Internal problem ID [6821]
Internal file name [OUTPUT/6068_Thursday_July_28_2022_04_29_02_AM_22805052/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: 1.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_y"
Maple gives the following as the ode type
[[_2nd_order, _missing_y], [_2nd_order, _reducible, _mu_y_y1]]
\[ \boxed {y^{\prime \prime }-x {y^{\prime }}^{3}=0} \]
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*} p^{\prime }\left (x \right )-x p \left (x \right )^{3} = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. In canonical form the ODE is \begin {align*} p' &= F(x,p)\\ &= f( x) g(p)\\ &= x \,p^{3} \end {align*}
Where \(f(x)=x\) and \(g(p)=p^{3}\). Integrating both sides gives \begin{align*} \frac {1}{p^{3}} \,dp &= x \,d x \\ \int { \frac {1}{p^{3}} \,dp} &= \int {x \,d x} \\ -\frac {1}{2 p^{2}}&=\frac {x^{2}}{2}+c_{1} \\ \end{align*} The solution is \[ -\frac {1}{2 p \left (x \right )^{2}}-\frac {x^{2}}{2}-c_{1} = 0 \] For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} -\frac {1}{2 {y^{\prime }}^{2}}-\frac {x^{2}}{2}-c_{1} = 0 \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 {1}{\sqrt {-x^{2}-2 c_{1}}} \tag {1} \\ y^{\prime }&=\frac {1}{\sqrt {-x^{2}-2 c_{1}}} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin {align*} y &= \int { -\frac {1}{\sqrt {-x^{2}-2 c_{1}}}\,\mathop {\mathrm {d}x}}\\ &= \arctan \left (\frac {\sqrt {-x^{2}-2 c_{1}}}{x}\right )+c_{2} \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} y &= \int { \frac {1}{\sqrt {-x^{2}-2 c_{1}}}\,\mathop {\mathrm {d}x}}\\ &= \arctan \left (\frac {x}{\sqrt {-x^{2}-2 c_{1}}}\right )+c_{3} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \arctan \left (\frac {\sqrt {-x^{2}-2 c_{1}}}{x}\right )+c_{2} \\ \tag{2} y &= \arctan \left (\frac {x}{\sqrt {-x^{2}-2 c_{1}}}\right )+c_{3} \\ \end{align*}
Verification of solutions
\[ y = \arctan \left (\frac {\sqrt {-x^{2}-2 c_{1}}}{x}\right )+c_{2} \] Verified OK.
\[ y = \arctan \left (\frac {x}{\sqrt {-x^{2}-2 c_{1}}}\right )+c_{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime }-x {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} \\ {} & {} & u^{\prime }\left (x \right )-x u \left (x \right )^{3}=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=x u \left (x \right )^{3} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{u \left (x \right )^{3}}=x \\ \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 x d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{2 u \left (x \right )^{2}}=\frac {x^{2}}{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & \left \{u \left (x \right )=\frac {1}{\sqrt {-x^{2}-2 c_{1}}}, u \left (x \right )=-\frac {1}{\sqrt {-x^{2}-2 c_{1}}}\right \} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\frac {1}{\sqrt {-x^{2}-2 c_{1}}} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=\frac {1}{\sqrt {-x^{2}-2 c_{1}}} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int \frac {1}{\sqrt {-x^{2}-2 c_{1}}}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=\arctan \left (\frac {x}{\sqrt {-x^{2}-2 c_{1}}}\right )+c_{2} \\ \bullet & {} & \textrm {Solve 2nd ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {1}{\sqrt {-x^{2}-2 c_{1}}} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=-\frac {1}{\sqrt {-x^{2}-2 c_{1}}} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int -\frac {1}{\sqrt {-x^{2}-2 c_{1}}}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=-\arctan \left (\frac {x}{\sqrt {-x^{2}-2 c_{1}}}\right )+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) = _a*_b(_a)^3, _b(_a), HINT = [[_a, -_b]]` *** Sublevel 2 *** symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[_a, -_b]
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 37
dsolve(diff(y(x),x$2)=x*(diff(y(x),x))^3,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \arctan \left (\frac {x}{\sqrt {-x^{2}+c_{1}}}\right )+c_{2} \\ y \left (x \right ) &= -\arctan \left (\frac {x}{\sqrt {-x^{2}+c_{1}}}\right )+c_{2} \\ \end{align*}
✓ Solution by Mathematica
Time used: 10.922 (sec). Leaf size: 57
DSolve[y''[x]==x*(y'[x])^3,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to c_2-\arctan \left (\frac {x}{\sqrt {-x^2-2 c_1}}\right ) \\ y(x)\to \arctan \left (\frac {x}{\sqrt {-x^2-2 c_1}}\right )+c_2 \\ y(x)\to c_2 \\ \end{align*}