1.62 problem 62

Internal problem ID [7754]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 62
Date solved : Monday, October 21, 2024 at 04:02:47 PM
CAS classification : [[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]

Solve

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

1.62.1 Solved as second order missing x ode

Time used: 0.412 (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*} y p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right ) = 1 \end{align*}

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

The ode \(p^{\prime } = \frac {1}{p y}\) is separable as it can be written as

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

Where

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

Integrating gives

\begin{align*} \int { \frac {1}{g(p)} \,dp} &= \int { f(y) \,dy}\\ \int { p\,dp} &= \int { \frac {1}{y} \,dy}\\ \frac {p^{2}}{2}&=\ln \left (y \right )+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&=\sqrt {2 \ln \left (y \right )+2 c_1}\\ p&=-\sqrt {2 \ln \left (y \right )+2 c_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 {2 \ln \left (y\right )+2 c_1} \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

\[ \int _{}^{y}\frac {1}{\sqrt {2 \ln \left (\tau \right )+2 c_1}}d \tau = x +c_2 \]

Singular solutions are found by solving

\begin{align*} \sqrt {2 \ln \left (y \right )+2 c_1}&= 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 = {\mathrm e}^{-c_1} \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 {2 \ln \left (y\right )+2 c_1} \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

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

Singular solutions are found by solving

\begin{align*} -\sqrt {2 \ln \left (y \right )+2 c_1}&= 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 = {\mathrm e}^{-c_1} \end{align*}

Will add steps showing solving for IC soon.

The solution

\[ y = {\mathrm e}^{-c_1} \]

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

1.62.2 Maple step by step solution

1.62.3 Maple trace

Methods for second order ODEs:
 

1.62.4 Maple dsolve solution

Solving time : 0.020 (sec)
Leaf size : 51

dsolve(y(x)*diff(diff(y(x),x),x) = 1, 
       y(x),singsol=all)
 

\begin{align*} \int _{}^{y}\frac {1}{\sqrt {2 \ln \left (\textit {\_a} \right )-c_1}}d \textit {\_a} -x -c_2 &= 0 \\ -\left (\int _{}^{y}\frac {1}{\sqrt {2 \ln \left (\textit {\_a} \right )-c_1}}d \textit {\_a} \right )-x -c_2 &= 0 \\ \end{align*}

1.62.5 Mathematica DSolve solution

Solving time : 60.072 (sec)
Leaf size : 93

DSolve[{y[x]*D[y[x],{x,2}]==1,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)\to \exp \left (-\text {erf}^{-1}\left (-i \sqrt {\frac {2}{\pi }} \sqrt {e^{c_1} (x+c_2){}^2}\right ){}^2-\frac {c_1}{2}\right ) \\ y(x)\to \exp \left (-\text {erf}^{-1}\left (i \sqrt {\frac {2}{\pi }} \sqrt {e^{c_1} (x+c_2){}^2}\right ){}^2-\frac {c_1}{2}\right ) \\ \end{align*}