2.1.70 Problem 70

Internal problem ID [10056]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 70
Date solved : Thursday, November 27, 2025 at 10:07:43 AM
CAS classification : [[_2nd_order, _quadrature]]

Solved by factoring the differential equation

Time used: 0.018 (sec)

Solve

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

Writing the ode as

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

Therefore we need to solve the following equations

\begin{align*} \tag{1} y &= 0 \\ \tag{2} y^{\prime \prime } a +b &= 0 \\ \end{align*}

Now each of the above equations is solved in turn.

Solving equation (1)

Solving for \(y\) from

\begin{align*} y = 0 \end{align*}

Solving gives

\begin{align*} y &= 0 \\ \end{align*}

Solving equation (2)

The ODE can be written as

\[ y^{\prime \prime } = -\frac {b}{a} \]

Integrating once gives

\[ y^{\prime }= -\frac {b x}{a} + c_1 \]

Integrating again gives

\[ y= -\frac {b \,x^{2}}{2 a} + c_1 x + c_2 \]

Summary of solutions found

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

✓ Maple. Time used: 0.001 (sec). Leaf size: 22

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

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

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
<- quadrature successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & a y \left (x \right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+b y \left (x \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )=-\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 \left (x \right )=\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 \left (x \right )=\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 ) \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 +y_{2}\left (x \right ) \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 , 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 (\int 1d x x -\int x d x \right )}{a} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (x \right )=-\frac {b \,x^{2}}{2 a} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C1} +\mathit {C2} x -\frac {b \,x^{2}}{2 a} \end {array} \]

✓ Mathematica. Time used: 0.002 (sec). Leaf size: 28

ode=a*y[x]*D[y[x],{x,2}]+b*y[x]==0; 
ic={}; 
DSolve[{ode,ic},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*}

✓ Sympy. Time used: 0.141 (sec). Leaf size: 19

from sympy import * 
x = symbols("x") 
a = symbols("a") 
b = symbols("b") 
y = Function("y") 
ode = Eq(a*y(x)*Derivative(y(x), (x, 2)) + b*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

\[ \left [ y{\left (x \right )} = C_{1} + C_{2} x - \frac {b x^{2}}{2 a}, \ y{\left (x \right )} = 0\right ] \]