Internal problem ID [4839]
Internal file name [OUTPUT/4332_Sunday_June_05_2022_01_01_33_PM_90742386/index.tex]
Book: Mathematical Methods in the Physical Sciences. third edition. Mary L. Boas. John Wiley.
2006
Section: Chapter 8, Ordinary differential equations. Section 7. Other second-Order equations.
page 435
Problem number: 1 (a).
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]]
\[ \boxed {y^{\prime \prime }+y y^{\prime }=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 5, y^{\prime }\left (0\right ) = 0] \end {align*}
Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime \prime }+y y^{\prime }\right )d x &= 0 \\ \frac {y^{2}}{2}+y^{\prime } = c_{1} \end {align*}
Which is now solved for \(y\). Integrating both sides gives \begin {align*} \int \frac {1}{-\frac {y^{2}}{2}+c_{1}}d y &= x +c_{2}\\ \frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {y \sqrt {2}}{2 \sqrt {c_{1}}}\right )}{\sqrt {c_{1}}}&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=\tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}}{2}\right ) \sqrt {c_{1}}\, \sqrt {2} \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = \tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}}{2}\right ) \sqrt {c_{1}}\, \sqrt {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 \(y = 5\) and \(x = 0\) in the above gives \begin {align*} 5 = \frac {\left ({\mathrm e}^{c_{2} \sqrt {2}\, \sqrt {c_{1}}}-1\right ) \sqrt {2}\, \sqrt {c_{1}}}{{\mathrm e}^{c_{2} \sqrt {2}\, \sqrt {c_{1}}}+1}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = c_{1} \left (1-\tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}}{2}\right )^{2}\right ) \end {align*}
substituting \(y^{\prime } = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = \frac {4 \,{\mathrm e}^{c_{2} \sqrt {2}\, \sqrt {c_{1}}} c_{1}}{{\mathrm e}^{2 c_{2} \sqrt {2}\, \sqrt {c_{1}}}+2 \,{\mathrm e}^{c_{2} \sqrt {2}\, \sqrt {c_{1}}}+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.
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 )+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 { -y\,\mathop {\mathrm {d}y}}\\ &= -\frac {y^{2}}{2}+c_{1} \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(y=5\) and \(p=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 0 = -\frac {25}{2}+c_{1} \end {align*}
The solutions are \begin {align*} c_{1} = {\frac {25}{2}} \end {align*}
Trying the constant \begin {align*} c_{1} = {\frac {25}{2}} \end {align*}
Substituting this in the general solution gives \begin {align*} p \left (y \right )&=-\frac {y^{2}}{2}+\frac {25}{2} \end {align*}
The constant \(c_{1} = {\frac {25}{2}}\) 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*} y^{\prime } = -\frac {y^{2}}{2}+\frac {25}{2} \end {align*}
Since ode has form \(y^{\prime }= f(y)\) and initial conditions \(y = 5\) is verified to satisfy the ode, then the solution is \begin {align*} y&=y_0 \\ &=5 \end {align*}
Initial conditions are used to solve for the constants of integration.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= 5 \\ \end{align*}
Verification of solutions
\[ y = 5 \] Verified OK.
Writing the ode as \[ y^{\prime \prime }+y y^{\prime } = 0 \] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime \prime }+y y^{\prime }\right )d x &= 0 \\ \frac {y^{2}}{2}+y^{\prime } = c_{1} \end {align*}
Which is now solved for \(y\). Integrating both sides gives \begin {align*} \int \frac {1}{-\frac {y^{2}}{2}+c_{1}}d y &= x +c_{2}\\ \frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {y \sqrt {2}}{2 \sqrt {c_{1}}}\right )}{\sqrt {c_{1}}}&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=\tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}}{2}\right ) \sqrt {c_{1}}\, \sqrt {2} \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = \tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}}{2}\right ) \sqrt {c_{1}}\, \sqrt {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 \(y = 5\) and \(x = 0\) in the above gives \begin {align*} 5 = \frac {\left ({\mathrm e}^{c_{2} \sqrt {2}\, \sqrt {c_{1}}}-1\right ) \sqrt {2}\, \sqrt {c_{1}}}{{\mathrm e}^{c_{2} \sqrt {2}\, \sqrt {c_{1}}}+1}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = c_{1} \left (1-\tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}}{2}\right )^{2}\right ) \end {align*}
substituting \(y^{\prime } = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = \frac {4 \,{\mathrm e}^{c_{2} \sqrt {2}\, \sqrt {c_{1}}} c_{1}}{{\mathrm e}^{2 c_{2} \sqrt {2}\, \sqrt {c_{1}}}+2 \,{\mathrm e}^{c_{2} \sqrt {2}\, \sqrt {c_{1}}}+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.
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 &= 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 {y\,d y} + \int {0\,d x} &= c_{1} \end {align*}
Which results in \begin {align*} \frac {y^{2}}{2}+y^{\prime } = c_{1} \end {align*}
Which is now solved Integrating both sides gives \begin {align*} \int \frac {1}{-\frac {y^{2}}{2}+c_{1}}d y &= x +c_{2}\\ \frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {y \sqrt {2}}{2 \sqrt {c_{1}}}\right )}{\sqrt {c_{1}}}&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=\tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}}{2}\right ) \sqrt {c_{1}}\, \sqrt {2} \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = \tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}}{2}\right ) \sqrt {c_{1}}\, \sqrt {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 \(y = 5\) and \(x = 0\) in the above gives \begin {align*} 5 = \frac {\left ({\mathrm e}^{c_{2} \sqrt {2}\, \sqrt {c_{1}}}-1\right ) \sqrt {2}\, \sqrt {c_{1}}}{{\mathrm e}^{c_{2} \sqrt {2}\, \sqrt {c_{1}}}+1}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = c_{1} \left (1-\tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}}{2}\right )^{2}\right ) \end {align*}
substituting \(y^{\prime } = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = \frac {4 \,{\mathrm e}^{c_{2} \sqrt {2}\, \sqrt {c_{1}}} c_{1}}{{\mathrm e}^{2 c_{2} \sqrt {2}\, \sqrt {c_{1}}}+2 \,{\mathrm e}^{c_{2} \sqrt {2}\, \sqrt {c_{1}}}+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.
Verification of solutions N/A
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }+y y^{\prime }=0, y \left (0\right )=5, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0\right ] \\ \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} \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )+y u \left (y \right )=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=-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 -y d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & u \left (y \right )=-\frac {y^{2}}{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-\frac {y^{2}}{2}+c_{1} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-\frac {y^{2}}{2}+c_{1} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=-\frac {y^{2}}{2}+c_{1} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {y^{2}}{2}+c_{1} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{-\frac {y^{2}}{2}+c_{1}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{-\frac {y^{2}}{2}+c_{1}}d x =\int 1d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\sqrt {2}\, \mathrm {arctanh}\left (\frac {y \sqrt {2}}{2 \sqrt {c_{1}}}\right )}{\sqrt {c_{1}}}=x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}}{2}\right ) \sqrt {c_{1}}\, \sqrt {2} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=\tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}}{2}\right ) \sqrt {c_{1}}\, \sqrt {2} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=5 \\ {} & {} & 5=\tanh \left (\frac {c_{2} \sqrt {2}\, \sqrt {c_{1}}}{2}\right ) \sqrt {c_{1}}\, \sqrt {2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=c_{1} \left (1-\tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}}{2}\right )^{2}\right ) \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=c_{1} \left (1-\tanh \left (\frac {c_{2} \sqrt {2}\, \sqrt {c_{1}}}{2}\right )^{2}\right ) \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & \circ & \textrm {The solution does not satisfy the initial conditions}\hspace {3pt} \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*_b(_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: 0.032 (sec). Leaf size: 5
dsolve([diff(y(x),x$2)+y(x)*diff(y(x),x)=0,y(0) = 5, D(y)(0) = 0],y(x), singsol=all)
\[ y \left (x \right ) = 5 \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[{y''[x]+y[x]*y'[x]==0,{y[0]==5,y'[0]==0}},y[x],x,IncludeSingularSolutions -> True]
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