Internal problem ID [6247]
Internal file name [OUTPUT/5495_Sunday_June_05_2022_03_41_33_PM_45489118/index.tex]
Book: Differential Equations: Theory, Technique, and Practice by George Simmons, Steven
Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 1. What is a differential equation. Section 1.9. Reduction of Order. Page
38
Problem number: 3(a).
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]]
\[ \boxed {y^{\prime \prime }-{y^{\prime }}^{2}=1} \]
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 )-1-p \left (x \right )^{2} = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. Integrating both sides gives \begin {align*} \int \frac {1}{p^{2}+1}d p &= x +c_{1}\\ \arctan \left (p \right )&=x +c_{1} \end {align*}
Solving for \(p\) gives these solutions \begin {align*} p_1&=\tan \left (x +c_{1} \right ) \end {align*}
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = \tan \left (x +c_{1} \right ) \end {align*}
Integrating both sides gives \begin {align*} y &= \int { \tan \left (x +c_{1} \right )\,\mathop {\mathrm {d}x}}\\ &= \frac {\ln \left (\tan \left (x +c_{1} \right )^{2}+1\right )}{2}+c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\ln \left (\tan \left (x +c_{1} \right )^{2}+1\right )}{2}+c_{2} \\ \end{align*}
Verification of solutions
\[ y = \frac {\ln \left (\tan \left (x +c_{1} \right )^{2}+1\right )}{2}+c_{2} \] 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 )-p \left (y \right )^{2} = 1 \end {align*}
Which is now solved as first order ode for \(p(y)\). Integrating both sides gives \begin {align*} \int \frac {p}{p^{2}+1}d p &= y +c_{1}\\ \frac {\ln \left (p^{2}+1\right )}{2}&=y +c_{1} \end {align*}
Solving for \(p\) gives these solutions \begin {align*} p_1&=\sqrt {-1+{\mathrm e}^{2 y +2 c_{1}}}\\ &=\sqrt {{\mathrm e}^{2 y} c_{1}^{2}-1}\\ p_2&=-\sqrt {-1+{\mathrm e}^{2 y +2 c_{1}}}\\ &=-\sqrt {{\mathrm e}^{2 y} c_{1}^{2}-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 } = \sqrt {{\mathrm e}^{2 y} c_{1}^{2}-1} \end {align*}
Integrating both sides gives \begin{align*} \int \frac {1}{\sqrt {{\mathrm e}^{2 y} c_{1}^{2}-1}}d y &= \int d x \\ -\arctan \left (\frac {1}{\sqrt {{\mathrm e}^{2 y} c_{1}^{2}-1}}\right )&=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 {{\mathrm e}^{2 y} c_{1}^{2}-1} \end {align*}
Integrating both sides gives \begin{align*} \int -\frac {1}{\sqrt {{\mathrm e}^{2 y} c_{1}^{2}-1}}d y &= \int d x \\ \arctan \left (\frac {1}{\sqrt {{\mathrm e}^{2 y} c_{1}^{2}-1}}\right )&=x +c_{3} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} -\arctan \left (\frac {1}{\sqrt {{\mathrm e}^{2 y} c_{1}^{2}-1}}\right ) &= x +c_{2} \\ \tag{2} \arctan \left (\frac {1}{\sqrt {{\mathrm e}^{2 y} c_{1}^{2}-1}}\right ) &= x +c_{3} \\ \end{align*}
Verification of solutions
\[ -\arctan \left (\frac {1}{\sqrt {{\mathrm e}^{2 y} c_{1}^{2}-1}}\right ) = x +c_{2} \] Verified OK.
\[ \arctan \left (\frac {1}{\sqrt {{\mathrm e}^{2 y} c_{1}^{2}-1}}\right ) = x +c_{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime }-{y^{\prime }}^{2}=1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime }\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )-u \left (x \right )^{2}=1 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=u \left (x \right )^{2}+1 \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{u \left (x \right )^{2}+1}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {u^{\prime }\left (x \right )}{u \left (x \right )^{2}+1}d x =\int 1d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \arctan \left (u \left (x \right )\right )=x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\tan \left (x +c_{1} \right ) \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\tan \left (x +c_{1} \right ) \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=\tan \left (x +c_{1} \right ) \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int \tan \left (x +c_{1} \right )d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=\frac {\ln \left (\tan \left (x +c_{1} \right )^{2}+1\right )}{2}+c_{2} \end {array} \]
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) <- 2nd order, 2 integrating factors of the form mu(x,y) successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 17
dsolve(diff(y(x),x$2)=1+(diff(y(x),x))^2,y(x), singsol=all)
\[ y \left (x \right ) = -\ln \left (-\cos \left (x \right ) c_{2} +c_{1} \sin \left (x \right )\right ) \]
✓ Solution by Mathematica
Time used: 1.938 (sec). Leaf size: 16
DSolve[y''[x]==1+(y'[x])^2,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_2-\log (\cos (x+c_1)) \]