30.35 problem 895

30.35.1 Solving as first order nonlinear p but separable ode
30.35.2 Maple step by step solution

Internal problem ID [4131]
Internal file name [OUTPUT/3624_Sunday_June_05_2022_09_50_08_AM_96497582/index.tex]

Book: Ordinary differential equations and their solutions. By George Moseley Murphy. 1960
Section: Various 30
Problem number: 895.
ODE order: 1.
ODE degree: 2.

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

Maple gives the following as the ode type

[_separable]

\[ \boxed {x^{2} {y^{\prime }}^{2}+y^{2}-y^{4}=0} \]

30.35.1 Solving as first order nonlinear p but separable ode

The ode has the form \begin {align*} (y')^{\frac {n}{m}} &= f(x) g(y)\tag {1} \end {align*}

Where \(n=2, m=1, f=\frac {1}{x^{2}} , g=y^{4}-y^{2}\). Hence the ode is \begin {align*} (y')^{2} &= \frac {y^{4}-y^{2}}{x^{2}} \end {align*}

Solving for \(y^{\prime }\) from (1) gives \begin {align*} y^{\prime } &=\sqrt {f g}\\ y^{\prime } &=-\sqrt {f g} \end {align*}

To be able to solve as separable ode, we have to now assume that \(f>0,g>0\). \begin {align*} \frac {1}{x^{2}} &> 0\\ y^{4}-y^{2} &> 0 \end {align*}

Under the above assumption the differential equations become separable and can be written as \begin {align*} y^{\prime } &=\sqrt {f}\, \sqrt {g}\\ y^{\prime } &=-\sqrt {f}\, \sqrt {g} \end {align*}

Therefore \begin {align*} \frac {1}{\sqrt {g}} \, dy &= \left (\sqrt {f}\right )\,dx\\ -\frac {1}{\sqrt {g}} \, dy &= \left (\sqrt {f}\right )\,dx \end {align*}

Replacing \(f(x),g(y)\) by their values gives \begin {align*} \frac {1}{\sqrt {y^{4}-y^{2}}} \, dy &= \left (\sqrt {\frac {1}{x^{2}}}\right )\,dx\\ -\frac {1}{\sqrt {y^{4}-y^{2}}} \, dy &= \left (\sqrt {\frac {1}{x^{2}}}\right )\,dx \end {align*}

Integrating now gives the solutions. \begin {align*} \int \frac {1}{\sqrt {y^{4}-y^{2}}}d y &= \int \sqrt {\frac {1}{x^{2}}}d x +c_{1}\\ \int -\frac {1}{\sqrt {y^{4}-y^{2}}}d y &= \int \sqrt {\frac {1}{x^{2}}}d x +c_{1} \end {align*}

Integrating gives \begin {align*} -\arcsin \left (\frac {1}{y}\right ) &= \sqrt {\frac {1}{x^{2}}}\, x \ln \left (x \right )+c_{1}\\ \arcsin \left (\frac {1}{y}\right ) &= \sqrt {\frac {1}{x^{2}}}\, x \ln \left (x \right )+c_{1} \end {align*}

Therefore \begin{align*} y &= -\frac {1}{\sin \left (\sqrt {\frac {1}{x^{2}}}\, x \ln \left (x \right )+c_{1} \right )} \\ y &= \frac {1}{\sin \left (\sqrt {\frac {1}{x^{2}}}\, x \ln \left (x \right )+c_{1} \right )} \\ \end{align*}

Summary

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

Verification of solutions

\[ y = -\frac {1}{\sin \left (\sqrt {\frac {1}{x^{2}}}\, x \ln \left (x \right )+c_{1} \right )} \] Verified OK.

\[ y = \frac {1}{\sin \left (\sqrt {\frac {1}{x^{2}}}\, x \ln \left (x \right )+c_{1} \right )} \] Verified OK.

30.35.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} {y^{\prime }}^{2}+y^{2}-y^{4}=0 \\ \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 {\sqrt {y^{2}-1}\, y}{x}, y^{\prime }=-\frac {\sqrt {y^{2}-1}\, y}{x}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {\sqrt {y^{2}-1}\, y}{x} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sqrt {y^{2}-1}\, y}=\frac {1}{x} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{\sqrt {y^{2}-1}\, y}d x =\int \frac {1}{x}d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\arctan \left (\frac {1}{\sqrt {y^{2}-1}}\right )=\ln \left (x \right )+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=\frac {\sqrt {\tan \left (\ln \left (x \right )+c_{1} \right )^{2}+1}}{\tan \left (\ln \left (x \right )+c_{1} \right )}, y=-\frac {\sqrt {\tan \left (\ln \left (x \right )+c_{1} \right )^{2}+1}}{\tan \left (\ln \left (x \right )+c_{1} \right )}\right \} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {\sqrt {y^{2}-1}\, y}{x} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sqrt {y^{2}-1}\, y}=-\frac {1}{x} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{\sqrt {y^{2}-1}\, y}d x =\int -\frac {1}{x}d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\arctan \left (\frac {1}{\sqrt {y^{2}-1}}\right )=-\ln \left (x \right )+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=\frac {\sqrt {\tan \left (-\ln \left (x \right )+c_{1} \right )^{2}+1}}{\tan \left (-\ln \left (x \right )+c_{1} \right )}, y=-\frac {\sqrt {\tan \left (-\ln \left (x \right )+c_{1} \right )^{2}+1}}{\tan \left (-\ln \left (x \right )+c_{1} \right )}\right \} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\left \{y=\frac {\sqrt {\tan \left (-\ln \left (x \right )+c_{1} \right )^{2}+1}}{\tan \left (-\ln \left (x \right )+c_{1} \right )}, y=-\frac {\sqrt {\tan \left (-\ln \left (x \right )+c_{1} \right )^{2}+1}}{\tan \left (-\ln \left (x \right )+c_{1} \right )}\right \}, \left \{y=\frac {\sqrt {\tan \left (\ln \left (x \right )+c_{1} \right )^{2}+1}}{\tan \left (\ln \left (x \right )+c_{1} \right )}, y=-\frac {\sqrt {\tan \left (\ln \left (x \right )+c_{1} \right )^{2}+1}}{\tan \left (\ln \left (x \right )+c_{1} \right )}\right \}\right \} \end {array} \]

Maple trace

`Methods for first order ODEs: 
   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   -> Solving 1st order ODE of high degree, 1st attempt 
   trying 1st order WeierstrassP solution for high degree ODE 
   trying 1st order WeierstrassPPrime solution for high degree ODE 
   trying 1st order JacobiSN solution for high degree ODE 
   trying 1st order ODE linearizable_by_differentiation 
   trying differential order: 1; missing variables 
   trying simple symmetries for implicit equations 
   <- symmetries for implicit equations successful`
 

Solution by Maple

Time used: 0.156 (sec). Leaf size: 52

dsolve(x^2*diff(y(x),x)^2+y(x)^2-y(x)^4 = 0,y(x), singsol=all)
 

\begin{align*} y \left (x \right ) &= -1 \\ y \left (x \right ) &= 1 \\ y \left (x \right ) &= 0 \\ y \left (x \right ) &= \csc \left (-\ln \left (x \right )+c_{1} \right ) \operatorname {csgn}\left (\sec \left (-\ln \left (x \right )+c_{1} \right )\right ) \\ y \left (x \right ) &= -\csc \left (-\ln \left (x \right )+c_{1} \right ) \operatorname {csgn}\left (\sec \left (-\ln \left (x \right )+c_{1} \right )\right ) \\ \end{align*}

Solution by Mathematica

Time used: 1.724 (sec). Leaf size: 88

DSolve[x^2 (y'[x])^2+y[x]^2-y[x]^4==0,y[x],x,IncludeSingularSolutions -> True]
 

\begin{align*} y(x)\to -\sqrt {\sec ^2(-\log (x)+c_1)} \\ y(x)\to \sqrt {\sec ^2(-\log (x)+c_1)} \\ y(x)\to -\sqrt {\sec ^2(\log (x)+c_1)} \\ y(x)\to \sqrt {\sec ^2(\log (x)+c_1)} \\ y(x)\to -1 \\ y(x)\to 0 \\ y(x)\to 1 \\ \end{align*}