1.70 problem 70

Internal problem ID [7762]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 70
Date solved : Monday, October 21, 2024 at 04:16:09 PM
CAS classification : [[_2nd_order, _quadrature]]

Solve

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

1.70.1 Solved as second order missing x ode

Time used: 0.493 (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 = 0 \end{align*}

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

Integrating gives

\begin{align*} \int -\frac {p a}{b}d p &= dy\\ -\frac {p^{2} a}{2 b}&= y +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 a b \left (y +c_1 \right )}}{a}\\ p&=-\frac {\sqrt {-2 a b \left (y +c_1 \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 a b \left (y+c_1 \right )}}{a} \end{align*}

Integrating gives

\begin{align*} \int \frac {a}{\sqrt {-2 a b \left (y +c_1 \right )}}d y &= dx\\ -\frac {\sqrt {2}\, \sqrt {-a b \left (y +c_1 \right )}}{b}&= x +c_2 \end{align*}

Singular solutions are found by solving

\begin{align*} \frac {\sqrt {-2 a b \left (y +c_1 \right )}}{a}&= 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 = -c_1 \end{align*}

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

\begin{align*} y&=-c_1\\ y&=-\frac {c_2^{2} b +2 c_2 b x +b \,x^{2}+2 c_1 a}{2 a} \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 a b \left (y+c_1 \right )}}{a} \end{align*}

Integrating gives

\begin{align*} \int -\frac {a}{\sqrt {-2 a b \left (y +c_1 \right )}}d y &= dx\\ \frac {\sqrt {2}\, \sqrt {-a b \left (y +c_1 \right )}}{b}&= x +c_3 \end{align*}

Singular solutions are found by solving

\begin{align*} -\frac {\sqrt {-2 a b \left (y +c_1 \right )}}{a}&= 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 = -c_1 \end{align*}

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

\begin{align*} y&=-c_1\\ y&=-\frac {c_3^{2} b +2 c_3 b x +b \,x^{2}+2 c_1 a}{2 a} \end{align*}

Will add steps showing solving for IC soon.

The solution

\[ y = -c_1 \]

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

1.70.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & a y \left (\frac {d}{d x}y^{\prime }\right )+b y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=-\frac {b}{a} \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {0\pm \left (\sqrt {0}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (x \right )=1 \\ \bullet & {} & \textrm {Repeated root, multiply}\hspace {3pt} y_{1}\left (x \right )\hspace {3pt}\textrm {by}\hspace {3pt} x \hspace {3pt}\textrm {to ensure linear independence}\hspace {3pt} \\ {} & {} & y_{2}\left (x \right )=x \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y=\mathit {C1} y_{1}\left (x \right )+\mathit {C2} y_{2}\left (x \right )+y_{p}\left (x \right ) \\ \bullet & {} & \textrm {Substitute in solutions of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y=\mathit {C1} +\mathit {C2} x +y_{p}\left (x \right ) \\ \square & {} & \textrm {Find a particular solution}\hspace {3pt} y_{p}\left (x \right )\hspace {3pt}\textrm {of the ODE}\hspace {3pt} \\ {} & \circ & \textrm {Use variation of parameters to find}\hspace {3pt} y_{p}\hspace {3pt}\textrm {here}\hspace {3pt} f \left (x \right )\hspace {3pt}\textrm {is the forcing function}\hspace {3pt} \\ {} & {} & \left [y_{p}\left (x \right )=-y_{1}\left (x \right ) \left (\int \frac {y_{2}\left (x \right ) f \left (x \right )}{W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )}d x \right )+y_{2}\left (x \right ) \left (\int \frac {y_{1}\left (x \right ) f \left (x \right )}{W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )}d x \right ), f \left (x \right )=-\frac {b}{a}\right ] \\ {} & \circ & \textrm {Wronskian of solutions of the homogeneous equation}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )=\left [\begin {array}{cc} 1 & x \\ 0 & 1 \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )=1 \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (x \right ) \\ {} & {} & y_{p}\left (x \right )=-\frac {b \left (\left (\int 1d x \right ) x -\left (\int x d x \right )\right )}{a} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (x \right )=-\frac {x^{2} b}{2 a} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=\mathit {C1} +\mathit {C2} x -\frac {x^{2} b}{2 a} \end {array} \]

1.70.3 Maple trace

Methods for second order ODEs:
 

1.70.4 Maple dsolve solution

Solving time : 0.003 (sec)
Leaf size : 22

dsolve(a*y(x)*diff(diff(y(x),x),x)+b*y(x) = 0, 
       y(x),singsol=all)
 

\begin{align*} y &= 0 \\ y &= -\frac {x^{2} b}{2 a}+c_1 x +c_2 \\ \end{align*}

1.70.5 Mathematica DSolve solution

Solving time : 0.003 (sec)
Leaf size : 28

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

\begin{align*} y(x)\to 0 \\ y(x)\to -\frac {b x^2}{2 a}+c_2 x+c_1 \\ \end{align*}