Internal problem ID [10132]
Internal file name [OUTPUT/9079_Monday_June_06_2022_06_24_59_AM_87561219/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1811 (book 6.220).
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]]
\[ \boxed {\sqrt {y}\, y^{\prime \prime }=a} \]
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*} \sqrt {y}\, p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right ) = a \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 {a}{\sqrt {y}\, p} \end {align*}
Where \(f(y)=\frac {a}{\sqrt {y}}\) and \(g(p)=\frac {1}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{p}} \,dp &= \frac {a}{\sqrt {y}} \,d y \\ \int { \frac {1}{\frac {1}{p}} \,dp} &= \int {\frac {a}{\sqrt {y}} \,d y} \\ \frac {p^{2}}{2}&=2 \sqrt {y}\, a +c_{1} \\ \end{align*} The solution is \[ \frac {p \left (y \right )^{2}}{2}-2 \sqrt {y}\, a -c_{1} = 0 \] 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 {{y^{\prime }}^{2}}{2}-2 \sqrt {y}\, a -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 }&=\sqrt {4 \sqrt {y}\, a +2 c_{1}} \tag {1} \\ y^{\prime }&=-\sqrt {4 \sqrt {y}\, a +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*} \int \frac {1}{\sqrt {4 \sqrt {y}\, a +2 c_{1}}}d y &= \int d x \\ \frac {\frac {\left (4 \sqrt {y}\, a +2 c_{1} \right )^{\frac {3}{2}}}{3}-2 c_{1} \sqrt {4 \sqrt {y}\, a +2 c_{1}}}{4 a^{2}}&=x +c_{2} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {1}{\sqrt {4 \sqrt {y}\, a +2 c_{1}}}d y &= \int d x \\ -\frac {\frac {\left (4 \sqrt {y}\, a +2 c_{1} \right )^{\frac {3}{2}}}{3}-2 c_{1} \sqrt {4 \sqrt {y}\, a +2 c_{1}}}{4 a^{2}}&=x +c_{3} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} \frac {\frac {\left (4 \sqrt {y}\, a +2 c_{1} \right )^{\frac {3}{2}}}{3}-2 c_{1} \sqrt {4 \sqrt {y}\, a +2 c_{1}}}{4 a^{2}} &= x +c_{2} \\ \tag{2} -\frac {\frac {\left (4 \sqrt {y}\, a +2 c_{1} \right )^{\frac {3}{2}}}{3}-2 c_{1} \sqrt {4 \sqrt {y}\, a +2 c_{1}}}{4 a^{2}} &= x +c_{3} \\ \end{align*}
Verification of solutions
\[ \frac {\frac {\left (4 \sqrt {y}\, a +2 c_{1} \right )^{\frac {3}{2}}}{3}-2 c_{1} \sqrt {4 \sqrt {y}\, a +2 c_{1}}}{4 a^{2}} = x +c_{2} \] Verified OK.
\[ -\frac {\frac {\left (4 \sqrt {y}\, a +2 c_{1} \right )^{\frac {3}{2}}}{3}-2 c_{1} \sqrt {4 \sqrt {y}\, a +2 c_{1}}}{4 a^{2}} = x +c_{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \sqrt {y}\, \left (\frac {d}{d x}y^{\prime }\right )=a \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (x \right )=y^{\prime } \\ \bullet & {} & \textrm {Compute}\hspace {3pt} \frac {d}{d x}y^{\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 ),\frac {d}{d x}y^{\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} \\ {} & {} & \sqrt {y}\, u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=a \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=\frac {a}{\sqrt {y}\, u \left (y \right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=\frac {a}{\sqrt {y}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )d y =\int \frac {a}{\sqrt {y}}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {u \left (y \right )^{2}}{2}=2 \sqrt {y}\, a +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & \left \{u \left (y \right )=\sqrt {4 \sqrt {y}\, a +2 c_{1}}, u \left (y \right )=-\sqrt {4 \sqrt {y}\, a +2 c_{1}}\right \} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\sqrt {4 \sqrt {y}\, a +2 c_{1}} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=\sqrt {4 \sqrt {y}\, a +2 c_{1}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\sqrt {4 \sqrt {y}\, a +2 c_{1}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sqrt {4 \sqrt {y}\, a +2 c_{1}}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{\sqrt {4 \sqrt {y}\, a +2 c_{1}}}d x =\int 1d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\frac {\left (4 \sqrt {y}\, a +2 c_{1} \right )^{\frac {3}{2}}}{3}-2 c_{1} \sqrt {4 \sqrt {y}\, a +2 c_{1}}}{4 a^{2}}=x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {6 c_{2} a^{2} \left (\left (6 c_{2} a^{2}+6 x \,a^{2}+2 \sqrt {9 c_{2}^{2} a^{4}+18 c_{2} a^{4} x +9 a^{4} x^{2}-2 c_{1}^{3}}\right )^{\frac {1}{3}}+\frac {2 c_{1}}{\left (6 c_{2} a^{2}+6 x \,a^{2}+2 \sqrt {9 c_{2}^{2} a^{4}+18 c_{2} a^{4} x +9 a^{4} x^{2}-2 c_{1}^{3}}\right )^{\frac {1}{3}}}\right )+6 a^{2} x \left (\left (6 c_{2} a^{2}+6 x \,a^{2}+2 \sqrt {9 c_{2}^{2} a^{4}+18 c_{2} a^{4} x +9 a^{4} x^{2}-2 c_{1}^{3}}\right )^{\frac {1}{3}}+\frac {2 c_{1}}{\left (6 c_{2} a^{2}+6 x \,a^{2}+2 \sqrt {9 c_{2}^{2} a^{4}+18 c_{2} a^{4} x +9 a^{4} x^{2}-2 c_{1}^{3}}\right )^{\frac {1}{3}}}\right )+c_{1} {\left (\left (6 c_{2} a^{2}+6 x \,a^{2}+2 \sqrt {9 c_{2}^{2} a^{4}+18 c_{2} a^{4} x +9 a^{4} x^{2}-2 c_{1}^{3}}\right )^{\frac {1}{3}}+\frac {2 c_{1}}{\left (6 c_{2} a^{2}+6 x \,a^{2}+2 \sqrt {9 c_{2}^{2} a^{4}+18 c_{2} a^{4} x +9 a^{4} x^{2}-2 c_{1}^{3}}\right )^{\frac {1}{3}}}\right )}^{2}+2 c_{1}^{2}}{8 a^{2}} \\ \bullet & {} & \textrm {Solve 2nd ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-\sqrt {4 \sqrt {y}\, a +2 c_{1}} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=-\sqrt {4 \sqrt {y}\, a +2 c_{1}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\sqrt {4 \sqrt {y}\, a +2 c_{1}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sqrt {4 \sqrt {y}\, a +2 c_{1}}}=-1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{\sqrt {4 \sqrt {y}\, a +2 c_{1}}}d x =\int \left (-1\right )d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\frac {\left (4 \sqrt {y}\, a +2 c_{1} \right )^{\frac {3}{2}}}{3}-2 c_{1} \sqrt {4 \sqrt {y}\, a +2 c_{1}}}{4 a^{2}}=-x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {6 c_{2} a^{2} \left (\left (6 c_{2} a^{2}-6 x \,a^{2}+2 \sqrt {9 c_{2}^{2} a^{4}-18 c_{2} a^{4} x +9 a^{4} x^{2}-2 c_{1}^{3}}\right )^{\frac {1}{3}}+\frac {2 c_{1}}{\left (6 c_{2} a^{2}-6 x \,a^{2}+2 \sqrt {9 c_{2}^{2} a^{4}-18 c_{2} a^{4} x +9 a^{4} x^{2}-2 c_{1}^{3}}\right )^{\frac {1}{3}}}\right )-6 a^{2} x \left (\left (6 c_{2} a^{2}-6 x \,a^{2}+2 \sqrt {9 c_{2}^{2} a^{4}-18 c_{2} a^{4} x +9 a^{4} x^{2}-2 c_{1}^{3}}\right )^{\frac {1}{3}}+\frac {2 c_{1}}{\left (6 c_{2} a^{2}-6 x \,a^{2}+2 \sqrt {9 c_{2}^{2} a^{4}-18 c_{2} a^{4} x +9 a^{4} x^{2}-2 c_{1}^{3}}\right )^{\frac {1}{3}}}\right )+c_{1} {\left (\left (6 c_{2} a^{2}-6 x \,a^{2}+2 \sqrt {9 c_{2}^{2} a^{4}-18 c_{2} a^{4} x +9 a^{4} x^{2}-2 c_{1}^{3}}\right )^{\frac {1}{3}}+\frac {2 c_{1}}{\left (6 c_{2} a^{2}-6 x \,a^{2}+2 \sqrt {9 c_{2}^{2} a^{4}-18 c_{2} a^{4} x +9 a^{4} x^{2}-2 c_{1}^{3}}\right )^{\frac {1}{3}}}\right )}^{2}+2 c_{1}^{2}}{8 a^{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))*_b(_a)-a/_a^(1/2) = 0, _b(_a), HINT = [[_a, (1/4)*_b]]` *** Sublevel 2 *** symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[_a, 1/4*_b]
✓ Solution by Maple
Time used: 0.062 (sec). Leaf size: 81
dsolve(y(x)^(1/2)*diff(diff(y(x),x),x)-a=0,y(x), singsol=all)
\begin{align*} \frac {\left (-2 a \sqrt {y \left (x \right )}-c_{1} \right ) \sqrt {4 a \sqrt {y \left (x \right )}-c_{1}}-6 a^{2} \left (c_{2} +x \right )}{6 a^{2}} &= 0 \\ \frac {\left (2 a \sqrt {y \left (x \right )}+c_{1} \right ) \sqrt {4 a \sqrt {y \left (x \right )}-c_{1}}-6 a^{2} \left (c_{2} +x \right )}{6 a^{2}} &= 0 \\ \end{align*}
✓ Solution by Mathematica
Time used: 60.111 (sec). Leaf size: 1881
DSolve[-a + Sqrt[y[x]]*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {288 a^4 c_1 x^2+576 a^4 c_1 c_2 x+288 a^4 c_1 c_2{}^2+a^4 \left (\frac {10368 a^8 x^4+41472 a^8 c_2 x^3+62208 a^8 c_2{}^2 x^2+41472 a^8 c_2{}^3 x+10368 a^8 c_2{}^4+720 a^4 c_1{}^3 x^2+1440 a^4 c_1{}^3 c_2 x+720 a^4 c_1{}^3 c_2{}^2+48 a^6 \sqrt {\frac {(x+c_2){}^2 \left (36 a^4 (x+c_2){}^2-c_1{}^3\right ){}^3}{a^8}}-c_1{}^6}{a^6}\right ){}^{2/3}+3 a^2 c_1{}^2 \sqrt [3]{\frac {10368 a^8 x^4+41472 a^8 c_2 x^3+62208 a^8 c_2{}^2 x^2+41472 a^8 c_2{}^3 x+10368 a^8 c_2{}^4+720 a^4 c_1{}^3 x^2+1440 a^4 c_1{}^3 c_2 x+720 a^4 c_1{}^3 c_2{}^2+48 a^6 \sqrt {\frac {(x+c_2){}^2 \left (36 a^4 (x+c_2){}^2-c_1{}^3\right ){}^3}{a^8}}-c_1{}^6}{a^6}}+c_1{}^4}{16 a^4 \sqrt [3]{\frac {10368 a^8 x^4+41472 a^8 c_2 x^3+62208 a^8 c_2{}^2 x^2+41472 a^8 c_2{}^3 x+10368 a^8 c_2{}^4+720 a^4 c_1{}^3 x^2+1440 a^4 c_1{}^3 c_2 x+720 a^4 c_1{}^3 c_2{}^2+48 a^6 \sqrt {\frac {(x+c_2){}^2 \left (36 a^4 (x+c_2){}^2-c_1{}^3\right ){}^3}{a^8}}-c_1{}^6}{a^6}}} \\ y(x)\to \frac {-288 i \sqrt {3} a^4 c_1 x^2-288 a^4 c_1 x^2-576 i \sqrt {3} a^4 c_1 c_2 x-576 a^4 c_1 c_2 x-288 i \sqrt {3} a^4 c_1 c_2{}^2-288 a^4 c_1 c_2{}^2+i \sqrt {3} a^4 \left (\frac {10368 a^8 x^4+41472 a^8 c_2 x^3+62208 a^8 c_2{}^2 x^2+41472 a^8 c_2{}^3 x+10368 a^8 c_2{}^4+720 a^4 c_1{}^3 x^2+1440 a^4 c_1{}^3 c_2 x+720 a^4 c_1{}^3 c_2{}^2+48 a^6 \sqrt {\frac {(x+c_2){}^2 \left (36 a^4 (x+c_2){}^2-c_1{}^3\right ){}^3}{a^8}}-c_1{}^6}{a^6}\right ){}^{2/3}-a^4 \left (\frac {10368 a^8 x^4+41472 a^8 c_2 x^3+62208 a^8 c_2{}^2 x^2+41472 a^8 c_2{}^3 x+10368 a^8 c_2{}^4+720 a^4 c_1{}^3 x^2+1440 a^4 c_1{}^3 c_2 x+720 a^4 c_1{}^3 c_2{}^2+48 a^6 \sqrt {\frac {(x+c_2){}^2 \left (36 a^4 (x+c_2){}^2-c_1{}^3\right ){}^3}{a^8}}-c_1{}^6}{a^6}\right ){}^{2/3}+6 a^2 c_1{}^2 \sqrt [3]{\frac {10368 a^8 x^4+41472 a^8 c_2 x^3+62208 a^8 c_2{}^2 x^2+41472 a^8 c_2{}^3 x+10368 a^8 c_2{}^4+720 a^4 c_1{}^3 x^2+1440 a^4 c_1{}^3 c_2 x+720 a^4 c_1{}^3 c_2{}^2+48 a^6 \sqrt {\frac {(x+c_2){}^2 \left (36 a^4 (x+c_2){}^2-c_1{}^3\right ){}^3}{a^8}}-c_1{}^6}{a^6}}-i \sqrt {3} c_1{}^4-c_1{}^4}{32 a^4 \sqrt [3]{\frac {10368 a^8 x^4+41472 a^8 c_2 x^3+62208 a^8 c_2{}^2 x^2+41472 a^8 c_2{}^3 x+10368 a^8 c_2{}^4+720 a^4 c_1{}^3 x^2+1440 a^4 c_1{}^3 c_2 x+720 a^4 c_1{}^3 c_2{}^2+48 a^6 \sqrt {\frac {(x+c_2){}^2 \left (36 a^4 (x+c_2){}^2-c_1{}^3\right ){}^3}{a^8}}-c_1{}^6}{a^6}}} \\ y(x)\to \frac {288 i \sqrt {3} a^4 c_1 x^2-288 a^4 c_1 x^2+576 i \sqrt {3} a^4 c_1 c_2 x-576 a^4 c_1 c_2 x+288 i \sqrt {3} a^4 c_1 c_2{}^2-288 a^4 c_1 c_2{}^2-i \sqrt {3} a^4 \left (\frac {10368 a^8 x^4+41472 a^8 c_2 x^3+62208 a^8 c_2{}^2 x^2+41472 a^8 c_2{}^3 x+10368 a^8 c_2{}^4+720 a^4 c_1{}^3 x^2+1440 a^4 c_1{}^3 c_2 x+720 a^4 c_1{}^3 c_2{}^2+48 a^6 \sqrt {\frac {(x+c_2){}^2 \left (36 a^4 (x+c_2){}^2-c_1{}^3\right ){}^3}{a^8}}-c_1{}^6}{a^6}\right ){}^{2/3}-a^4 \left (\frac {10368 a^8 x^4+41472 a^8 c_2 x^3+62208 a^8 c_2{}^2 x^2+41472 a^8 c_2{}^3 x+10368 a^8 c_2{}^4+720 a^4 c_1{}^3 x^2+1440 a^4 c_1{}^3 c_2 x+720 a^4 c_1{}^3 c_2{}^2+48 a^6 \sqrt {\frac {(x+c_2){}^2 \left (36 a^4 (x+c_2){}^2-c_1{}^3\right ){}^3}{a^8}}-c_1{}^6}{a^6}\right ){}^{2/3}+6 a^2 c_1{}^2 \sqrt [3]{\frac {10368 a^8 x^4+41472 a^8 c_2 x^3+62208 a^8 c_2{}^2 x^2+41472 a^8 c_2{}^3 x+10368 a^8 c_2{}^4+720 a^4 c_1{}^3 x^2+1440 a^4 c_1{}^3 c_2 x+720 a^4 c_1{}^3 c_2{}^2+48 a^6 \sqrt {\frac {(x+c_2){}^2 \left (36 a^4 (x+c_2){}^2-c_1{}^3\right ){}^3}{a^8}}-c_1{}^6}{a^6}}+i \sqrt {3} c_1{}^4-c_1{}^4}{32 a^4 \sqrt [3]{\frac {10368 a^8 x^4+41472 a^8 c_2 x^3+62208 a^8 c_2{}^2 x^2+41472 a^8 c_2{}^3 x+10368 a^8 c_2{}^4+720 a^4 c_1{}^3 x^2+1440 a^4 c_1{}^3 c_2 x+720 a^4 c_1{}^3 c_2{}^2+48 a^6 \sqrt {\frac {(x+c_2){}^2 \left (36 a^4 (x+c_2){}^2-c_1{}^3\right ){}^3}{a^8}}-c_1{}^6}{a^6}}} \\ \end{align*}