Internal problem ID [6827]
Internal file name [OUTPUT/6074_Thursday_July_28_2022_04_29_17_AM_48951552/index.tex]
Book: Elementary differential equations. By Earl D. Rainville, Phillip E. Bedient. Macmilliam
Publishing Co. NY. 6th edition. 1981.
Section: CHAPTER 16. Nonlinear equations. Section 101. Independent variable missing.
EXERCISES Page 324
Problem number: 7.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_x", "second_order_ode_missing_y"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_y_y1]]
\[ \boxed {2 a y^{\prime \prime }+{y^{\prime }}^{3}=0} \]
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 a p^{\prime }\left (x \right )+p \left (x \right )^{3} = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. Integrating both sides gives \begin {align*} \int -\frac {2 a}{p^{3}}d p &= x +c_{1}\\ \frac {a}{p^{2}}&=x +c_{1} \end {align*}
Solving for \(p\) gives these solutions \begin {align*} p_1&=\frac {\sqrt {\left (x +c_{1} \right ) a}}{x +c_{1}}\\ p_2&=-\frac {\sqrt {\left (x +c_{1} \right ) a}}{x +c_{1}} \end {align*}
For solution (1) found earlier, since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = \frac {\sqrt {\left (x +c_{1} \right ) a}}{x +c_{1}} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { \frac {\sqrt {\left (x +c_{1} \right ) a}}{x +c_{1}}\,\mathop {\mathrm {d}x}}\\ &= 2 \sqrt {\left (x +c_{1} \right ) a}+c_{2} \end {align*}
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = -\frac {\sqrt {\left (x +c_{1} \right ) a}}{x +c_{1}} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { -\frac {\sqrt {\left (x +c_{1} \right ) a}}{x +c_{1}}\,\mathop {\mathrm {d}x}}\\ &= -2 \sqrt {\left (x +c_{1} \right ) a}+c_{3} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= 2 \sqrt {\left (x +c_{1} \right ) a}+c_{2} \\ \tag{2} y &= -2 \sqrt {\left (x +c_{1} \right ) a}+c_{3} \\ \end{align*}
Verification of solutions
\[ y = 2 \sqrt {\left (x +c_{1} \right ) a}+c_{2} \] Verified OK.
\[ y = -2 \sqrt {\left (x +c_{1} \right ) a}+c_{3} \] 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 a p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+p \left (y \right )^{3} = 0 \end {align*}
Which is now solved as first order ode for \(p(y)\). Integrating both sides gives \begin {align*} \int -\frac {2 a}{p^{2}}d p &= y +c_{1}\\ \frac {2 a}{p}&=y +c_{1} \end {align*}
Solving for \(p\) gives these solutions \begin {align*} p_1&=\frac {2 a}{y +c_{1}} \end {align*}
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 {2 a}{y+c_{1}} \end {align*}
Integrating both sides gives \begin {align*} \int \frac {y +c_{1}}{2 a}d y &= x +c_{2}\\ \frac {y \left (y +2 c_{1} \right )}{4 a}&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=-c_{1} -\sqrt {4 a c_{2} +4 x a +c_{1}^{2}}\\ y_2&=-c_{1} +\sqrt {4 a c_{2} +4 x a +c_{1}^{2}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -c_{1} -\sqrt {4 a c_{2} +4 x a +c_{1}^{2}} \\ \tag{2} y &= -c_{1} +\sqrt {4 a c_{2} +4 x a +c_{1}^{2}} \\ \end{align*}
Verification of solutions
\[ y = -c_{1} -\sqrt {4 a c_{2} +4 x a +c_{1}^{2}} \] Verified OK.
\[ y = -c_{1} +\sqrt {4 a c_{2} +4 x a +c_{1}^{2}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 a y^{\prime \prime }+{y^{\prime }}^{3}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime }\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & 2 a u^{\prime }\left (x \right )+u \left (x \right )^{3}=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=-\frac {u \left (x \right )^{3}}{2 a} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{u \left (x \right )^{3}}=-\frac {1}{2 a} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {u^{\prime }\left (x \right )}{u \left (x \right )^{3}}d x =\int -\frac {1}{2 a}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{2 u \left (x \right )^{2}}=-\frac {x}{2 a}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & \left \{u \left (x \right )=\frac {\sqrt {-\left (2 c_{1} a -x \right ) a}}{2 c_{1} a -x}, u \left (x \right )=-\frac {\sqrt {-\left (2 c_{1} a -x \right ) a}}{2 c_{1} a -x}\right \} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\frac {\sqrt {-\left (2 c_{1} a -x \right ) a}}{2 c_{1} a -x} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=\frac {\sqrt {-\left (2 c_{1} a -x \right ) a}}{2 c_{1} a -x} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int \frac {\sqrt {-\left (2 c_{1} a -x \right ) a}}{2 c_{1} a -x}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=-2 \sqrt {-\left (2 c_{1} a -x \right ) a}+c_{2} \\ \bullet & {} & \textrm {Solve 2nd ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {\sqrt {-\left (2 c_{1} a -x \right ) a}}{2 c_{1} a -x} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=-\frac {\sqrt {-\left (2 c_{1} a -x \right ) a}}{2 c_{1} a -x} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int -\frac {\sqrt {-\left (2 c_{1} a -x \right ) a}}{2 c_{1} a -x}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=2 \sqrt {-\left (2 c_{1} a -x \right ) a}+c_{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 -> Calling odsolve with the ODE`, diff(_b(_a), _a) = -(1/2)*_b(_a)^3/a, _b(_a), HINT = [[1, 0], [y, -_b^2], [_a, -(1/2)*_b]]` *** symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[1, 0], [y, -_b^2], [_a, -1/2*_b]
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 29
dsolve(2*a*diff(y(x),x$2)+diff(y(x),x)^3=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= 2 \sqrt {\left (x +c_{1} \right ) a}+c_{2} \\ y \left (x \right ) &= -2 \sqrt {\left (x +c_{1} \right ) a}+c_{2} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.33 (sec). Leaf size: 51
DSolve[2*a*y''[x]+(y'[x])^3==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to c_2-2 \sqrt {a} \sqrt {x-2 a c_1} \\ y(x)\to 2 \sqrt {a} \sqrt {x-2 a c_1}+c_2 \\ \end{align*}