1.68 problem 68

Internal problem ID [7760]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 68
Date solved : Monday, October 21, 2024 at 04:16:04 PM
CAS classification : [[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]

Solve

\begin{align*} a y y^{\prime \prime }+b y&=c \end{align*}

1.68.1 Solved as second order missing x ode

Time used: 1.928 (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*} a y p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+b y = c \end{align*}

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

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

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

Where

\begin{align*} f(y) &= -\frac {b y -c}{a 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 {b y -c}{a y} \,dy}\\ \frac {p^{2}}{2}&=\frac {c \ln \left (y \right )-b y}{a}+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&=\frac {\sqrt {2}\, \sqrt {a \left (c \ln \left (y \right )+c_1 a -b y \right )}}{a}\\ p&=-\frac {\sqrt {2}\, \sqrt {a \left (c \ln \left (y \right )+c_1 a -b y \right )}}{a} \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}\, \sqrt {a \left (c \ln \left (y\right )+c_1 a -b y\right )}}{a} \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

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

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}\, \sqrt {a \left (c \ln \left (y\right )+c_1 a -b y\right )}}{a} \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

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

Will add steps showing solving for IC soon.

1.68.2 Maple step by step solution

1.68.3 Maple trace

Methods for second order ODEs:
 

1.68.4 Maple dsolve solution

Solving time : 0.019 (sec)
Leaf size : 68

dsolve(a*y(x)*diff(diff(y(x),x),x)+b*y(x) = c, 
       y(x),singsol=all)
 

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

1.68.5 Mathematica DSolve solution

Solving time : 0.47 (sec)
Leaf size : 43

DSolve[{a*y[x]*D[y[x],{x,2}]+b*y[x]==c,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

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