Internal problem ID [6860]
Internal file name [OUTPUT/6107_Friday_July_29_2022_03_09_43_AM_35066495/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: 43.
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, _reducible, _mu_x_y1]]
\[ \boxed {4 y {y^{\prime }}^{2} y^{\prime \prime }-{y^{\prime }}^{4}=3} \]
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*} 4 y p \left (y \right )^{3} \left (\frac {d}{d y}p \left (y \right )\right )-p \left (y \right )^{4} = 3 \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^{4}+3}{4 y \,p^{3}} \end {align*}
Where \(f(y)=\frac {1}{4 y}\) and \(g(p)=\frac {p^{4}+3}{p^{3}}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {p^{4}+3}{p^{3}}} \,dp &= \frac {1}{4 y} \,d y \\ \int { \frac {1}{\frac {p^{4}+3}{p^{3}}} \,dp} &= \int {\frac {1}{4 y} \,d y} \\ \frac {\ln \left (p^{4}+3\right )}{4}&=\frac {\ln \left (y \right )}{4}+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} \left (p^{4}+3\right )^{\frac {1}{4}} &= {\mathrm e}^{\frac {\ln \left (y \right )}{4}+c_{1}} \end {align*}
Which simplifies to \begin {align*} \left (p^{4}+3\right )^{\frac {1}{4}} &= c_{2} y^{\frac {1}{4}} \end {align*}
Which simplifies to \[ \left (p \left (y \right )^{4}+3\right )^{\frac {1}{4}} = c_{2} y^{\frac {1}{4}} {\mathrm e}^{c_{1}} \] The solution is \[ \left (p \left (y \right )^{4}+3\right )^{\frac {1}{4}} = c_{2} y^{\frac {1}{4}} {\mathrm e}^{c_{1}} \] For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} \left ({y^{\prime }}^{4}+3\right )^{\frac {1}{4}} = c_{2} y^{\frac {1}{4}} {\mathrm e}^{c_{1}} \end {align*}
Solving the given ode for \(y^{\prime }\) results in \(4\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=\left (c_{2}^{4} y \,{\mathrm e}^{4 c_{1}}-3\right )^{\frac {1}{4}} \tag {1} \\ y^{\prime }&=i \left (c_{2}^{4} y \,{\mathrm e}^{4 c_{1}}-3\right )^{\frac {1}{4}} \tag {2} \\ y^{\prime }&=-\left (c_{2}^{4} y \,{\mathrm e}^{4 c_{1}}-3\right )^{\frac {1}{4}} \tag {3} \\ y^{\prime }&=-i \left (c_{2}^{4} y \,{\mathrm e}^{4 c_{1}}-3\right )^{\frac {1}{4}} \tag {4} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin{align*} \int \frac {1}{\left (c_{2}^{4} y \,{\mathrm e}^{4 c_{1}}-3\right )^{\frac {1}{4}}}d y &= \int d x \\ \frac {4 \left (c_{2}^{4} y \,{\mathrm e}^{4 c_{1}}-3\right )^{\frac {3}{4}} {\mathrm e}^{-4 c_{1}}}{3 c_{2}^{4}}&=x +c_{3} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {i}{\left (c_{2}^{4} y \,{\mathrm e}^{4 c_{1}}-3\right )^{\frac {1}{4}}}d y &= \int d x \\ -\frac {4 i \left (c_{2}^{4} y \,{\mathrm e}^{4 c_{1}}-3\right )^{\frac {3}{4}} {\mathrm e}^{-4 c_{1}}}{3 c_{2}^{4}}&=x +c_{4} \\ \end{align*} Solving equation (3)
Integrating both sides gives \begin{align*} \int -\frac {1}{\left (c_{2}^{4} y \,{\mathrm e}^{4 c_{1}}-3\right )^{\frac {1}{4}}}d y &= \int d x \\ -\frac {4 \left (c_{2}^{4} y \,{\mathrm e}^{4 c_{1}}-3\right )^{\frac {3}{4}} {\mathrm e}^{-4 c_{1}}}{3 c_{2}^{4}}&=x +c_{5} \\ \end{align*} Solving equation (4)
Integrating both sides gives \begin{align*} \int \frac {i}{\left (c_{2}^{4} y \,{\mathrm e}^{4 c_{1}}-3\right )^{\frac {1}{4}}}d y &= \int d x \\ \frac {4 i \left (c_{2}^{4} y \,{\mathrm e}^{4 c_{1}}-3\right )^{\frac {3}{4}} {\mathrm e}^{-4 c_{1}}}{3 c_{2}^{4}}&=x +c_{6} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (3 \left ({\mathrm e}^{4 c_{1}} c_{2}^{4} \left (x +c_{3} \right )\right )^{\frac {4}{3}}+4 \,6^{\frac {2}{3}}\right ) 6^{\frac {1}{3}} {\mathrm e}^{-4 c_{1}}}{8 c_{2}^{4}} \\ \tag{2} y &= \frac {\left (\frac {3 \,3^{\frac {1}{3}} 4^{\frac {2}{3}} \left (i {\mathrm e}^{4 c_{1}} c_{2}^{4} \left (x +c_{4} \right )\right )^{\frac {4}{3}}}{16}+3\right ) {\mathrm e}^{-4 c_{1}}}{c_{2}^{4}} \\ \tag{3} y &= \frac {\left (\left (-\frac {3 c_{5} {\mathrm e}^{4 c_{1}} c_{2}^{4}}{4}-\frac {3 \,{\mathrm e}^{4 c_{1}} c_{2}^{4} x}{4}\right )^{\frac {4}{3}}+3\right ) {\mathrm e}^{-4 c_{1}}}{c_{2}^{4}} \\ \tag{4} y &= \frac {\left (\left (-\frac {3 i {\mathrm e}^{4 c_{1}} c_{2}^{4} \left (x +c_{6} \right )}{4}\right )^{\frac {4}{3}}+3\right ) {\mathrm e}^{-4 c_{1}}}{c_{2}^{4}} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (3 \left ({\mathrm e}^{4 c_{1}} c_{2}^{4} \left (x +c_{3} \right )\right )^{\frac {4}{3}}+4 \,6^{\frac {2}{3}}\right ) 6^{\frac {1}{3}} {\mathrm e}^{-4 c_{1}}}{8 c_{2}^{4}} \] Verified OK.
\[ y = \frac {\left (\frac {3 \,3^{\frac {1}{3}} 4^{\frac {2}{3}} \left (i {\mathrm e}^{4 c_{1}} c_{2}^{4} \left (x +c_{4} \right )\right )^{\frac {4}{3}}}{16}+3\right ) {\mathrm e}^{-4 c_{1}}}{c_{2}^{4}} \] Verified OK.
\[ y = \frac {\left (\left (-\frac {3 c_{5} {\mathrm e}^{4 c_{1}} c_{2}^{4}}{4}-\frac {3 \,{\mathrm e}^{4 c_{1}} c_{2}^{4} x}{4}\right )^{\frac {4}{3}}+3\right ) {\mathrm e}^{-4 c_{1}}}{c_{2}^{4}} \] Verified OK.
\[ y = \frac {\left (\left (-\frac {3 i {\mathrm e}^{4 c_{1}} c_{2}^{4} \left (x +c_{6} \right )}{4}\right )^{\frac {4}{3}}+3\right ) {\mathrm e}^{-4 c_{1}}}{c_{2}^{4}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 y {y^{\prime }}^{2} y^{\prime \prime }-{y^{\prime }}^{4}=3 \\ \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} \\ {} & {} & 4 y u \left (y \right )^{3} \left (\frac {d}{d y}u \left (y \right )\right )-u \left (y \right )^{4}=3 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=\frac {u \left (y \right )^{4}+3}{4 y u \left (y \right )^{3}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\left (\frac {d}{d y}u \left (y \right )\right ) u \left (y \right )^{3}}{u \left (y \right )^{4}+3}=\frac {1}{4 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 )^{3}}{u \left (y \right )^{4}+3}d y =\int \frac {1}{4 y}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (u \left (y \right )^{4}+3\right )}{4}=\frac {\ln \left (y \right )}{4}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & \left \{u \left (y \right )={\left (y \left ({\mathrm e}^{c_{1}}\right )^{4}-3\right )}^{\frac {1}{4}}, u \left (y \right )=-{\left (y \left ({\mathrm e}^{c_{1}}\right )^{4}-3\right )}^{\frac {1}{4}}\right \} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )={\left (y \left ({\mathrm e}^{c_{1}}\right )^{4}-3\right )}^{\frac {1}{4}} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }={\left (y \left ({\mathrm e}^{c_{1}}\right )^{4}-3\right )}^{\frac {1}{4}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }={\left (y \left ({\mathrm e}^{c_{1}}\right )^{4}-3\right )}^{\frac {1}{4}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{{\left (y \left ({\mathrm e}^{c_{1}}\right )^{4}-3\right )}^{\frac {1}{4}}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{{\left (y \left ({\mathrm e}^{c_{1}}\right )^{4}-3\right )}^{\frac {1}{4}}}d x =\int 1d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {4 {\left (y \left ({\mathrm e}^{c_{1}}\right )^{4}-3\right )}^{\frac {3}{4}}}{3 \left ({\mathrm e}^{c_{1}}\right )^{4}}=x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\left (3 {\left (\left ({\mathrm e}^{c_{1}}\right )^{4} \left (x +c_{2} \right )\right )}^{\frac {4}{3}}+4 \,6^{\frac {2}{3}}\right ) 6^{\frac {1}{3}}}{8 \left ({\mathrm e}^{c_{1}}\right )^{4}} \\ \bullet & {} & \textrm {Solve 2nd ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-{\left (y \left ({\mathrm e}^{c_{1}}\right )^{4}-3\right )}^{\frac {1}{4}} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=-{\left (y \left ({\mathrm e}^{c_{1}}\right )^{4}-3\right )}^{\frac {1}{4}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-{\left (y \left ({\mathrm e}^{c_{1}}\right )^{4}-3\right )}^{\frac {1}{4}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{{\left (y \left ({\mathrm e}^{c_{1}}\right )^{4}-3\right )}^{\frac {1}{4}}}=-1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{{\left (y \left ({\mathrm e}^{c_{1}}\right )^{4}-3\right )}^{\frac {1}{4}}}d x =\int \left (-1\right )d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {4 {\left (y \left ({\mathrm e}^{c_{1}}\right )^{4}-3\right )}^{\frac {3}{4}}}{3 \left ({\mathrm e}^{c_{1}}\right )^{4}}=-x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\left (3 {\left (\left ({\mathrm e}^{c_{1}}\right )^{4} \left (-x +c_{2} \right )\right )}^{\frac {4}{3}}+4 \,6^{\frac {2}{3}}\right ) 6^{\frac {1}{3}}}{8 \left ({\mathrm e}^{c_{1}}\right )^{4}} \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))*_b(_a)-(1/4)*(_b(_a)^4+3)/(_a*_b(_a)^2) = 0, _b(_a), HINT = [[_a, 0]]` *** Su symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[_a, 0]
✓ Solution by Maple
Time used: 0.079 (sec). Leaf size: 111
dsolve(4*y(x)*diff(y(x),x)^2*diff(y(x),x$2)=diff(y(x),x)^4+3,y(x), singsol=all)
\begin{align*} \frac {-4 \left (c_{1} y \left (x \right )-3\right )^{\frac {3}{4}}+\left (-3 x -3 c_{2} \right ) c_{1}}{3 c_{1}} &= 0 \\ \frac {4 \left (c_{1} y \left (x \right )-3\right )^{\frac {3}{4}}+\left (-3 x -3 c_{2} \right ) c_{1}}{3 c_{1}} &= 0 \\ \frac {-4 i \left (c_{1} y \left (x \right )-3\right )^{\frac {3}{4}}+\left (-3 x -3 c_{2} \right ) c_{1}}{3 c_{1}} &= 0 \\ \frac {4 i \left (c_{1} y \left (x \right )-3\right )^{\frac {3}{4}}+\left (-3 x -3 c_{2} \right ) c_{1}}{3 c_{1}} &= 0 \\ \end{align*}
✓ Solution by Mathematica
Time used: 60.242 (sec). Leaf size: 156
DSolve[4*y[x]*(y'[x])^2*y''[x]==(y'[x])^4+3,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {3}{8} e^{-4 c_1} \left (8+\sqrt [3]{6} \left (-e^{4 c_1} (x+c_2)\right ){}^{4/3}\right ) \\ y(x)\to \frac {3}{8} e^{-4 c_1} \left (8+\sqrt [3]{6} \left (-i e^{4 c_1} (x+c_2)\right ){}^{4/3}\right ) \\ y(x)\to \frac {3}{8} e^{-4 c_1} \left (8+\sqrt [3]{6} \left (i e^{4 c_1} (x+c_2)\right ){}^{4/3}\right ) \\ y(x)\to \frac {3}{8} e^{-4 c_1} \left (8+\sqrt [3]{6} \left (e^{4 c_1} (x+c_2)\right ){}^{4/3}\right ) \\ \end{align*}