2.5 problem 5

Internal problem ID [8094]
Book : Second order enumerated odes
Section : section 2
Problem number : 5
Date solved : Monday, October 21, 2024 at 04:49:28 PM
CAS classification : [[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]

Solve

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

2.5.1 Solved as second order missing x ode

Time used: 1.063 (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 )^{2} \left (\frac {d}{d y}p \left (y \right )\right )+y^{2} = 0 \end{align*}

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

The ode \(p^{\prime } = -\frac {y^{2}}{p^{2}}\) is separable as it can be written as

\begin{align*} p^{\prime }&= -\frac {y^{2}}{p^{2}}\\ &= f(y) g(p) \end{align*}

Where

\begin{align*} f(y) &= -y^{2}\\ g(p) &= \frac {1}{p^{2}} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(p)} \,dp} &= \int { f(y) \,dy}\\ \int { p^{2}\,dp} &= \int { -y^{2} \,dy}\\ \frac {p^{3}}{3}&=-\frac {y^{3}}{3}+c_1 \end{align*}

Solving for \(p\) from the above solution(s) gives (after possible removing of solutions that do not verify)

\begin{align*} p&=\left (-y^{3}+3 c_1 \right )^{{1}/{3}}\\ p&=-\frac {\left (-y^{3}+3 c_1 \right )^{{1}/{3}}}{2}-\frac {i \sqrt {3}\, \left (-y^{3}+3 c_1 \right )^{{1}/{3}}}{2}\\ p&=-\frac {\left (-y^{3}+3 c_1 \right )^{{1}/{3}}}{2}+\frac {i \sqrt {3}\, \left (-y^{3}+3 c_1 \right )^{{1}/{3}}}{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 } = \left (-y^{3}+3 c_1 \right )^{{1}/{3}} \end{align*}

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}\frac {1}{\left (-\tau ^{3}+3 c_1 \right )^{{1}/{3}}}d \tau = x +c_2 \]

Singular solutions are found by solving

\begin{align*} \left (-y^{3}+3 c_1 \right )^{{1}/{3}}&= 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 = 3^{{1}/{3}} c_1^{{1}/{3}}\\ y = -\frac {3^{{1}/{3}} c_1^{{1}/{3}}}{2}+\frac {i 3^{{5}/{6}} c_1^{{1}/{3}}}{2}\\ y = -\frac {3^{{1}/{3}} c_1^{{1}/{3}}}{2}-\frac {i 3^{{5}/{6}} c_1^{{1}/{3}}}{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 {\left (-y^{3}+3 c_1 \right )^{{1}/{3}}}{2}-\frac {i \sqrt {3}\, \left (-y^{3}+3 c_1 \right )^{{1}/{3}}}{2} \end{align*}

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}\frac {1}{-\frac {\left (-\tau ^{3}+3 c_1 \right )^{{1}/{3}}}{2}-\frac {i \sqrt {3}\, \left (-\tau ^{3}+3 c_1 \right )^{{1}/{3}}}{2}}d \tau = x +c_3 \]

Singular solutions are found by solving

\begin{align*} -\frac {\left (-y^{3}+3 c_1 \right )^{{1}/{3}}}{2}-\frac {i \sqrt {3}\, \left (-y^{3}+3 c_1 \right )^{{1}/{3}}}{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 = 3^{{1}/{3}} c_1^{{1}/{3}}\\ y = -\frac {3^{{1}/{3}} c_1^{{1}/{3}}}{2}+\frac {i 3^{{5}/{6}} c_1^{{1}/{3}}}{2}\\ y = -\frac {3^{{1}/{3}} c_1^{{1}/{3}}}{2}-\frac {i 3^{{5}/{6}} c_1^{{1}/{3}}}{2} \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 } = -\frac {\left (-y^{3}+3 c_1 \right )^{{1}/{3}}}{2}+\frac {i \sqrt {3}\, \left (-y^{3}+3 c_1 \right )^{{1}/{3}}}{2} \end{align*}

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}\frac {1}{-\frac {\left (-\tau ^{3}+3 c_1 \right )^{{1}/{3}}}{2}+\frac {i \sqrt {3}\, \left (-\tau ^{3}+3 c_1 \right )^{{1}/{3}}}{2}}d \tau = x +c_4 \]

Singular solutions are found by solving

\begin{align*} -\frac {\left (-y^{3}+3 c_1 \right )^{{1}/{3}}}{2}+\frac {i \sqrt {3}\, \left (-y^{3}+3 c_1 \right )^{{1}/{3}}}{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 = 3^{{1}/{3}} c_1^{{1}/{3}}\\ y = -\frac {3^{{1}/{3}} c_1^{{1}/{3}}}{2}+\frac {i 3^{{5}/{6}} c_1^{{1}/{3}}}{2}\\ y = -\frac {3^{{1}/{3}} c_1^{{1}/{3}}}{2}-\frac {i 3^{{5}/{6}} c_1^{{1}/{3}}}{2} \end{align*}

Will add steps showing solving for IC soon.

The solution

\[ y = 3^{{1}/{3}} c_1^{{1}/{3}} \]

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

\[ y = -\frac {3^{{1}/{3}} c_1^{{1}/{3}}}{2}-\frac {i 3^{{5}/{6}} c_1^{{1}/{3}}}{2} \]

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

\[ y = -\frac {3^{{1}/{3}} c_1^{{1}/{3}}}{2}+\frac {i 3^{{5}/{6}} c_1^{{1}/{3}}}{2} \]

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

2.5.2 Solved as second order can be made integrable

Time used: 0.542 (sec)

Multiplying the ode by \(y^{\prime }\) gives

\[ y^{2} y^{\prime }+{y^{\prime }}^{2} y^{\prime \prime } = 0 \]

Integrating the above w.r.t \(x\) gives

\begin{align*} \int \left (y^{2} y^{\prime }+{y^{\prime }}^{2} y^{\prime \prime }\right )d x &= 0 \\ \frac {y^{3}}{3}+\frac {{y^{\prime }}^{3}}{3} &= c_1 \end{align*}

Which is now solved for \(y\). Solving for the derivative gives these ODE’s to solve

\begin{align*} \tag{1} y^{\prime }&=\left (-y^{3}+3 c_1 \right )^{{1}/{3}} \\ \tag{2} y^{\prime }&=-\frac {\left (-y^{3}+3 c_1 \right )^{{1}/{3}}}{2}-\frac {i \sqrt {3}\, \left (-y^{3}+3 c_1 \right )^{{1}/{3}}}{2} \\ \tag{3} y^{\prime }&=-\frac {\left (-y^{3}+3 c_1 \right )^{{1}/{3}}}{2}+\frac {i \sqrt {3}\, \left (-y^{3}+3 c_1 \right )^{{1}/{3}}}{2} \\ \end{align*}

Now each of the above is solved separately.

Solving Eq. (1)

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}\frac {1}{\left (-\tau ^{3}+3 c_1 \right )^{{1}/{3}}}d \tau = x +c_2 \]

Singular solutions are found by solving

\begin{align*} \left (-y^{3}+3 c_1 \right )^{{1}/{3}}&= 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 = 3^{{1}/{3}} c_1^{{1}/{3}}\\ y = -\frac {3^{{1}/{3}} c_1^{{1}/{3}}}{2}+\frac {i 3^{{5}/{6}} c_1^{{1}/{3}}}{2}\\ y = -\frac {3^{{1}/{3}} c_1^{{1}/{3}}}{2}-\frac {i 3^{{5}/{6}} c_1^{{1}/{3}}}{2} \end{align*}

Solving Eq. (2)

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}\frac {1}{-\frac {\left (-\tau ^{3}+3 c_1 \right )^{{1}/{3}}}{2}-\frac {i \sqrt {3}\, \left (-\tau ^{3}+3 c_1 \right )^{{1}/{3}}}{2}}d \tau = x +c_3 \]

Singular solutions are found by solving

\begin{align*} -\frac {\left (-y^{3}+3 c_1 \right )^{{1}/{3}}}{2}-\frac {i \sqrt {3}\, \left (-y^{3}+3 c_1 \right )^{{1}/{3}}}{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 = 3^{{1}/{3}} c_1^{{1}/{3}}\\ y = -\frac {3^{{1}/{3}} c_1^{{1}/{3}}}{2}+\frac {i 3^{{5}/{6}} c_1^{{1}/{3}}}{2}\\ y = -\frac {3^{{1}/{3}} c_1^{{1}/{3}}}{2}-\frac {i 3^{{5}/{6}} c_1^{{1}/{3}}}{2} \end{align*}

Solving Eq. (3)

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}\frac {1}{-\frac {\left (-\tau ^{3}+3 c_1 \right )^{{1}/{3}}}{2}+\frac {i \sqrt {3}\, \left (-\tau ^{3}+3 c_1 \right )^{{1}/{3}}}{2}}d \tau = x +c_4 \]

Singular solutions are found by solving

\begin{align*} -\frac {\left (-y^{3}+3 c_1 \right )^{{1}/{3}}}{2}+\frac {i \sqrt {3}\, \left (-y^{3}+3 c_1 \right )^{{1}/{3}}}{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 = 3^{{1}/{3}} c_1^{{1}/{3}}\\ y = -\frac {3^{{1}/{3}} c_1^{{1}/{3}}}{2}+\frac {i 3^{{5}/{6}} c_1^{{1}/{3}}}{2}\\ y = -\frac {3^{{1}/{3}} c_1^{{1}/{3}}}{2}-\frac {i 3^{{5}/{6}} c_1^{{1}/{3}}}{2} \end{align*}

Will add steps showing solving for IC soon.

The solution

\[ y = 3^{{1}/{3}} c_1^{{1}/{3}} \]

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

\[ y = -\frac {3^{{1}/{3}} c_1^{{1}/{3}}}{2}-\frac {i 3^{{5}/{6}} c_1^{{1}/{3}}}{2} \]

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

\[ y = -\frac {3^{{1}/{3}} c_1^{{1}/{3}}}{2}+\frac {i 3^{{5}/{6}} c_1^{{1}/{3}}}{2} \]

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

2.5.3 Maple step by step solution

2.5.4 Maple trace

Methods for second order ODEs:
 

2.5.5 Maple dsolve solution

Solving time : 0.029 (sec)
Leaf size : 61

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

\begin{align*} y &= 0 \\ y &= {\mathrm e}^{\frac {\sqrt {3}\, \left (\int \tan \left (\operatorname {RootOf}\left (-\sqrt {3}\, \ln \left (\cos \left (\textit {\_Z} \right )^{2}\right )-2 \sqrt {3}\, \ln \left (\tan \left (\textit {\_Z} \right )+\sqrt {3}\right )+6 \sqrt {3}\, c_1 +6 \sqrt {3}\, x +6 \textit {\_Z} \right )\right )d x \right )}{2}+c_2 +\frac {x}{2}} \\ \end{align*}

2.5.6 Mathematica DSolve solution

Solving time : 1.855 (sec)
Leaf size : 180

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

\[ y(x)\to \frac {c_2}{\sqrt [3]{1+\text {InverseFunction}\left [\frac {1}{6} \log \left (\text {$\#$1}^2-\text {$\#$1}+1\right )+\frac {\arctan \left (\frac {2 \text {$\#$1}-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \log (\text {$\#$1}+1)\&\right ][-x+c_1]} \sqrt [3]{\text {InverseFunction}\left [\frac {1}{6} \log \left (\text {$\#$1}^2-\text {$\#$1}+1\right )+\frac {\arctan \left (\frac {2 \text {$\#$1}-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \log (\text {$\#$1}+1)\&\right ][-x+c_1]{}^2-\text {InverseFunction}\left [\frac {1}{6} \log \left (\text {$\#$1}^2-\text {$\#$1}+1\right )+\frac {\arctan \left (\frac {2 \text {$\#$1}-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \log (\text {$\#$1}+1)\&\right ][-x+c_1]+1}} \]