Internal problem ID [6245]
Internal file name [OUTPUT/5493_Sunday_June_05_2022_03_41_28_PM_68194147/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(b).
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], [_2nd_order, _with_potential_symmetries], [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {y y^{\prime \prime }-y^{\prime } y^{2}-{y^{\prime }}^{2}=0} \] With initial conditions \begin {align*} \left [y \left (0\right ) = -{\frac {1}{2}}, y^{\prime }\left (0\right ) = 1\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*} y p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+\left (-p \left (y \right )-y^{2}\right ) p \left (y \right ) = 0 \end {align*}
Which is now solved as first order ode for \(p(y)\).
Entering Linear first order ODE solver. The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int -\frac {1}{y}d y} \\ &= \frac {1}{y} \\ \end{align*} The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}y}}\left ( \mu p\right ) &= \left (\mu \right ) \left (y\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}y}} \left (\frac {p}{y}\right ) &= \left (\frac {1}{y}\right ) \left (y\right )\\ \mathrm {d} \left (\frac {p}{y}\right ) &= \mathrm {d} y \end {align*}
Integrating gives \begin {align*} \frac {p}{y} &= \int {\mathrm {d} y}\\ \frac {p}{y} &= y + c_{1} \end {align*}
Dividing both sides by the integrating factor \(\mu =\frac {1}{y}\) results in \begin {align*} p \left (y \right ) &= c_{1} y +y^{2} \end {align*}
which simplifies to \begin {align*} p \left (y \right ) &= y \left (y +c_{1} \right ) \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(y=-{\frac {1}{2}}\) and \(p=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 1 = \frac {1}{4}-\frac {c_{1}}{2} \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 this in the general solution gives \begin {align*} p \left (y \right )&=\frac {y \left (2 y -3\right )}{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*} y^{\prime } = \frac {y \left (2 y-3\right )}{2} \end {align*}
Integrating both sides gives \begin {align*} \int \frac {2}{y \left (2 y -3\right )}d y &= \int {dx}\\ -\frac {2 \ln \left (y \right )}{3}+\frac {2 \ln \left (y -\frac {3}{2}\right )}{3}&= x +c_{2} \end {align*}
Initial conditions are used to solve for \(c_{2}\). Substituting \(x=0\) and \(y=-{\frac {1}{2}}\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} \frac {4 \ln \left (2\right )}{3} = c_{2} \end {align*}
The solutions are \begin {align*} c_{2} = \frac {4 \ln \left (2\right )}{3} \end {align*}
Trying the constant \begin {align*} c_{2} = \frac {4 \ln \left (2\right )}{3} \end {align*}
Substituting \(c_{2}\) found above in the general solution gives \begin {align*} -\frac {2 \ln \left (y \right )}{3}+\frac {2 \ln \left (y -\frac {3}{2}\right )}{3} = x +\frac {4 \ln \left (2\right )}{3} \end {align*}
The constant \(c_{2} = \frac {4 \ln \left (2\right )}{3}\) 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 \left (4 \,{\mathrm e}^{\frac {3 x}{2}}-1\right )} \\ \end{align*}
Verification of solutions
\[ y = -\frac {3}{2 \left (4 \,{\mathrm e}^{\frac {3 x}{2}}-1\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y y^{\prime \prime }+\left (-y^{\prime }-y^{2}\right ) y^{\prime }=0, y \left (0\right )=-\frac {1}{2}, 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 \\ {} & {} & 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 )+\left (-u \left (y \right )-y^{2}\right ) u \left (y \right )=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=-\frac {-u \left (y \right )-y^{2}}{y} \\ \bullet & {} & \textrm {Collect w.r.t.}\hspace {3pt} u \left (y \right )\hspace {3pt}\textrm {and simplify}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=\frac {u \left (y \right )}{y}+y \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} u \left (y \right )\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )-\frac {u \left (y \right )}{y}=y \\ \bullet & {} & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (y \right ) \\ {} & {} & \mu \left (y \right ) \left (\frac {d}{d y}u \left (y \right )-\frac {u \left (y \right )}{y}\right )=\mu \left (y \right ) y \\ \bullet & {} & \textrm {Assume the lhs of the ODE is the total derivative}\hspace {3pt} \frac {d}{d y}\left (u \left (y \right ) \mu \left (y \right )\right ) \\ {} & {} & \mu \left (y \right ) \left (\frac {d}{d y}u \left (y \right )-\frac {u \left (y \right )}{y}\right )=\left (\frac {d}{d y}u \left (y \right )\right ) \mu \left (y \right )+u \left (y \right ) \left (\frac {d}{d y}\mu \left (y \right )\right ) \\ \bullet & {} & \textrm {Isolate}\hspace {3pt} \frac {d}{d y}\mu \left (y \right ) \\ {} & {} & \frac {d}{d y}\mu \left (y \right )=-\frac {\mu \left (y \right )}{y} \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (y \right )=\frac {1}{y} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int \left (\frac {d}{d y}\left (u \left (y \right ) \mu \left (y \right )\right )\right )d y =\int \mu \left (y \right ) y d y +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & u \left (y \right ) \mu \left (y \right )=\int \mu \left (y \right ) y d y +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\frac {\int \mu \left (y \right ) y d y +c_{1}}{\mu \left (y \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (y \right )=\frac {1}{y} \\ {} & {} & u \left (y \right )=y \left (\int 1d y +c_{1} \right ) \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & u \left (y \right )=y \left (y +c_{1} \right ) \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=y \left (y +c_{1} \right ) \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=y \left (y+c_{1} \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y \left (y+c_{1} \right ) \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y \left (y+c_{1} \right )}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y \left (y+c_{1} \right )}d x =\int 1d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (y\right )}{c_{1}}-\frac {\ln \left (y+c_{1} \right )}{c_{1}}=x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=-\frac {c_{1} {\mathrm e}^{c_{2} c_{1} +c_{1} x}}{-1+{\mathrm e}^{c_{2} c_{1} +c_{1} x}} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=-\frac {c_{1} {\mathrm e}^{c_{2} c_{1} +c_{1} x}}{-1+{\mathrm e}^{c_{2} c_{1} +c_{1} x}} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=-\frac {1}{2} \\ {} & {} & -\frac {1}{2}=-\frac {c_{1} {\mathrm e}^{c_{2} c_{1}}}{-1+{\mathrm e}^{c_{2} c_{1}}} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {c_{1}^{2} \left ({\mathrm e}^{c_{2} c_{1} +c_{1} x}\right )^{2}}{\left (-1+{\mathrm e}^{c_{2} c_{1} +c_{1} x}\right )^{2}}-\frac {c_{1}^{2} {\mathrm e}^{c_{2} c_{1} +c_{1} x}}{-1+{\mathrm e}^{c_{2} c_{1} +c_{1} x}} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=\frac {c_{1}^{2} \left ({\mathrm e}^{c_{2} c_{1}}\right )^{2}}{\left (-1+{\mathrm e}^{c_{2} c_{1}}\right )^{2}}-\frac {c_{1}^{2} {\mathrm e}^{c_{2} c_{1}}}{-1+{\mathrm e}^{c_{2} c_{1}}} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-\frac {3}{2}, c_{2} =\frac {4 \ln \left (2\right )}{3}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {3 \,{\mathrm e}^{-\frac {3 x}{2}}}{-8+2 \,{\mathrm e}^{-\frac {3 x}{2}}} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {3 \,{\mathrm e}^{-\frac {3 x}{2}}}{-8+2 \,{\mathrm e}^{-\frac {3 x}{2}}} \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 <- differential order: 2; canonical coordinates successful <- differential order 2; missing variables successful`
✓ Solution by Maple
Time used: 0.203 (sec). Leaf size: 16
dsolve([y(x)*diff(y(x),x$2)=y(x)^2*diff(y(x),x)+(diff(y(x),x))^2,y(0) = -1/2, D(y)(0) = 1],y(x), singsol=all)
\[ y \left (x \right ) = -\frac {3}{8 \,{\mathrm e}^{\frac {3 x}{2}}-2} \]
✓ Solution by Mathematica
Time used: 1.982 (sec). Leaf size: 20
DSolve[{y[x]*y''[x]==y[x]^2*y'[x]+(y'[x])^2,{y[0]==-1/2,y'[0]==1}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {3}{2-8 e^{3 x/2}} \]