Internal problem ID [4754]
Internal file name [OUTPUT/4247_Sunday_June_05_2022_12_47_26_PM_51583505/index.tex]
Book: Mathematical Methods in the Physical Sciences. third edition. Mary L. Boas. John Wiley.
2006
Section: Chapter 8, Ordinary differential equations. Section 2. Separable equations. page
398
Problem number: 6.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "exact", "bernoulli", "separable", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[_separable]
\[ \boxed {y^{\prime }-\frac {2 x y^{2}+x}{x^{2} y-y}=0} \] With initial conditions \begin {align*} \left [y \left (\sqrt {2}\right ) = 0\right ] \end {align*}
This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(x,y)\\ &= \frac {x \left (2 y^{2}+1\right )}{y \left (x^{2}-1\right )} \end {align*}
\(f(x,y)\) is not defined at \(y = 0\) therefore existence and uniqueness theorem do not apply.
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {x \left (2 y^{2}+1\right )}{y \left (x^{2}-1\right )} \end {align*}
Where \(f(x)=\frac {x}{x^{2}-1}\) and \(g(y)=\frac {2 y^{2}+1}{y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {2 y^{2}+1}{y}} \,dy &= \frac {x}{x^{2}-1} \,d x \\ \int { \frac {1}{\frac {2 y^{2}+1}{y}} \,dy} &= \int {\frac {x}{x^{2}-1} \,d x} \\ \frac {\ln \left (y^{2}+\frac {1}{2}\right )}{4}&=\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x +1\right )}{2}+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} \frac {\sqrt {2}\, \left (4 y^{2}+2\right )^{\frac {1}{4}}}{2} &= {\mathrm e}^{\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x +1\right )}{2}+c_{1}} \end {align*}
Which simplifies to \begin {align*} \frac {\sqrt {2}\, \left (4 y^{2}+2\right )^{\frac {1}{4}}}{2} &= c_{2} {\mathrm e}^{\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x +1\right )}{2}} \end {align*}
Which can be simplified to become \[ \frac {\sqrt {2}\, \left (4 y^{2}+2\right )^{\frac {1}{4}}}{2} = c_{2} \sqrt {x -1}\, \sqrt {x +1}\, {\mathrm e}^{c_{1}} \] The solution is \[ \frac {\sqrt {2}\, \left (4 y^{2}+2\right )^{\frac {1}{4}}}{2} = c_{2} \sqrt {x -1}\, \sqrt {x +1}\, {\mathrm e}^{c_{1}} \] Initial conditions are used to solve for \(c_{1}\). Substituting \(x=\sqrt {2}\) and \(y=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} \frac {2^{\frac {3}{4}}}{2} = \sqrt {1+\sqrt {2}}\, \sqrt {\sqrt {2}-1}\, c_{2} {\mathrm e}^{c_{1}} \end {align*}
The solutions are \begin {align*} c_{1} = \frac {\ln \left (\frac {1}{2 c_{2}^{4}}\right )}{4} \end {align*}
Trying the constant \begin {align*} c_{1} = \frac {\ln \left (\frac {1}{2 c_{2}^{4}}\right )}{4} \end {align*}
Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \frac {\sqrt {2}\, \left (4 y^{2}+2\right )^{\frac {1}{4}}}{2} = \frac {c_{2} \sqrt {x +1}\, \sqrt {x -1}\, \sqrt {2}\, \sqrt {\sqrt {2}\, \sqrt {\frac {1}{c_{2}^{4}}}}}{2} \end {align*}
The constant \(c_{1} = \frac {\ln \left (\frac {1}{2 c_{2}^{4}}\right )}{4}\) gives valid solution.
Summary
The solution(s) found are the following \begin{align*} \tag{1} \frac {\sqrt {2}\, \left (4 y^{2}+2\right )^{\frac {1}{4}}}{2} &= \frac {c_{2} \sqrt {x +1}\, \sqrt {x -1}\, 2^{\frac {3}{4}} \left (\frac {1}{c_{2}^{4}}\right )^{\frac {1}{4}}}{2} \\ \end{align*}
Verification of solutions
\[ \frac {\sqrt {2}\, \left (4 y^{2}+2\right )^{\frac {1}{4}}}{2} = \frac {c_{2} \sqrt {x +1}\, \sqrt {x -1}\, 2^{\frac {3}{4}} \left (\frac {1}{c_{2}^{4}}\right )^{\frac {1}{4}}}{2} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-\frac {2 x y^{2}+x}{x^{2} y-y}=0, y \left (\sqrt {2}\right )=0\right ] \\ \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} \\ {} & {} & y^{\prime }=\frac {2 x y^{2}+x}{x^{2} y-y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime } y}{2 y^{2}+1}=\frac {x}{\left (x -1\right ) \left (x +1\right )} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime } y}{2 y^{2}+1}d x =\int \frac {x}{\left (x -1\right ) \left (x +1\right )}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (2 y^{2}+1\right )}{4}=\frac {\ln \left (\left (x -1\right ) \left (x +1\right )\right )}{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=-\frac {\sqrt {-2+2 \,{\mathrm e}^{4 c_{1}} x^{4}-4 \,{\mathrm e}^{4 c_{1}} x^{2}+2 \,{\mathrm e}^{4 c_{1}}}}{2}, y=\frac {\sqrt {-2+2 \,{\mathrm e}^{4 c_{1}} x^{4}-4 \,{\mathrm e}^{4 c_{1}} x^{2}+2 \,{\mathrm e}^{4 c_{1}}}}{2}\right \} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (\sqrt {2}\right )=0 \\ {} & {} & 0=-\frac {\sqrt {-2+2 \,{\mathrm e}^{4 c_{1}}}}{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\frac {\sqrt {2}\, \sqrt {x^{2} \left (x^{2}-2\right )}}{2} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (\sqrt {2}\right )=0 \\ {} & {} & 0=\frac {\sqrt {-2+2 \,{\mathrm e}^{4 c_{1}}}}{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {\sqrt {2}\, \sqrt {x^{2} \left (x^{2}-2\right )}}{2} \\ \bullet & {} & \textrm {Solutions to the IVP}\hspace {3pt} \\ {} & {} & \left \{y=-\frac {\sqrt {2}\, \sqrt {x^{2} \left (x^{2}-2\right )}}{2}, y=\frac {\sqrt {2}\, \sqrt {x^{2} \left (x^{2}-2\right )}}{2}\right \} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli <- Bernoulli successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 31
dsolve([diff(y(x),x)=(2*x*y(x)^2+x)/(x^2*y(x)-y(x)),y(sqrt(2)) = 0],y(x), singsol=all)
\begin{align*} y \left (x \right ) &= -\frac {\sqrt {2 x^{2}-4}\, x}{2} \\ y \left (x \right ) &= \frac {\sqrt {2 x^{2}-4}\, x}{2} \\ \end{align*}
✓ Solution by Mathematica
Time used: 3.88 (sec). Leaf size: 48
DSolve[{y'[x]==(2*x*y[x]^2+x)/(x^2*y[x]-y[x]),{y[Sqrt[2]]==0}},y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {\sqrt {x^2 \left (x^2-2\right )}}{\sqrt {2}} \\ y(x)\to \frac {\sqrt {x^2 \left (x^2-2\right )}}{\sqrt {2}} \\ \end{align*}