Internal problem ID [12180]
Internal file name [OUTPUT/10833_Thursday_September_21_2023_05_47_36_AM_66756878/index.tex]
Book: Differential equations and the calculus of variations by L. ElSGOLTS. MIR PUBLISHERS,
MOSCOW, Third printing 1977.
Section: Chapter 2, DIFFERENTIAL EQUATIONS OF THE SECOND ORDER AND HIGHER.
Problems page 172
Problem number: Problem 18.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_x", "second_order_ode_can_be_made_integrable"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]
\[ \boxed {y^{\prime \prime }-3 \sqrt {y}=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = 2] \end {align*}
Multiplying the ode by \(y^{\prime }\) gives \[ y^{\prime } y^{\prime \prime }-3 \sqrt {y}\, y^{\prime } = 0 \] Integrating the above w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime } y^{\prime \prime }-3 \sqrt {y}\, y^{\prime }\right )d x &= 0 \\ \frac {{y^{\prime }}^{2}}{2}-2 y^{\frac {3}{2}} = c_2 \end {align*}
Which is now solved for \(y\). 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 y^{\frac {3}{2}}+2 c_{1}} \tag {1} \\ y^{\prime }&=-\sqrt {4 y^{\frac {3}{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*} \int \frac {1}{\sqrt {4 y^{\frac {3}{2}}+2 c_{1}}}d y &= \int {dx}\\ \int _{}^{y}\frac {1}{\sqrt {4 \textit {\_a}^{\frac {3}{2}}+2 c_{1}}}d \textit {\_a}&= x +c_{2} \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} \int -\frac {1}{\sqrt {4 y^{\frac {3}{2}}+2 c_{1}}}d y &= \int {dx}\\ \int _{}^{y}-\frac {1}{\sqrt {4 \textit {\_a}^{\frac {3}{2}}+2 c_{1}}}d \textit {\_a}&= x +c_{3} \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the First solution \begin {align*} \int _{}^{y}\frac {1}{\sqrt {4 \textit {\_a}^{\frac {3}{2}}+2 c_{1}}}d \textit {\_a} = x +c_{2} \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = 1\) and \(x = 0\) in the above gives \begin {align*} \int _{}^{1}\frac {1}{\sqrt {4 \textit {\_a}^{\frac {3}{2}}+2 c_{1}}}d \textit {\_a} = c_{2}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \sqrt {4 {\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {1}{\sqrt {4 \textit {\_a}^{\frac {3}{2}}+2 c_{1}}}d \textit {\_a} \right )+x +c_{2} \right )}^{\frac {3}{2}}+2 c_{1}} \end {align*}
substituting \(y^{\prime } = 2\) and \(x = 0\) in the above gives \begin {align*} 2 = \sqrt {4 {\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {1}{\sqrt {4 \textit {\_a}^{\frac {3}{2}}+2 c_{1}}}d \textit {\_a} \right )+c_{2} \right )}^{\frac {3}{2}}+2 c_{1}}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). There is no solution for the constants of integrations. This solution is removed.
Looking at the Second solution \begin {align*} \int _{}^{y}-\frac {1}{\sqrt {4 \textit {\_a}^{\frac {3}{2}}+2 c_{1}}}d \textit {\_a} = x +c_{3} \tag {2} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = 1\) and \(x = 0\) in the above gives \begin {align*} -\left (\int _{}^{1}\frac {1}{\sqrt {4 \textit {\_a}^{\frac {3}{2}}+2 c_{1}}}d \textit {\_a} \right ) = c_{3}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = -\sqrt {4 {\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}-\frac {1}{\sqrt {4 \textit {\_a}^{\frac {3}{2}}+2 c_{1}}}d \textit {\_a} \right )+x +c_{3} \right )}^{\frac {3}{2}}+2 c_{1}} \end {align*}
substituting \(y^{\prime } = 2\) and \(x = 0\) in the above gives \begin {align*} 2 = -\sqrt {4 \operatorname {RootOf}\left (\int _{}^{\textit {\_Z}}\frac {1}{\sqrt {4 \textit {\_a}^{\frac {3}{2}}+2 c_{1}}}d \textit {\_a} +c_{3} \right )^{\frac {3}{2}}+2 c_{1}}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{3}\}\). There is no solution for the constants of integrations. This solution is removed.
Verification of solutions N/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*} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )-3 \sqrt {y} = 0 \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 {3 \sqrt {y}}{p} \end {align*}
Where \(f(y)=3 \sqrt {y}\) and \(g(p)=\frac {1}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{p}} \,dp &= 3 \sqrt {y} \,d y \\ \int { \frac {1}{\frac {1}{p}} \,dp} &= \int {3 \sqrt {y} \,d y} \\ \frac {p^{2}}{2}&=2 y^{\frac {3}{2}}+c_{1} \\ \end{align*} The solution is \[ \frac {p \left (y \right )^{2}}{2}-2 y^{\frac {3}{2}}-c_{1} = 0 \] Initial conditions are used to solve for \(c_{1}\). Substituting \(y=1\) and \(p=2\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -c_{1} = 0 \end {align*}
The solutions are \begin {align*} c_{1} = 0 \end {align*}
Trying the constant \begin {align*} c_{1} = 0 \end {align*}
Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \frac {p^{2}}{2}-2 y^{\frac {3}{2}} = 0 \end {align*}
The constant \(c_{1} = 0\) 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*} \frac {{y^{\prime }}^{2}}{2}-2 y^{\frac {3}{2}} = 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 }&=2 \sqrt {y^{\frac {3}{2}}} \tag {1} \\ y^{\prime }&=-2 \sqrt {y^{\frac {3}{2}}} \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}{2 \sqrt {y^{\frac {3}{2}}}}d y &= \int {dx}\\ \frac {2 y}{\sqrt {y^{\frac {3}{2}}}}&= x +c_{2} \end {align*}
Initial conditions are used to solve for \(c_{2}\). Substituting \(x=0\) and \(y=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 2 = c_{2} \end {align*}
The solutions are \begin {align*} c_{2} = 2 \end {align*}
Trying the constant \begin {align*} c_{2} = 2 \end {align*}
Substituting \(c_{2}\) found above in the general solution gives \begin {align*} \frac {2 y}{\sqrt {y^{\frac {3}{2}}}} = x +2 \end {align*}
The constant \(c_{2} = 2\) gives valid solution.
Solving equation (2)
Integrating both sides gives \begin {align*} \int -\frac {1}{2 \sqrt {y^{\frac {3}{2}}}}d y &= \int {dx}\\ -\frac {2 y}{\sqrt {y^{\frac {3}{2}}}}&= x +c_{3} \end {align*}
Initial conditions are used to solve for \(c_{3}\). Substituting \(x=0\) and \(y=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -2 = c_{3} \end {align*}
The solutions are \begin {align*} c_{3} = -2 \end {align*}
Trying the constant \begin {align*} c_{3} = -2 \end {align*}
Substituting \(c_{3}\) found above in the general solution gives \begin {align*} -\frac {2 y}{\sqrt {y^{\frac {3}{2}}}} = x -2 \end {align*}
The constant \(c_{3} = -2\) 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 {\left (x^{2}+4 x +4\right )^{2}}{16} \\ \tag{2} y &= \frac {\left (x^{2}-4 x +4\right )^{2}}{16} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (x^{2}+4 x +4\right )^{2}}{16} \] Verified OK.
\[ y = \frac {\left (x^{2}-4 x +4\right )^{2}}{16} \] Warning, solution could not be verified
Maple trace
`Methods for second order ODEs: *** Sublevel 2 *** 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)-3*_a^(1/2) = 0, _b(_a), HINT = [[_a, (3/4)*_b]]` *** Sublevel 3 symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[_a, 3/4*_b]
✓ Solution by Maple
Time used: 0.281 (sec). Leaf size: 11
dsolve([diff(y(x),x$2)=3*sqrt(y(x)),y(0) = 1, D(y)(0) = 2],y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (x +2\right )^{4}}{16} \]
✓ Solution by Mathematica
Time used: 0.1 (sec). Leaf size: 14
DSolve[{y''[x]==3*Sqrt[y[x]],{y[0]==1,y'[0]==2}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{16} (x+2)^4 \]