2.1.16 problem 16

Internal problem ID [8499]
Book : Second order enumerated odes
Section : section 1
Problem number : 16
Date solved : Sunday, November 10, 2024 at 03:55:10 AM
CAS classification : [[_2nd_order, _missing_x]]

Solve

\begin{align*} {y^{\prime \prime }}^{2}+y^{\prime }&=1 \end{align*}

Solved as second order missing y ode

Time used: 0.589 (sec)

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 )-1 = 0 \end{align*}

Which is now solve for \(p(x)\) as first order ode. Solving for the derivative gives these ODE’s to solve

\begin{align*} \tag{1} p^{\prime }\left (x \right )&=\sqrt {1-p \left (x \right )} \\ \tag{2} p^{\prime }\left (x \right )&=-\sqrt {1-p \left (x \right )} \\ \end{align*}

Now each of the above is solved separately.

Solving Eq. (1)

Integrating gives

\begin{align*} \int \frac {1}{\sqrt {1-p}}d p &= dx\\ -2 \sqrt {1-p}&= x +c_1 \end{align*}

Singular solutions are found by solving

\begin{align*} \sqrt {1-p}&= 0 \end{align*}

for \(p \left (x \right )\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} p \left (x \right ) = 1 \end{align*}

Solving Eq. (2)

Integrating gives

\begin{align*} \int -\frac {1}{\sqrt {1-p}}d p &= dx\\ 2 \sqrt {1-p}&= x +c_2 \end{align*}

Singular solutions are found by solving

\begin{align*} -\sqrt {1-p}&= 0 \end{align*}

for \(p \left (x \right )\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} p \left (x \right ) = 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*} -2 \sqrt {1-y^{\prime }} = x +c_1 \end{align*}

Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).

\begin{align*} \int {dy} &= \int {-\frac {1}{4} c_1^{2}-\frac {1}{2} c_1 x -\frac {1}{4} x^{2}+1\, dx}\\ y &= -\frac {x^{3}}{12}-\frac {c_1 \,x^{2}}{4}-\frac {\left (c_1 +2\right ) \left (c_1 -2\right ) x}{4} + c_3 \end{align*}

\begin{align*} y&= -\frac {1}{12} x^{3}-\frac {1}{4} c_1 \,x^{2}-\frac {1}{4} c_1^{2} x +x +c_3 \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*} 2 \sqrt {1-y^{\prime }} = x +c_2 \end{align*}

Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).

\begin{align*} \int {dy} &= \int {-\frac {1}{4} c_2^{2}-\frac {1}{2} c_2 x -\frac {1}{4} x^{2}+1\, dx}\\ y &= -\frac {x^{3}}{12}-\frac {c_2 \,x^{2}}{4}-\frac {\left (c_2 +2\right ) \left (c_2 -2\right ) x}{4} + c_4 \end{align*}

\begin{align*} y&= -\frac {1}{12} x^{3}-\frac {1}{4} c_2 \,x^{2}-\frac {1}{4} c_2^{2} x +x +c_4 \end{align*}

For solution (3) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = 1 \end{align*}

Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).

\begin{align*} \int {dy} &= \int {1\, dx}\\ y &= x + c_5 \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= x +c_5 \\ y &= -\frac {1}{12} x^{3}-\frac {1}{4} c_1 \,x^{2}-\frac {1}{4} c_1^{2} x +x +c_3 \\ y &= -\frac {1}{12} x^{3}-\frac {1}{4} c_2 \,x^{2}-\frac {1}{4} c_2^{2} x +x +c_4 \\ \end{align*}

Maple step by step solution

Maple trace

`Methods for second order ODEs: 
   *** Sublevel 2 *** 
   Methods for second order ODEs: 
   Successful isolation of d^2y/dx^2: 2 solutions were found. Trying to solve each resulting ODE. 
      *** Sublevel 3 *** 
      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 
      <- 2nd order ODE linearizable_by_differentiation successful 
   ------------------- 
   * Tackling next ODE. 
      *** Sublevel 3 *** 
      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 
      <- 2nd order ODE linearizable_by_differentiation successful 
-> Calling odsolve with the ODE`, diff(y(x), x) = 1, y(x), singsol = none`   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying 1st order linear 
   <- 1st order linear successful`
 

Maple dsolve solution

Solving time : 0.306 (sec)
Leaf size : 30

dsolve(diff(diff(y(x),x),x)^2+diff(y(x),x) = 1, 
       y(x),singsol=all)
 

\begin{align*} y &= x +c_{1} \\ y &= -\frac {1}{12} x^{3}+\frac {1}{2} c_{1} x^{2}-c_{1}^{2} x +x +c_{2} \\ \end{align*}

Mathematica DSolve solution

Solving time : 0.024 (sec)
Leaf size : 67

DSolve[{(D[y[x],{x,2}])^2+D[y[x],x]==1,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)\to -\frac {x^3}{12}-\frac {c_1 x^2}{4}+x-\frac {c_1{}^2 x}{4}+c_2 \\ y(x)\to -\frac {x^3}{12}+\frac {c_1 x^2}{4}+x-\frac {c_1{}^2 x}{4}+c_2 \\ \end{align*}