Internal problem ID [9916]
Internal file name [OUTPUT/8863_Monday_June_06_2022_05_42_01_AM_17675308/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1594 (6.4).
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]]
\[ \boxed {y^{\prime \prime }-6 y^{2}+4 y=0} \]
Multiplying the ode by \(y^{\prime }\) gives \[ y^{\prime } y^{\prime \prime }+\left (-6 y+4\right ) y^{\prime } y = 0 \] Integrating the above w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime } y^{\prime \prime }+\left (-6 y+4\right ) y^{\prime } y\right )d x &= 0 \\ \frac {{y^{\prime }}^{2}}{2}-2 y^{3}+2 y^{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^{3}-4 y^{2}+2 c_{1}} \tag {1} \\ y^{\prime }&=-\sqrt {4 y^{3}-4 y^{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^{3}-4 y^{2}+2 c_{1}}}d y &= \int {dx}\\ \int _{}^{y}\frac {1}{\sqrt {4 \textit {\_a}^{3}-4 \textit {\_a}^{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^{3}-4 y^{2}+2 c_{1}}}d y &= \int {dx}\\ \int _{}^{y}-\frac {1}{\sqrt {4 \textit {\_a}^{3}-4 \textit {\_a}^{2}+2 c_{1}}}d \textit {\_a}&= x +c_{3} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}\frac {1}{\sqrt {4 \textit {\_a}^{3}-4 \textit {\_a}^{2}+2 c_{1}}}d \textit {\_a} &= x +c_{2} \\ \tag{2} \int _{}^{y}-\frac {1}{\sqrt {4 \textit {\_a}^{3}-4 \textit {\_a}^{2}+2 c_{1}}}d \textit {\_a} &= x +c_{3} \\ \end{align*}
Verification of solutions
\[ \int _{}^{y}\frac {1}{\sqrt {4 \textit {\_a}^{3}-4 \textit {\_a}^{2}+2 c_{1}}}d \textit {\_a} = x +c_{2} \] Verified OK.
\[ \int _{}^{y}-\frac {1}{\sqrt {4 \textit {\_a}^{3}-4 \textit {\_a}^{2}+2 c_{1}}}d \textit {\_a} = x +c_{3} \] 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*} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+\left (-6 y +4\right ) 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 {2 y \left (3 y -2\right )}{p} \end {align*}
Where \(f(y)=2 y \left (3 y -2\right )\) and \(g(p)=\frac {1}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{p}} \,dp &= 2 y \left (3 y -2\right ) \,d y \\ \int { \frac {1}{\frac {1}{p}} \,dp} &= \int {2 y \left (3 y -2\right ) \,d y} \\ \frac {p^{2}}{2}&=2 y^{3}-2 y^{2}+c_{1} \\ \end{align*} The solution is \[ \frac {p \left (y \right )^{2}}{2}-2 y^{3}+2 y^{2}-c_{1} = 0 \] 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 {{y^{\prime }}^{2}}{2}-2 y^{3}+2 y^{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 }&=\sqrt {4 y^{3}-4 y^{2}+2 c_{1}} \tag {1} \\ y^{\prime }&=-\sqrt {4 y^{3}-4 y^{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^{3}-4 y^{2}+2 c_{1}}}d y &= \int {dx}\\ \int _{}^{y}\frac {1}{\sqrt {4 \textit {\_a}^{3}-4 \textit {\_a}^{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^{3}-4 y^{2}+2 c_{1}}}d y &= \int {dx}\\ \int _{}^{y}-\frac {1}{\sqrt {4 \textit {\_a}^{3}-4 \textit {\_a}^{2}+2 c_{1}}}d \textit {\_a}&= x +c_{3} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}\frac {1}{\sqrt {4 \textit {\_a}^{3}-4 \textit {\_a}^{2}+2 c_{1}}}d \textit {\_a} &= x +c_{2} \\ \tag{2} \int _{}^{y}-\frac {1}{\sqrt {4 \textit {\_a}^{3}-4 \textit {\_a}^{2}+2 c_{1}}}d \textit {\_a} &= x +c_{3} \\ \end{align*}
Verification of solutions
\[ \int _{}^{y}\frac {1}{\sqrt {4 \textit {\_a}^{3}-4 \textit {\_a}^{2}+2 c_{1}}}d \textit {\_a} = x +c_{2} \] Verified OK.
\[ \int _{}^{y}-\frac {1}{\sqrt {4 \textit {\_a}^{3}-4 \textit {\_a}^{2}+2 c_{1}}}d \textit {\_a} = x +c_{3} \] Verified OK.
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 `, `-> Computing symmetries using: way = exp_sym -> Calling odsolve with the ODE`, (diff(_b(_a), _a))*_b(_a)-6*_a^2+4*_a = 0, _b(_a)` *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli <- Bernoulli successful <- differential order: 2; canonical coordinates successful <- differential order 2; missing variables successful`
✓ Solution by Maple
Time used: 0.453 (sec). Leaf size: 59
dsolve(diff(diff(y(x),x),x)-6*y(x)^2+4*y(x)=0,y(x), singsol=all)
\begin{align*} \int _{}^{y \left (x \right )}\frac {1}{\sqrt {4 \textit {\_a}^{3}-4 \textit {\_a}^{2}+c_{1}}}d \textit {\_a} -x -c_{2} &= 0 \\ -\left (\int _{}^{y \left (x \right )}\frac {1}{\sqrt {4 \textit {\_a}^{3}-4 \textit {\_a}^{2}+c_{1}}}d \textit {\_a} \right )-x -c_{2} &= 0 \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.486 (sec). Leaf size: 373
DSolve[4*y[x] - 6*y[x]^2 + y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\frac {4 \left (\text {Root}\left [4 \text {$\#$1}^3-4 \text {$\#$1}^2+c_1\&,2\right ]-\text {Root}\left [4 \text {$\#$1}^3-4 \text {$\#$1}^2+c_1\&,3\right ]\right ) \left (y(x)-\text {Root}\left [4 \text {$\#$1}^3-4 \text {$\#$1}^2+c_1\&,1\right ]\right ) \left (y(x)-\text {Root}\left [4 \text {$\#$1}^3-4 \text {$\#$1}^2+c_1\&,2\right ]\right ) \left (y(x)-\text {Root}\left [4 \text {$\#$1}^3-4 \text {$\#$1}^2+c_1\&,3\right ]\right ) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {\text {Root}\left [4 \text {$\#$1}^3-4 \text {$\#$1}^2+c_1\&,3\right ]-y(x)}{\text {Root}\left [4 \text {$\#$1}^3-4 \text {$\#$1}^2+c_1\&,3\right ]-\text {Root}\left [4 \text {$\#$1}^3-4 \text {$\#$1}^2+c_1\&,2\right ]}}\right ),\frac {\text {Root}\left [4 \text {$\#$1}^3-4 \text {$\#$1}^2+c_1\&,2\right ]-\text {Root}\left [4 \text {$\#$1}^3-4 \text {$\#$1}^2+c_1\&,3\right ]}{\text {Root}\left [4 \text {$\#$1}^3-4 \text {$\#$1}^2+c_1\&,1\right ]-\text {Root}\left [4 \text {$\#$1}^3-4 \text {$\#$1}^2+c_1\&,3\right ]}\right ){}^2}{\left (4 y(x)^3-4 y(x)^2+c_1\right ) \left (\text {Root}\left [4 \text {$\#$1}^3-4 \text {$\#$1}^2+c_1\&,3\right ]-\text {Root}\left [4 \text {$\#$1}^3-4 \text {$\#$1}^2+c_1\&,1\right ]\right ) \left (\text {Root}\left [4 \text {$\#$1}^3-4 \text {$\#$1}^2+c_1\&,3\right ]-\text {Root}\left [4 \text {$\#$1}^3-4 \text {$\#$1}^2+c_1\&,2\right ]\right )}=(x+c_2){}^2,y(x)\right ] \]