2.1.67 problem 67

Internal problem ID [8455]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 67
Date solved : Thursday, December 12, 2024 at 09:10:08 AM
CAS classification : [[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]

Solve

\begin{align*} 3 y y^{\prime \prime }+y&=5 \end{align*}

Solved as second order missing x ode

Time used: 483.685 (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*} 3 y p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+y = 5 \end{align*}

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

The ode \(p^{\prime } = -\frac {y -5}{3 p y}\) is separable as it can be written as

\begin{align*} p^{\prime }&= -\frac {y -5}{3 p y}\\ &= f(y) g(p) \end{align*}

Where

\begin{align*} f(y) &= -\frac {y -5}{3 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 {y -5}{3 y} \,dy}\\ \frac {p^{2}}{2}&=-\frac {y}{3}+\ln \left (y^{{5}/{3}}\right )+c_1 \end{align*}

Solving for \(p\) gives

\begin{align*} p &= -\frac {\sqrt {18 \ln \left (y^{{5}/{3}}\right )+18 c_1 -6 y}}{3} \\ p &= \frac {\sqrt {18 \ln \left (y^{{5}/{3}}\right )+18 c_1 -6 y}}{3} \\ \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 {18 \ln \left (y^{{5}/{3}}\right )+18 c_1 -6 y}}{3} \end{align*}

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

\[ \int _{}^{y}-\frac {3}{\sqrt {18 \ln \left (\tau ^{{5}/{3}}\right )+18 c_1 -6 \tau }}d \tau = x +c_2 \]

Singular solutions are found by solving

\begin{align*} -\frac {\sqrt {18 \ln \left (y^{{5}/{3}}\right )+18 c_1 -6 y}}{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 = \left ({\mathrm e}^{-\frac {5 \operatorname {LambertW}\left (-\frac {243^{{4}/{5}} {\mathrm e}^{-\frac {3 c_1}{5}}}{405}\right )}{3}-c_1}\right )^{{3}/{5}} \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 {18 \ln \left (y^{{5}/{3}}\right )+18 c_1 -6 y}}{3} \end{align*}

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

\[ \int _{}^{y}\frac {3}{\sqrt {18 \ln \left (\tau ^{{5}/{3}}\right )+18 c_1 -6 \tau }}d \tau = x +c_3 \]

Singular solutions are found by solving

\begin{align*} \frac {\sqrt {18 \ln \left (y^{{5}/{3}}\right )+18 c_1 -6 y}}{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 = \left ({\mathrm e}^{-\frac {5 \operatorname {LambertW}\left (-\frac {243^{{4}/{5}} {\mathrm e}^{-\frac {3 c_1}{5}}}{405}\right )}{3}-c_1}\right )^{{3}/{5}} \end{align*}

Will add steps showing solving for IC soon.

The solution

\[ \int _{}^{y}-\frac {3}{\sqrt {18 \ln \left (\tau ^{{5}/{3}}\right )+18 c_1 -6 \tau }}d \tau = x +c_2 \]

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

\[ y = \left ({\mathrm e}^{-\frac {5 \operatorname {LambertW}\left (-\frac {243^{{4}/{5}} {\mathrm e}^{-\frac {3 c_1}{5}}}{405}\right )}{3}-c_1}\right )^{{3}/{5}} \]

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

Summary of solutions found

\begin{align*} \int _{}^{y}\frac {3}{\sqrt {18 \ln \left (\tau ^{{5}/{3}}\right )+18 c_1 -6 \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)+(1/3)*(_a-5)/_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 : 0.062 (sec)
Leaf size : 59

dsolve(3*y(x)*diff(diff(y(x),x),x)+y(x) = 5, 
       y(x),singsol=all)
 

\begin{align*} -3 \left (\int _{}^{y}\frac {1}{\sqrt {30 \ln \left (\textit {\_a} \right )+9 c_{1} -6 \textit {\_a}}}d \textit {\_a} \right )-x -c_{2} &= 0 \\ 3 \left (\int _{}^{y}\frac {1}{\sqrt {30 \ln \left (\textit {\_a} \right )+9 c_{1} -6 \textit {\_a}}}d \textit {\_a} \right )-x -c_{2} &= 0 \\ \end{align*}

Mathematica DSolve solution

Solving time : 0.324 (sec)
Leaf size : 41

DSolve[{3*y[x]*D[y[x],{x,2}]+y[x]==5,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\[ \text {Solve}\left [\int _1^{y(x)}\frac {1}{\sqrt {c_1+\frac {2}{3} (5 \log (K[1])-K[1])}}dK[1]{}^2=(x+c_2){}^2,y(x)\right ] \]