30.31 problem 891

30.31.1 Maple step by step solution

Internal problem ID [4127]
Internal file name [OUTPUT/3620_Sunday_June_05_2022_09_49_37_AM_89150540/index.tex]

Book: Ordinary differential equations and their solutions. By George Moseley Murphy. 1960
Section: Various 30
Problem number: 891.
ODE order: 1.
ODE degree: 2.

The type(s) of ODE detected by this program : "quadrature"

Maple gives the following as the ode type

[_quadrature]

\[ \boxed {x^{2} {y^{\prime }}^{2}=a^{2}} \] The ode \begin {align*} x^{2} {y^{\prime }}^{2} = a^{2} \end {align*}

is factored to \begin {align*} \left (y^{\prime } x -a \right ) \left (y^{\prime } x +a \right ) = 0 \end {align*}

Which gives the following equations \begin {align*} y^{\prime } x -a = 0\tag {1} \\ y^{\prime } x +a = 0\tag {2} \\ \end {align*}

Each of the above equations is now solved.

Solving ODE (1) Integrating both sides gives \begin {align*} y &= \int { \frac {a}{x}\,\mathop {\mathrm {d}x}}\\ &= a \ln \left (x \right )+c_{1} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= a \ln \left (x \right )+c_{1} \\ \end{align*}

Verification of solutions

\[ y = a \ln \left (x \right )+c_{1} \] Verified OK.

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= a \ln \left (x \right )+c_{1} \\ \end{align*}

Verification of solutions

\[ y = a \ln \left (x \right )+c_{1} \] Verified OK.

Solving ODE (2) Integrating both sides gives \begin {align*} y &= \int { -\frac {a}{x}\,\mathop {\mathrm {d}x}}\\ &= -a \ln \left (x \right )+c_{2} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -a \ln \left (x \right )+c_{2} \\ \end{align*}

Verification of solutions

\[ y = -a \ln \left (x \right )+c_{2} \] Verified OK.

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -a \ln \left (x \right )+c_{2} \\ \end{align*}

Verification of solutions

\[ y = -a \ln \left (x \right )+c_{2} \] Verified OK.

30.31.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} {y^{\prime }}^{2}=a^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=\frac {a}{x}, y^{\prime }=-\frac {a}{x}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {a}{x} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime }d x =\int \frac {a}{x}d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y=a \ln \left (x \right )+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=a \ln \left (x \right )+c_{1} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {a}{x} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime }d x =\int -\frac {a}{x}d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y=-a \ln \left (x \right )+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=-a \ln \left (x \right )+c_{1} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{y=-a \ln \left (x \right )+c_{1} , y=a \ln \left (x \right )+c_{1} \right \} \end {array} \]

Maple trace

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

Solution by Maple

Time used: 0.0 (sec). Leaf size: 20

dsolve(x^2*diff(y(x),x)^2 = a^2,y(x), singsol=all)
 

\begin{align*} y \left (x \right ) &= a \ln \left (x \right )+c_{1} \\ y \left (x \right ) &= -a \ln \left (x \right )+c_{1} \\ \end{align*}

Solution by Mathematica

Time used: 0.003 (sec). Leaf size: 24

DSolve[x^2 (y'[x])^2==a^2,y[x],x,IncludeSingularSolutions -> True]
 

\begin{align*} y(x)\to -a \log (x)+c_1 \\ y(x)\to a \log (x)+c_1 \\ \end{align*}