Internal problem ID [9954]
Internal file name [OUTPUT/8901_Monday_June_06_2022_05_48_05_AM_90440633/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1632 (6.42).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_integrable_as_is", "second_order_ode_missing_x", "exact nonlinear second order ode"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], [_2nd_order, _exact, _nonlinear], _Lagerstrom, [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {y^{\prime \prime }-2 a y y^{\prime }=0} \]
Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime \prime }-2 a y y^{\prime }\right )d x &= 0 \\ -a y^{2}+y^{\prime } = c_{1} \end {align*}
Which is now solved for \(y\). Integrating both sides gives \begin {align*} \int \frac {1}{a \,y^{2}+c_{1}}d y &= x +c_{2}\\ \frac {\arctan \left (\frac {a y}{\sqrt {c_{1} a}}\right )}{\sqrt {c_{1} a}}&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=\frac {\tan \left (c_{2} \sqrt {c_{1} a}+x \sqrt {c_{1} a}\right ) \sqrt {c_{1} a}}{a} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\tan \left (c_{2} \sqrt {c_{1} a}+x \sqrt {c_{1} a}\right ) \sqrt {c_{1} a}}{a} \\ \end{align*}
Verification of solutions
\[ y = \frac {\tan \left (c_{2} \sqrt {c_{1} a}+x \sqrt {c_{1} a}\right ) \sqrt {c_{1} a}}{a} \] 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 )-2 a y p \left (y \right ) = 0 \end {align*}
Which is now solved as first order ode for \(p(y)\). Integrating both sides gives \begin {align*} p \left (y \right ) &= \int { 2 a y\,\mathop {\mathrm {d}y}}\\ &= a \,y^{2}+c_{1} \end {align*}
For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} y^{\prime } = a y^{2}+c_{1} \end {align*}
Integrating both sides gives \begin {align*} \int \frac {1}{a \,y^{2}+c_{1}}d y &= x +c_{2}\\ \frac {\arctan \left (\frac {a y}{\sqrt {c_{1} a}}\right )}{\sqrt {c_{1} a}}&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=\frac {\tan \left (c_{2} \sqrt {c_{1} a}+x \sqrt {c_{1} a}\right ) \sqrt {c_{1} a}}{a} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\tan \left (c_{2} \sqrt {c_{1} a}+x \sqrt {c_{1} a}\right ) \sqrt {c_{1} a}}{a} \\ \end{align*}
Verification of solutions
\[ y = \frac {\tan \left (c_{2} \sqrt {c_{1} a}+x \sqrt {c_{1} a}\right ) \sqrt {c_{1} a}}{a} \] Verified OK.
Writing the ode as \[ y^{\prime \prime }-2 a y y^{\prime } = 0 \] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime \prime }-2 a y y^{\prime }\right )d x &= 0 \\ -a y^{2}+y^{\prime } = c_{1} \end {align*}
Which is now solved for \(y\). Integrating both sides gives \begin {align*} \int \frac {1}{a \,y^{2}+c_{1}}d y &= x +c_{2}\\ \frac {\arctan \left (\frac {a y}{\sqrt {c_{1} a}}\right )}{\sqrt {c_{1} a}}&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=\frac {\tan \left (c_{2} \sqrt {c_{1} a}+x \sqrt {c_{1} a}\right ) \sqrt {c_{1} a}}{a} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\tan \left (c_{2} \sqrt {c_{1} a}+x \sqrt {c_{1} a}\right ) \sqrt {c_{1} a}}{a} \\ \end{align*}
Verification of solutions
\[ y = \frac {\tan \left (c_{2} \sqrt {c_{1} a}+x \sqrt {c_{1} a}\right ) \sqrt {c_{1} a}}{a} \] Verified OK.
An exact non-linear second order ode has the form \begin {align*} a_{2} \left (x , y, y^{\prime }\right ) y^{\prime \prime }+a_{1} \left (x , y, y^{\prime }\right ) y^{\prime }+a_{0} \left (x , y, y^{\prime }\right )&=0 \end {align*}
Where the following conditions are satisfied \begin {align*} \frac {\partial a_2}{\partial y} &= \frac {\partial a_1}{\partial y'}\\ \frac {\partial a_2}{\partial x} &= \frac {\partial a_0}{\partial y'}\\ \frac {\partial a_1}{\partial x} &= \frac {\partial a_0}{\partial y} \end {align*}
Looking at the the ode given we see that \begin {align*} a_2 &= 1\\ a_1 &= -2 a y\\ a_0 &= 0 \end {align*}
Applying the conditions to the above shows this is a nonlinear exact second order ode. Therefore it can be reduced to first order ode given by \begin {align*} \int {a_2\,d y'} + \int {a_1\,d y} + \int {a_0\,d x} &= c_{1}\\ \int {1\,d y'} + \int {-2 a y\,d y} + \int {0\,d x} &= c_{1} \end {align*}
Which results in \begin {align*} -a y^{2}+y^{\prime } = c_{1} \end {align*}
Which is now solved Integrating both sides gives \begin {align*} \int \frac {1}{a \,y^{2}+c_{1}}d y &= x +c_{2}\\ \frac {\arctan \left (\frac {a y}{\sqrt {c_{1} a}}\right )}{\sqrt {c_{1} a}}&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=\frac {\tan \left (c_{2} \sqrt {c_{1} a}+x \sqrt {c_{1} a}\right ) \sqrt {c_{1} a}}{a} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\tan \left (c_{2} \sqrt {c_{1} a}+x \sqrt {c_{1} a}\right ) \sqrt {c_{1} a}}{a} \\ \end{align*}
Verification of solutions
\[ y = \frac {\tan \left (c_{2} \sqrt {c_{1} a}+x \sqrt {c_{1} a}\right ) \sqrt {c_{1} a}}{a} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }-2 a y y^{\prime }=0 \\ \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} \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )-2 a y u \left (y \right )=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=2 a y \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int \left (\frac {d}{d y}u \left (y \right )\right )d y =\int 2 a y d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & u \left (y \right )=a \,y^{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=a \,y^{2}+c_{1} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=a \,y^{2}+c_{1} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=a y^{2}+c_{1} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=a y^{2}+c_{1} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{a y^{2}+c_{1}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{a y^{2}+c_{1}}d x =\int 1d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\arctan \left (\frac {a y}{\sqrt {c_{1} a}}\right )}{\sqrt {c_{1} a}}=x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\tan \left (c_{2} \sqrt {c_{1} a}+x \sqrt {c_{1} a}\right ) \sqrt {c_{1} a}}{a} \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)-2*_a*_b(_a)*a = 0, _b(_a), HINT = [[_a, 2*_b]]` *** Sublevel 2 *** symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[_a, 2*_b]
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 23
dsolve(diff(diff(y(x),x),x)-2*a*y(x)*diff(y(x),x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\tan \left (\left (c_{2} +x \right ) \sqrt {c_{1} a}\right ) \sqrt {c_{1} a}}{a} \]
✓ Solution by Mathematica
Time used: 25.806 (sec). Leaf size: 34
DSolve[-2*a*y[x]*y'[x] + y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {\sqrt {c_1} \tan \left (\sqrt {a} \sqrt {c_1} (x+c_2)\right )}{\sqrt {a}} \]