Internal problem ID [2298]
Internal file name [OUTPUT/2298_Tuesday_February_27_2024_08_24_14_AM_17887916/index.tex]
Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath.
Boston. 1964
Section: Exercise 35, page 157
Problem number: 26.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_x", "second_order_ode_missing_y"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {y^{\prime \prime }+2 {y^{\prime }}^{2}=2} \]
This is second order ode with missing dependent variable \(y\). Let \begin {align*} p(x) &= y^{\prime } \end {align*}
Then \begin {align*} p'(x) &= y^{\prime \prime } \end {align*}
Hence the ode becomes \begin {align*} p^{\prime }\left (x \right )+2 p \left (x \right )^{2}-2 = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. Integrating both sides gives \begin {align*} \int \frac {1}{-2 p^{2}+2}d p &= x +c_{1}\\ \frac {\operatorname {arctanh}\left (p \right )}{2}&=x +c_{1} \end {align*}
Solving for \(p\) gives these solutions \begin {align*} p_1&=\tanh \left (2 x +2 c_{1} \right ) \end {align*}
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = \tanh \left (2 x +2 c_{1} \right ) \end {align*}
Integrating both sides gives \begin {align*} y &= \int { \tanh \left (2 x +2 c_{1} \right )\,\mathop {\mathrm {d}x}}\\ &= \frac {\ln \left (\cosh \left (2 x +2 c_{1} \right )\right )}{2}+c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\ln \left (\cosh \left (2 x +2 c_{1} \right )\right )}{2}+c_{2} \\ \end{align*}
Verification of solutions
\[ y = \frac {\ln \left (\cosh \left (2 x +2 c_{1} \right )\right )}{2}+c_{2} \] 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 p \left (y \right )^{2} = 2 \end {align*}
Which is now solved as first order ode for \(p(y)\). Integrating both sides gives \begin {align*} \int -\frac {p}{2 \left (p^{2}-1\right )}d p &= y +c_{1}\\ -\frac {\ln \left (p^{2}-1\right )}{4}&=y +c_{1} \end {align*}
Solving for \(p\) gives these solutions \begin {align*} p_1&=\sqrt {1+{\mathrm e}^{-4 y -4 c_{1}}}\\ &=\sqrt {1+\frac {{\mathrm e}^{-4 y}}{c_{1}^{4}}}\\ p_2&=-\sqrt {1+{\mathrm e}^{-4 y -4 c_{1}}}\\ &=-\sqrt {1+\frac {{\mathrm e}^{-4 y}}{c_{1}^{4}}} \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 } = \sqrt {1+\frac {{\mathrm e}^{-4 y}}{c_{1}^{4}}} \end {align*}
Integrating both sides gives \begin {align*} \int \frac {1}{\sqrt {1+\frac {{\mathrm e}^{-4 y}}{c_{1}^{4}}}}d y &= \int {dx}\\ \int _{}^{y}\frac {1}{\sqrt {1+\frac {{\mathrm e}^{-4 \textit {\_a}}}{c_{1}^{4}}}}d \textit {\_a}&= x +c_{2} \end {align*}
For solution (2) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} y^{\prime } = -\sqrt {1+\frac {{\mathrm e}^{-4 y}}{c_{1}^{4}}} \end {align*}
Integrating both sides gives \begin{align*} \int -\frac {1}{\sqrt {1+\frac {{\mathrm e}^{-4 y}}{c_{1}^{4}}}}d y &= \int d x \\ -\frac {{\mathrm e}^{-2 y} \sqrt {\frac {{\mathrm e}^{4 y} c_{1}^{4}+1}{c_{1}^{4}}}\, \ln \left ({\mathrm e}^{2 y}+\sqrt {\frac {{\mathrm e}^{4 y} c_{1}^{4}+1}{c_{1}^{4}}}\right )}{2 \sqrt {\frac {\left ({\mathrm e}^{4 y} c_{1}^{4}+1\right ) {\mathrm e}^{-4 y}}{c_{1}^{4}}}}&=x +c_{3} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}\frac {1}{\sqrt {1+\frac {{\mathrm e}^{-4 \textit {\_a}}}{c_{1}^{4}}}}d \textit {\_a} &= x +c_{2} \\ \tag{2} -\frac {{\mathrm e}^{-2 y} \sqrt {\frac {{\mathrm e}^{4 y} c_{1}^{4}+1}{c_{1}^{4}}}\, \ln \left ({\mathrm e}^{2 y}+\sqrt {\frac {{\mathrm e}^{4 y} c_{1}^{4}+1}{c_{1}^{4}}}\right )}{2 \sqrt {\frac {\left ({\mathrm e}^{4 y} c_{1}^{4}+1\right ) {\mathrm e}^{-4 y}}{c_{1}^{4}}}} &= x +c_{3} \\ \end{align*}
Verification of solutions
\[ \int _{}^{y}\frac {1}{\sqrt {1+\frac {{\mathrm e}^{-4 \textit {\_a}}}{c_{1}^{4}}}}d \textit {\_a} = x +c_{2} \] Verified OK.
\[ -\frac {{\mathrm e}^{-2 y} \sqrt {\frac {{\mathrm e}^{4 y} c_{1}^{4}+1}{c_{1}^{4}}}\, \ln \left ({\mathrm e}^{2 y}+\sqrt {\frac {{\mathrm e}^{4 y} c_{1}^{4}+1}{c_{1}^{4}}}\right )}{2 \sqrt {\frac {\left ({\mathrm e}^{4 y} c_{1}^{4}+1\right ) {\mathrm e}^{-4 y}}{c_{1}^{4}}}} = x +c_{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime }+2 {y^{\prime }}^{2}=2 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime }\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )+2 u \left (x \right )^{2}=2 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=-2 u \left (x \right )^{2}+2 \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{-2 u \left (x \right )^{2}+2}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {u^{\prime }\left (x \right )}{-2 u \left (x \right )^{2}+2}d x =\int 1d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\mathrm {arctanh}\left (u \left (x \right )\right )}{2}=x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\tanh \left (2 x +2 c_{1} \right ) \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\tanh \left (2 x +2 c_{1} \right ) \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=\tanh \left (2 x +2 c_{1} \right ) \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int \tanh \left (2 x +2 c_{1} \right )d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=\frac {\ln \left (\cosh \left (2 x +2 c_{1} \right )\right )}{2}+c_{2} \end {array} \]
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 a quadrature checking if the LODE has constant coefficients <- constant coefficients successful <- 2nd order, 2 integrating factors of the form mu(x,y) successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 24
dsolve(diff(y(x),x$2)+2*diff(y(x),x)^2=2,y(x), singsol=all)
\[ y \left (x \right ) = -x -\frac {\ln \left (2\right )}{2}+\frac {\ln \left (-{\mathrm e}^{4 x} c_{1} +c_{2} \right )}{2} \]
✓ Solution by Mathematica
Time used: 0.389 (sec). Leaf size: 62
DSolve[y''[x]+2*y'[x]^2==2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {1}{2} \left (-\log \left (e^{2 x}\right )+\log \left (e^{4 x}+e^{2 c_1}\right )+2 c_2\right ) \\ y(x)\to \frac {1}{2} \left (-\log \left (e^{2 x}\right )+\log \left (e^{4 x}\right )+2 c_2\right ) \\ \end{align*}