sqrt {-2 {\mathrm e}^{c_{1}} x -2 c_{2}}} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=0 \\ {} & {} & 0=-\frac {\sqrt {-2 \,{\mathrm e}^{c_{1}}-2 c_{2}}+1}{\sqrt {-2 \,{\mathrm e}^{c_{1}}-2 c_{2}}} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {{\mathrm e}^{c_{1}}}{-2 \,{\mathrm e}^{c_{1}} x -2 c_{2}}-\frac {\left (\sqrt {-2 \,{\mathrm e}^{c_{1}} x -2 c_{2}}+1\right ) {\mathrm e}^{c_{1}}}{\left (-2 \,{\mathrm e}^{c_{1}} x -2 c_{2} \right )^{\frac {3}{2}}} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=-\frac {1}{2} \\ {} & {} & -\frac {1}{2}=\frac {{\mathrm e}^{c_{1}}}{-2 \,{\mathrm e}^{c_{1}}-2 c_{2}}-\frac {\left (\sqrt {-2 \,{\mathrm e}^{c_{1}}-2 c_{2}}+1\right ) {\mathrm e}^{c_{1}}}{\left (-2 \,{\mathrm e}^{c_{1}}-2 c_{2} \right )^{\frac {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} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {\sqrt {x}-1}{\sqrt {x}} \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville <- 2nd_order Liouville successful`
✓ Solution by Maple
Time used: 0.344 (sec). Leaf size: 15
dsolve([(y(x)+1)*diff(y(x),x$2)=3*(diff(y(x),x))^2,y(1) = 0, D(y)(1) = -1/2],y(x), singsol=all)
\[ y \left (x \right ) = \frac {-x +\sqrt {x}}{x} \]
✓ Solution by Mathematica
Time used: 1.693 (sec). Leaf size: 572
DSolve[{(y[x]+1)*y''[x]==3*(y'[x])^2,{y[1]==0,y'[0]==-1/2}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {6 \left (\left (-12+3\ 2^{2/3} \sqrt [3]{27-3 \sqrt {69}}-\sqrt [3]{2} \left (27-3 \sqrt {69}\right )^{2/3}+3\ 2^{2/3} \sqrt [3]{3 \left (9+\sqrt {69}\right )}-\sqrt [3]{2} \left (3 \left (9+\sqrt {69}\right )\right )^{2/3}\right ) x+3 \sqrt {2} \sqrt {\left (12-3\ 2^{2/3} \sqrt [3]{27-3 \sqrt {69}}+\sqrt [3]{2} \left (27-3 \sqrt {69}\right )^{2/3}-3\ 2^{2/3} \sqrt [3]{3 \left (9+\sqrt {69}\right )}+\sqrt [3]{2} \left (3 \left (9+\sqrt {69}\right )\right )^{2/3}\right ) x-\sqrt [3]{2} \left (3 \left (9+\sqrt {69}\right )\right )^{2/3}+2\ 2^{2/3} \sqrt [3]{3 \left (9+\sqrt {69}\right )}+\sqrt [3]{9-\sqrt {69}} \left (9+\sqrt {69}\right )^{2/3}+\left (9-\sqrt {69}\right )^{2/3} \sqrt [3]{9+\sqrt {69}}-\sqrt [3]{2} \left (27-3 \sqrt {69}\right )^{2/3}+2\ 2^{2/3} \sqrt [3]{27-3 \sqrt {69}}+6}+\sqrt [3]{2} \left (3 \left (9+\sqrt {69}\right )\right )^{2/3}-2\ 2^{2/3} \sqrt [3]{3 \left (9+\sqrt {69}\right )}-\sqrt [3]{9-\sqrt {69}} \left (9+\sqrt {69}\right )^{2/3}-\left (9-\sqrt {69}\right )^{2/3} \sqrt [3]{9+\sqrt {69}}+\sqrt [3]{2} \left (27-3 \sqrt {69}\right )^{2/3}-2\ 2^{2/3} \sqrt [3]{27-3 \sqrt {69}}-6\right )}{\left (12-3\ 2^{2/3} \sqrt [3]{27-3 \sqrt {69}}+\sqrt [3]{2} \left (27-3 \sqrt {69}\right )^{2/3}-3\ 2^{2/3} \sqrt [3]{3 \left (9+\sqrt {69}\right )}+\sqrt [3]{2} \left (3 \left (9+\sqrt {69}\right )\right )^{2/3}\right ) \left (6 x+2^{2/3} \sqrt [3]{3} \left (\sqrt [3]{9-\sqrt {69}}+\sqrt [3]{9+\sqrt {69}}\right )\right )} \]