2.1.23 problem 23

Internal problem ID [8770]
Book : Second order enumerated odes
Section : section 1
Problem number : 23
Date solved : Thursday, December 12, 2024 at 09:43:13 AM
CAS classification : [[_2nd_order, _missing_x]]

Solve

\begin{align*} y^{\prime \prime }+{y^{\prime }}^{2}+y&=0 \end{align*}

Solved as second order missing x ode

Time used: 0.788 (sec)

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 \end{align*}

Then

\begin{align*} y'' &= \frac {dp}{dx}\\ &= \frac {dp}{dy}\frac {dy}{dx}\\ &= 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}+y = 0 \end{align*}

Which is now solved as first order ode for \(p(y)\).

In canonical form, the ODE is

\begin{align*} p' &= F(y,p)\\ &= -\frac {p^{2}+y}{p} \end{align*}

This is a Bernoulli ODE.

\[ p' = \left (-1\right ) p + \left (-y\right )\frac {1}{p} \tag {1} \]

The standard Bernoulli ODE has the form

\[ p' = f_0(y)p+f_1(y)p^n \tag {2} \]

Comparing this to (1) shows that

\begin{align*} f_0 &=-1\\ f_1 &=-y \end{align*}

The first step is to divide the above equation by \(p^n \) which gives

\[ \frac {p'}{p^n} = f_0(y) p^{1-n} +f_1(y) \tag {3} \]

The next step is use the substitution \(v = p^{1-n}\) in equation (3) which generates a new ODE in \(v \left (y \right )\) which will be linear and can be easily solved using an integrating factor. Backsubstitution then gives the solution \(p(y)\) which is what we want.

This method is now applied to the ODE at hand. Comparing the ODE (1) With (2) Shows that

\begin{align*} f_0(y)&=-1\\ f_1(y)&=-y\\ n &=-1 \end{align*}

Dividing both sides of ODE (1) by \(p^n=\frac {1}{p}\) gives

\begin{align*} p'p &= -p^{2} -y \tag {4} \end{align*}

Let

\begin{align*} v &= p^{1-n} \\ &= p^{2} \tag {5} \end{align*}

Taking derivative of equation (5) w.r.t \(y\) gives

\begin{align*} v' &= 2 pp' \tag {6} \end{align*}

Substituting equations (5) and (6) into equation (4) gives

\begin{align*} \frac {v^{\prime }\left (y \right )}{2}&= -v \left (y \right )-y\\ v' &= -2 v -2 y \tag {7} \end{align*}

The above now is a linear ODE in \(v \left (y \right )\) which is now solved.

In canonical form a linear first order is

\begin{align*} v^{\prime }\left (y \right ) + q(y)v \left (y \right ) &= p(y) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(y) &=2\\ p(y) &=-2 y \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dy}}\\ &= {\mathrm e}^{\int 2d y}\\ &= {\mathrm e}^{2 y} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}y}}\left ( \mu v\right ) &= \mu p \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}y}}\left ( \mu v\right ) &= \left (\mu \right ) \left (-2 y\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}y}} \left (v \,{\mathrm e}^{2 y}\right ) &= \left ({\mathrm e}^{2 y}\right ) \left (-2 y\right ) \\ \mathrm {d} \left (v \,{\mathrm e}^{2 y}\right ) &= \left (-2 y \,{\mathrm e}^{2 y}\right )\, \mathrm {d} y \\ \end{align*}

Integrating gives

\begin{align*} v \,{\mathrm e}^{2 y}&= \int {-2 y \,{\mathrm e}^{2 y} \,dy} \\ &=-\frac {\left (2 y -1\right ) {\mathrm e}^{2 y}}{2} + c_1 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{2 y}\) gives the final solution

\[ v \left (y \right ) = -y +\frac {1}{2}+c_1 \,{\mathrm e}^{-2 y} \]

The substitution \(v = p^{1-n}\) is now used to convert the above solution back to \(p\) which results in

\[ p^{2} = -y +\frac {1}{2}+c_1 \,{\mathrm e}^{-2 y} \]

Solving for \(p\) gives

\begin{align*} p &= -\frac {\sqrt {2+4 c_1 \,{\mathrm e}^{-2 y}-4 y}}{2} \\ p &= \frac {\sqrt {2+4 c_1 \,{\mathrm e}^{-2 y}-4 y}}{2} \\ \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 } = -\frac {\sqrt {2+4 c_1 \,{\mathrm e}^{-2 y}-4 y}}{2} \end{align*}

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

\[ \int _{}^{y}-\frac {2}{\sqrt {2+4 c_1 \,{\mathrm e}^{-2 \tau }-4 \tau }}d \tau = x +c_2 \]

Singular solutions are found by solving

\begin{align*} -\frac {\sqrt {2+4 c_1 \,{\mathrm e}^{-2 y}-4 y}}{2}&= 0 \end{align*}

for \(y\). 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*} y = \frac {\operatorname {LambertW}\left (2 c_1 \,{\mathrm e}^{-1}\right )}{2}+\frac {1}{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 } = \frac {\sqrt {2+4 c_1 \,{\mathrm e}^{-2 y}-4 y}}{2} \end{align*}

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

\[ \int _{}^{y}\frac {2}{\sqrt {2+4 c_1 \,{\mathrm e}^{-2 \tau }-4 \tau }}d \tau = x +c_3 \]

Singular solutions are found by solving

\begin{align*} \frac {\sqrt {2+4 c_1 \,{\mathrm e}^{-2 y}-4 y}}{2}&= 0 \end{align*}

for \(y\). 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*} y = \frac {\operatorname {LambertW}\left (2 c_1 \,{\mathrm e}^{-1}\right )}{2}+\frac {1}{2} \end{align*}

Will add steps showing solving for IC soon.

The solution

\[ y = \frac {\operatorname {LambertW}\left (2 c_1 \,{\mathrm e}^{-1}\right )}{2}+\frac {1}{2} \]

was found not to satisfy the ode or the IC. Hence it is removed.

Summary of solutions found

\begin{align*} \int _{}^{y}-\frac {2}{\sqrt {2+4 c_1 \,{\mathrm e}^{-2 \tau }-4 \tau }}d \tau &= x +c_2 \\ \int _{}^{y}\frac {2}{\sqrt {2+4 c_1 \,{\mathrm e}^{-2 \tau }-4 \tau }}d \tau &= x +c_3 \\ \end{align*}

Maple step by step solution

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 
`, `-> Computing symmetries using: way = exp_sym 
-> Calling odsolve with the ODE`, (diff(_b(_a), _a))*_b(_a)+_b(_a)^2+_a = 0, _b(_a)`   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying 1st order linear 
   trying Bernoulli 
   <- Bernoulli successful 
<- differential order: 2; canonical coordinates successful 
<- differential order 2; missing variables successful`
 

Maple dsolve solution

Solving time : 0.028 (sec)
Leaf size : 61

dsolve(diff(diff(y(x),x),x)+diff(y(x),x)^2+y(x) = 0, 
       y(x),singsol=all)
 

\begin{align*} -2 \left (\int _{}^{y}\frac {1}{\sqrt {2+4 \,{\mathrm e}^{-2 \textit {\_a}} c_{1} -4 \textit {\_a}}}d \textit {\_a} \right )-x -c_{2} &= 0 \\ 2 \left (\int _{}^{y}\frac {1}{\sqrt {2+4 \,{\mathrm e}^{-2 \textit {\_a}} c_{1} -4 \textit {\_a}}}d \textit {\_a} \right )-x -c_{2} &= 0 \\ \end{align*}

Mathematica DSolve solution

Solving time : 0.746 (sec)
Leaf size : 272

DSolve[{D[y[x],{x,2}]+(D[y[x],x])^2+y[x]==0,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}-\frac {\sqrt {2}}{\sqrt {2 e^{-2 K[1]} c_1-2 K[1]+1}}dK[1]\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {\sqrt {2}}{\sqrt {2 e^{-2 K[2]} c_1-2 K[2]+1}}dK[2]\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}-\frac {\sqrt {2}}{\sqrt {2 e^{-2 K[1]} (-c_1)-2 K[1]+1}}dK[1]\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}-\frac {\sqrt {2}}{\sqrt {2 e^{-2 K[1]} c_1-2 K[1]+1}}dK[1]\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {\sqrt {2}}{\sqrt {2 e^{-2 K[2]} (-c_1)-2 K[2]+1}}dK[2]\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {\sqrt {2}}{\sqrt {2 e^{-2 K[2]} c_1-2 K[2]+1}}dK[2]\&\right ][x+c_2] \\ \end{align*}